123: A basic example of creating an interactive plot with HoloViews and Bokeh

This is a very simple of example of producing an interactive visualisation using Holoviews (which calls on Bokeh). These visualisations can be viewed in Jupyter notebooks, or may be saved as a single html page which needs only a web browser to see. Here we show room temperature and humidity, with the plots allowing the choice of which room to show.

To create these interactive plots you will need to install pyviz, holoviews and bokeh as described below.

Install libraries if needed

From terminal run:

conda install -c pyviz holoviews bokeh

holoviews --install-examples

Import libraries

import numpy as np
import pandas as pd
import holoviews as hv
import panel as pn
from holoviews import opts

Create some dummy data

# Set length of data set to create
length = 25

# Build strings for location data
location1 = ['Window'] * length
location2 = ['Porch'] * length
location3 = ['Fridge'] * length

# Set temperature to normal distribution (mu, sigma, length)
temperature1 = np.random.normal(25 ,5, length)
temperature2 = np.random.normal(15 ,3, length)
temperature3 = np.random.normal(4, 0.5,length)

# Set temperature to uniform distribution (min, max, length)
humidity1 = np.random.uniform(30, 60, length)
humidity2 = np.random.uniform(60, 80, length)
humidity3 = np.random.uniform(80, 99, length)

# Record mean temperature/humidity (use np.repeat to repeata single value)
mean_temp1 = np.repeat(np.mean(temperature1), length)
mean_temp2 = np.repeat(np.mean(temperature2), length)
mean_temp3 = np.repeat(np.mean(temperature3), length)

mean_humidity1 = np.repeat(np.mean(humidity1), length)
mean_humidity2 = np.repeat(np.mean(humidity2), length)
mean_humidity3 = np.repeat(np.mean(humidity3), length)

# Concatenate three sets of data into single list/arrays
location = location1 + location2 + location3
temperature = np.concatenate((temperature1, temperature2, temperature3))
mean_temperature = np.concatenate((mean_temp1, mean_temp2, mean_temp3))
humidity = np.concatenate((humidity1, humidity2, humidity3))
mean_humidity = np.concatenate((mean_humidity1, mean_humidity2, mean_humidity3))

# Create list of days
days = list(range(1,length + 1))
day = days * 3 # times 3 as there are three locations

# Transfer data to pandas DataFrame
data = pd.DataFrame()
data['day'] = day
data['location'] = location
data['temperature'] = temperature
data['humidity'] = humidity
data['mean_temperature'] = mean_temperature
data['mean_humidity'] = mean_humidity



day	location	temperature	humidity	mean_temperature	mean_humidity
0	1	Window	26.081745	49.611333	25.222169	45.43133
1	2	Window	31.452276	39.027559	25.222169	45.43133
2	3	Window	19.031828	58.825912	25.222169	45.43133
3	4	Window	21.309825	52.741160	25.222169	45.43133
4	5	Window	13.529042	39.977335	25.222169	45.43133

Build bar chart

# Make holoviews data table
key_dimensions   = ['location']
value_dimensions = ['day', 'temperature', 'humidity', 'mean_temperature', 'mean_humidity']
hv_data = hv.Table(data, key_dimensions, value_dimensions)

# Build bar charts
bars1 = hv_data.to.bars(['day'], ['temperature'])
bars2 = hv_data.to.bars(['day'], ['humidity']).opts(color='Red')

# Compose plot
bar_plot = bars1 + bars2

# Show plot (only work in Jupyter notebook)

Build scatter chart

# Build scatter charts
scatter1 = hv_data.to.scatter(['day'], ['temperature'])
scatter2 = hv_data.to.scatter(['day'], ['humidity']).opts(color='Red')

# Compose plot
scatter_plot = scatter1 + scatter2

# Show plot

Build line chart for mean temperature and humidity

# Build line charts
line1 = hv_data.to.curve(['day'], ['mean_temperature'])
line2 = hv_data.to.curve(['day'], ['mean_humidity']).opts(color='r')

# Compose plot
line_chart = line1 + line2

# Show plot

Combine line and scatter charts

Here we combine the line and scatter charts. We viewed them individually before, though this is not actually necessary.

# Compose plot (* creates overlays of two or more plots)
combined_plot = line1 * scatter1 + line2 * scatter2

# Show plot

Save to html

The interactive plot may be saved as html which may be shared with, and viewed by, anyone (there is no need for anything other than a standard web browser to view the interactive plot).

hv.save(combined_plot, 'holoviews_example.html')

122: Oversampling to correct for imbalanced data using naive sampling or SMOTE

Machine learning can have poor performance for minority classes (where one or more classes represent only a small proportion of the overall data set compared with a dominant class). One method of improving performance is to balance out the number of examples between different classes. Here two methods are described:

  1. Resampling from the minority classes to give the same number of examples as the majority class.
  2. SMOTE (Synthetic Minority Over-sampling Technique): creating synthetic data based on creating new data points that are mid-way between two near neighbours in any particular class.

SMOTE uses imblearn See: https://imbalanced-learn.org

Install with: pip install -U imbalanced-learn, or conda install -c conda-forge imbalanced-learn


N. V. Chawla, K. W. Bowyer, L. O.Hall, W. P. Kegelmeyer, “SMOTE: synthetic minority over-sampling technique,” Journal of artificial intelligence research, 16, 321-357, 2002

Create dummy data

First we will create some unbalanced dummy data: Classes 0, 1 and 2 will represent 1%, 5% and 94% of the data respectively.

from sklearn.datasets import make_classification
X, y = make_classification(n_samples=5000, n_features=2, n_informative=2, n_redundant=0, n_repeated=0, n_classes=3,                            n_clusters_per_class=1, weights=[0.01, 0.05, 0.94],                            class_sep=0.8, random_state=0)

Count instances of each class.

from collections import Counter

[(0, 64), (1, 262), (2, 4674)]

Define function to plot data

import matplotlib.pyplot as plt

def plot_classes(X,y):
    colours = ['k','b','g']
    point_colours = [colours[val] for val in y]
    X1 = X[:,0]
    X2 = X[:,1]
    plt.scatter(X1, X2, facecolor = point_colours, edgecolor = 'k')

Plots data using function


Oversample with naive sampling to match numbers in each class

With naive resampling we repeatedly randomly sample from the minority classes and add that the new sample to the existing data set, leading to multiple instances of the minority classes.This builds up the number of minority class samples.

from imblearn.over_sampling import RandomOverSampler
ros = RandomOverSampler(random_state=0)
X_resampled, y_resampled = ros.fit_resample(X, y)

Count instances of each class in the augmented data.

from collections import Counter

[(0, 4674), (1, 4674), (2, 4674)]

Plot augmented data (it looks the same as the original as points are overlaid).


SMOTE with continuous variables

SMOTE (synthetic minority oversampling technique) works by finding two near neighbours in a minority class, producing a new point midway between the two existing points and adding that new point in to the sample. The example shown is in two dimensions, but SMOTE will work across multiple dimensions (features). SMOTE therefore helps to ‘fill in’ the feature space occupied by minority classes.

from imblearn.over_sampling import SMOTE
X_resampled, y_resampled = SMOTE().fit_resample(X, y)

# Count instances of each class
from collections import Counter

[(0, 4674), (1, 4674), (2, 4674)]

Plot augmented data (note minority class data points now exist in new spaces).

SMOTE with mixed continuous and binary/categorical values

It is not possible to calculate a ‘mid point’ between two points of binary or categorical data. An extension to the SMOTE method allows for use of binary or categorical data by taking the most common occurring category of nearest neighbours to a minority class point.

# create a synthetic data set with continuous and categorical features
import numpy as np
rng = np.random.RandomState(42)
n_samples = 50
X = np.empty((n_samples, 3), dtype=object)
X[:, 0] = rng.choice(['A', 'B', 'C'], size=n_samples).astype(object)
X[:, 1] = rng.randn(n_samples)
X[:, 2] = rng.randint(3, size=n_samples)
y = np.array([0] * 20 + [1] * 30)

Count instances of each class


[(0, 20), (1, 30)]

Show last 10 original data points

print (X[-10:])

[['A' 1.4689412854323924 2]
 ['C' -1.1238983345400366 0]
 ['C' 0.9500053955071801 2]
 ['A' 1.7265164685753638 1]
 ['A' 0.4578850770000152 0]
 ['C' -1.6842873783658814 0]
 ['B' 0.32684522397001387 0]
 ['A' -0.0811189541586873 2]
 ['B' 0.46779475326315173 1]
 ['B' 0.7361223506692577 0]]

Use SMOTENC to create new data points.

from imblearn.over_sampling import SMOTENC
smote_nc = SMOTENC(categorical_features=[0, 2], random_state=0)
X_resampled, y_resampled = smote_nc.fit_resample(X, y)

Count instances of each class


[(0, 30), (1, 30)]

Show last 10 values of X (SMOTE data points are added to the end of the original data set)

print (X_resampled[-10:])

[['C' -1.0600505672469849 1]
 ['C' -0.36965644259183145 1]
 ['A' 0.1453826708354494 2]
 ['C' -1.7442827953859052 2]
 ['C' -1.6278053447258838 2]
 ['A' 0.5246469549655818 2]
 ['B' -0.3657680728116921 2]
 ['A' 0.9344237230779993 2]
 ['B' 0.3710891618824609 2]
 ['B' 0.3327240726719727 2]]

121. More list comprehension examples

Example 1 – double the numbers

Standard loop approach:

foo = [1, 2, 3, 4]
bar = []

for x in foo:
    bar.append(x * 2)


[2, 4, 6, 8]

Using list comprehension:

foo = [1, 2, 3, 4]
bar = [x * 2 for x in foo]


[2, 4, 6, 8]

Example 2 – convert Celsius to Fahrenheit

This example calls a function from within the list comprehension.

Define the function:

def convert_celsius_to_fahrenheit(deg_celsius):
    Convert degress celsius to fahrenheit
    Returns float value - temp in fahrenheit
    Keyword arguments:
        def_celcius -- temp in degrees celsius
    return (9/5) * deg_celsius + 32

Standard loop approach:

#list of temps in degree celsius to convert to fahrenheit
celsius = [39.2, 36.5, 37.3, 41.0]

#standard for loop approach
fahrenheit = []
for x in celsius:

print('Using standard for loop: {}'.format(fahrenheit))


Using standard for loop: [102.56, 97.7, 99.14, 105.8]

Using list comprehension

fahrenheit = [convert_celsius_to_fahrenheit(x) for x in celsius]
print('Using list comprehension: {}'.format(fahrenheit))


Using list comprehension: [102.56, 97.7, 99.14, 105.8]

Example 3 – convert the strings to different data types

This example also make use of the zip function. Zip allows you to iterate through two lists at the same time.

inputs = ["1", "3.142", "True", "spam"]
converters = [int, float, bool, str]

values_with_correct_data_types = [t(s) for (s, t) in zip(inputs, converters)]


[1, 3.142, True, 'spam']

Example 4 – Using if statements within a list comprehension

The example filters a list of file names to the python files only

unfiltered_files = ['test.py', 'names.csv', 'fun_module.py', 'prog.config']

# Standard loop form
python_files = []
# filter the files using a standard for loop 
for file in unfiltered_files:
    if file[-2:] == 'py':
print('using standard for loop: {}'.format(python_files))

#list comprehension
python_files = [file for file in unfiltered_files if file[-2:] == 'py']

print('using list comprehension {}'.format(python_files))


using standard for loop: ['test.py', 'fun_module.py']
using list comprehension ['test.py', 'fun_module.py']

Example 5 – List comprehension to create a list of lists

list_of_lists = []

# Standard loop form
for i in range(5):
    sub_list = []
    for j in range(3):
        sub_list.append(i * j)


# List comprehension
list_of_lists = [[i * j for j in range(3)] for i in range(5)]



[[0, 0, 0], [0, 1, 2], [0, 2, 4], [0, 3, 6], [0, 4, 8]]
[[0, 0, 0], [0, 1, 2], [0, 2, 4], [0, 3, 6], [0, 4, 8]]

Example 6: Iterate over all items in a list of lists

The code converts a list of lists to a list of items
We call this flattening the list.

list_of_lists = [[8, 2, 1], [9, 1, 2], [4, 5, 100]]

# Standard loop form
flat_list = []
for row in list_of_lists:
    for col in row:


# List comprehension:
flat_list = [item for sublist in list_of_lists for item in sublist]


[8, 2, 1, 9, 1, 2, 4, 5, 100]
[8, 2, 1, 9, 1, 2, 4, 5, 100]

120. Generating log normal samples from provided arithmetic mean and standard deviation of original population

The log normal distribution is frequently a useful distribution for mimicking process times in healthcare pathways (or many other non-automated processes). The distribution has a right skew which may frequently occur when some clinical process step has some additional complexity to it compared to the ‘usual’ case.

To sample from a log normal distribution we need to convert the mean and standard deviation that was calculated from the original non-logged population into the mu and sigma of the underlying log normal population.

(For maximum computation effuiciency, when calling the function repeatedly using the same mean and standard deviation, you may wish to split this into two functions – one to calculate mu and sigma which needs only calling once, and the other to sample from the log normal distribution given mu and sigma).

For more on the maths see:


import numpy as np

def generate_lognormal_samples(mean, stdev, n=1):
    Returns n samples taken from a lognormal distribution, based on mean and
    standard deviation calaculated from the original non-logged population.
    Converts mean and standard deviation to underlying lognormal distribution
    mu and sigma based on calculations desribed at:
    Returns a numpy array of floats if n > 1, otherwise return a float
    # Calculate mu and sigma of underlying lognormal distribution
    phi = (stdev ** 2 + mean ** 2) ** 0.5
    mu = np.log(mean ** 2 / phi)
    sigma = (np.log(phi ** 2 / mean ** 2)) ** 0.5
    # Generate lognormal population
    generated_pop = np.random.lognormal(mu, sigma , n)
    # Convert single sample (if n=1) to a float, otherwise leave as array
    generated_pop = \
        generated_pop[0] if len(generated_pop) == 1 else generated_pop
    return generated_pop

Test the function

We will generate a population of 100,000 samples with a given mean and standard deviation (these would be calculated on the non-logged population), and test the resulting generated population has the same mean and standard deviation.

mean = 10
stdev = 10
generated_pop = generate_lognormal_samples(mean, stdev, 100000)
print ('Mean:', generated_pop.mean())
print ('Standard deviation:', generated_pop.std())


Mean: 10.043105926813356
Standard deviation: 9.99527575740651

Plot a histogram of the generated population:

import matplotlib.pyplot as plt
%matplotlib inline
bins = np.arange(0,51,1)
plt.hist(generated_pop, bins=bins)

Generating a single sample

The function will return a single number if no n is given in the function call:

print (generate_lognormal_samples(mean, stdev))

Out: 6.999376449335125

119: Optimising scikit-learn machine learning models with grid search or randomized search

Machine learning models have many hyper-parameters (parameters set before a model is fitted, and which remain constant throughout model fitting). Optimising model hyper-parameters may involve many model runs with alternative hyper-parameters. In SciKit-Learn, this may be performed in an automated fashion using Grid Search (which explores all combinations of provided hyper-parameters) or Randomized Search (which randomly selects combinations to test).

Grid search and randomized search will perform this optimisation using k-fold validation which avoids potential bias in training/test splits.

Here we will revisit a previous example of machine learning, using Random Forests to predict whether a person has breast cancer. We will then use Grid Search to optimise performance, using the ‘f1’ performance score (https://en.wikipedia.org/wiki/F1_score) as an accuracy score that balances the importance of false negatives and false positives.

First we will look at how we previously built the Random Forests model.

(See https://pythonhealthcare.org/2018/04/17/72-machine-learning-random-forests/ for previous post on Random Forest method)

# import required modules

from sklearn import datasets
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler
import numpy as np
import pandas as pd

def calculate_diagnostic_performance (actual_predicted):
    """ Calculate diagnostic performance.
    Takes a Numpy array of 1 and zero, two columns: actual and predicted
    Note that some statistics are repeats with different names
    (precision = positive_predictive_value and recall = sensitivity).
    Both names are returned
    Returns a dictionary of results:
    1) accuracy: proportion of test results that are correct    
    2) sensitivity: proportion of true +ve identified
    3) specificity: proportion of true -ve identified
    4) positive likelihood: increased probability of true +ve if test +ve
    5) negative likelihood: reduced probability of true +ve if test -ve
    6) false positive rate: proportion of false +ves in true -ve patients
    7) false negative rate:  proportion of false -ves in true +ve patients
    8) positive predictive value: chance of true +ve if test +ve
    9) negative predictive value: chance of true -ve if test -ve
    10) precision = positive predictive value 
    11) recall = sensitivity
    12) f1 = (2 * precision * recall) / (precision + recall)
    13) positive rate = rate of true +ve (not strictly a performance measure)
    # Calculate results
    actual_positives = actual_predicted[:, 0] == 1
    actual_negatives = actual_predicted[:, 0] == 0
    test_positives = actual_predicted[:, 1] == 1
    test_negatives = actual_predicted[:, 1] == 0
    test_correct = actual_predicted[:, 0] == actual_predicted[:, 1]
    accuracy = np.average(test_correct)
    true_positives = actual_positives & test_positives
    true_negatives = actual_negatives & test_negatives
    sensitivity = np.sum(true_positives) / np.sum(actual_positives)
    specificity = np.sum(true_negatives) / np.sum(actual_negatives)
    positive_likelihood = sensitivity / (1 - specificity)
    negative_likelihood = (1 - sensitivity) / specificity
    false_positive_rate = 1 - specificity
    false_negative_rate = 1 - sensitivity
    positive_predictive_value = np.sum(true_positives) / np.sum(test_positives)
    negative_predictive_value = np.sum(true_negatives) / np.sum(test_negatives)
    precision = positive_predictive_value
    recall = sensitivity
    f1 = (2 * precision * recall) / (precision + recall)
    positive_rate = np.mean(actual_predicted[:,1])
    # Add results to dictionary
    performance = {}
    performance['accuracy'] = accuracy
    performance['sensitivity'] = sensitivity
    performance['specificity'] = specificity
    performance['positive_likelihood'] = positive_likelihood
    performance['negative_likelihood'] = negative_likelihood
    performance['false_positive_rate'] = false_positive_rate
    performance['false_negative_rate'] = false_negative_rate
    performance['positive_predictive_value'] = positive_predictive_value
    performance['negative_predictive_value'] = negative_predictive_value
    performance['precision'] = precision
    performance['recall'] = recall
    performance['f1'] = f1
    performance['positive_rate'] = positive_rate

    return performance

def load_data ():
    """Load the data set. Here we load the Breast Cancer Wisconsin (Diagnostic)
    Data Set. Data could be loaded from other sources though the structure
    should be compatible with this data set, that is an object with the 
    following attributes:
        .data (holds feature data)
        .feature_names (holds feature titles)
        .target_names (holds outcome classification names)
        .target (holds classification as zero-based number)
        .DESCR (holds text-based description of data set)"""
    data_set = datasets.load_breast_cancer()
    return data_set

def normalise (X_train,X_test):
    """Normalise X data, so that training set has mean of zero and standard
    deviation of one"""
    # Initialise a new scaling object for normalising input data
    # Set up the scaler just on the training set
    # Apply the scaler to the training and test sets
    return X_train_std, X_test_std

def print_diagnostic_results (performance):
    """Iterate through, and print, the performance metrics dictionary"""
    print('\nMachine learning diagnostic performance measures:')
    for key, value in performance.items():
        print (key,'= %0.3f' %value) # print 3 decimal places

def print_feaure_importances (model, features):
    print ()
    print ('Feature importances:')
    print ('--------------------')
    df = pd.DataFrame()
    df['feature'] = features
    df['importance'] = model.feature_importances_
    df = df.sort_values('importance', ascending = False)
    print (df)

def split_data (data_set, split=0.25):
    """Extract X and y data from data_set object, and split into training and
    test data. Split defaults to 75% training, 25% test if not other value 
    passed to function"""
        X,y,test_size=split, random_state=0)
    return X_train,X_test,y_train,y_test

def test_model(model, X, y):
    """Return predicted y given X (attributes)"""
    y_pred = model.predict(X)
    test_results = np.vstack((y, y_pred)).T
    return test_results

def train_model (X, y):
    """Train the model. Note n_jobs=-1 uses all cores on a computer"""
    from sklearn.ensemble import RandomForestClassifier
    model = RandomForestClassifier(n_jobs=-1)
    model.fit (X,y)
    return model

###### Main code #######

# Load data
data_set = load_data()

# Split data into training and test sets
X_train,X_test,y_train,y_test = split_data(data_set, 0.25)

# Normalise data (not needed for Random Forests)
# X_train_std, X_test_std = normalise(X_train,X_test)

# Train model
model = train_model(X_train, y_train)

# Produce results for test set
test_results = test_model(model, X_test, y_test)

# Measure performance of test set predictions
performance = calculate_diagnostic_performance(test_results)

# Print performance metrics
Out: Machine learning diagnostic performance measures:

accuracy = 0.951
sensitivity = 0.944
specificity = 0.962
positive_likelihood = 25.028
negative_likelihood = 0.058
false_positive_rate = 0.038
false_negative_rate = 0.056
positive_predictive_value = 0.977
negative_predictive_value = 0.911
precision = 0.977
recall = 0.944
f1 = 0.960
positive_rate = 0.608

Optimise with grid search

NOTE: Grid search may take considerable time to run!

Grid search enables us to perform an exhaustive search of hyper-parameters (those model parameters that are constant in any one model). We define which hyper-parameters we wish to change, and what values we wish to try. All combinations are tested. Test are performed using k-fold validation which re-runs the model with different train/test splits (this avoids bias in our train/test split, but does increase the time required). You may wish to time small grid search first, so you have a better idea of how many parameter combinations you can realistically look at.

We pass four arguments to the grid search method:

1) The range of values for the hyper-parameters, defined in a dictionary 2) The machine learning model to use 3) The number of k-fold splits to use (cv); a value of 5 will give five 80:20 training/test splits with each sample being present in the test set once 4) The accuracy score to use. In a classification model ‘accuracy’ is common. For a regression model using scoring='neg_mean_squared_error' is common (for grid search an accuracy score must be a ‘utility function’ rather than a ‘cost function’, that is, higher values are better).

If the best model uses a value at one extreme of the provided hyper-paramter ranges then it is best to expand the range of that hyper-paraemter to be sure an optimum has been found.

More info on grid search: https://scikit-learn.org/stable/modules/grid_search.html

An alternative approach is randomised hyper-parameter searching. See https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.RandomizedSearchCV.html

# Use Grid search to optimise
# n_jobs is set to -1 to use all cores on the CPU

from sklearn.model_selection import GridSearchCV
param_grid = {'n_estimators': [10, 30, 100, 300, 1000, 3000],
              'bootstrap': [True, False],
              'min_samples_split': [2, 4, 6, 8, 10],
              'n_jobs': [-1]}

# Grid search will use k-fold cross-validation (CV is number of splits)
# Grid search also needs a ultility function (higher is better) rather than
# a cost function (lower is better) so use neg square mean error

from sklearn.ensemble import RandomForestClassifier
forest_grid = RandomForestClassifier()
grid_search = GridSearchCV(forest_grid, param_grid, cv=10,

grid_search.fit(X_train, y_train); #';' suppresses printed output

Show optimised model hyper-parameters:

# show best parameters
# If best parameters are at the extremes of the searches then extend the range



{'bootstrap': True, 'min_samples_split': 6, 'n_estimators': 30, 'n_jobs': -1}
# Or, show full description


RandomForestClassifier(bootstrap=True, class_weight=None, criterion='gini',
            max_depth=None, max_features='auto', max_leaf_nodes=None,
            min_impurity_decrease=0.0, min_impurity_split=None,
            min_samples_leaf=1, min_samples_split=6,
            min_weight_fraction_leaf=0.0, n_estimators=30, n_jobs=-1,
            oob_score=False, random_state=None, verbose=0,

Now we will use the optimised model. We could use the text above (from the output of grid_search.best_estimator_, or we can use grid_search.best_estimator_ directly.

# Use optimised model
model = grid_search.best_estimator_
model.fit (X_train, y_train);

Test optimised model:

test_results = test_model(model, X_test, y_test)

# Measure performance of test set predictions
performance = calculate_diagnostic_performance(test_results)

# Print performance metrics


Machine learning diagnostic performance measures:

accuracy = 0.972
sensitivity = 0.967
specificity = 0.981
positive_likelihood = 51.233
negative_likelihood = 0.034
false_positive_rate = 0.019
false_negative_rate = 0.033
positive_predictive_value = 0.989
negative_predictive_value = 0.945
precision = 0.989
recall = 0.967
f1 = 0.978
positive_rate = 0.615

Our accuracy has now increased from 95.1% to 97.2%.

When the number of parameter combinations because unreasonable large for grid search, and alternative is to use random search, which will select parameters randomly from the ranges given. The number of combinations tried is given by the argument n_iter.

Below is an example where we expand the number of arguments varied (becoming too large for grid search) and use random search to test 50 different samples.

For more details see https://scikit-learn.org/stable/modules/generated/sklearn.model_selection.RandomizedSearchCV.html

## Use Grid search to optimise
# n_jobs is set to -1 to use all cores on the CPU

from sklearn.model_selection import RandomizedSearchCV

param_grid = {'n_estimators': [10, 30, 100, 300, 1000, 3000],
              'bootstrap': [True, False],
              'min_samples_split': range(2,11),
              'max_depth': range(1,30),
              'min_samples_split': [2, 4, 6, 8, 10],
              'n_jobs': [-1]}

n_iter_search = 50

from sklearn.ensemble import RandomForestClassifier
forest_grid = RandomForestClassifier()
random_search = RandomizedSearchCV(forest_grid, param_grid, cv=10,
                           n_iter=n_iter_search, scoring='accuracy')

random_search.fit(X_train, y_train); #';' suppresses printed output
# show best parameters
# If best parameters are at the extremes of the searches then extend the range



{'n_jobs': -1,
 'n_estimators': 100,
 'min_samples_split': 2,
 'max_depth': 29,
 'bootstrap': False}
# Or, show full description


RandomForestClassifier(bootstrap=False, class_weight=None, criterion='gini',
            max_depth=29, max_features='auto', max_leaf_nodes=None,
            min_impurity_decrease=0.0, min_impurity_split=None,
            min_samples_leaf=1, min_samples_split=2,
            min_weight_fraction_leaf=0.0, n_estimators=100, n_jobs=-1,
            oob_score=False, random_state=None, verbose=0,

Now we will train a model with the optimized model hyper-parameters, and test against the test set.

# Use optimised model
model = random_search.best_estimator_
model.fit (X_train, y_train);

test_results = test_model(model, X_test, y_test)

# Measure performance of test set predictions
performance = calculate_diagnostic_performance(test_results)

# Print performance metrics


Machine learning diagnostic performance measures:

accuracy = 0.986
sensitivity = 0.989
specificity = 0.981
positive_likelihood = 52.411
negative_likelihood = 0.011
false_positive_rate = 0.019
false_negative_rate = 0.011
positive_predictive_value = 0.989
negative_predictive_value = 0.981
precision = 0.989
recall = 0.989
f1 = 0.989
positive_rate = 0.629

So though random search does not explore all combinations, because we can increase the number of parameters to explore, comapred with grid search, we have increased our accuracy to 98.6%

118: Python basics – saving python objects to disk with pickle

Sometimes we may wish to save Python objects to disc (for example if we have performed a lot of processing to get to a certain point). We can use Python’s pickle method to save and reload any Python object. Here we will save and reload a NumPy array, and then save and reload a collection of different objects.

Saving a single python object

Here we will use pickle to save a single object, a NumPy array.

import pickle 
import numpy as np

# Create array of random numbers:
my_array= np.random.rand(2,4)
print (my_array)


[[0.6383297  0.45250192 0.09882854 0.84896196]
 [0.97006917 0.29206495 0.92500062 0.52965801]]
# Save using pickle
filename = 'pickled_array.p'
with open(filename, 'wb') as filehandler:
    pickle.dump(my_array, filehandler)

Reload and print pickled array:

filename = 'pickled_array.p'
with open(filename, 'rb') as filehandler: 
    reloaded_array = pickle.load(filehandler)

print ('Reloaded array:')
print (reloaded_array)


Reloaded array:
[[0.6383297  0.45250192 0.09882854 0.84896196]
 [0.97006917 0.29206495 0.92500062 0.52965801]]

Using a tuple to save multiple objects

We can use pickle to save a collection of objects grouped together as a list, a dictionary, or a tuple. Here we will save a collection of objects as a tuple.

# Create an array, a list, and a dictionary
my_array = np.random.rand(2,4)
my_list =['A', 'B', 'C']
my_dictionary = {'name': 'Bob', 'Age': 42}
# Save all items in a tuple
items_to_save = (my_array, my_list, my_dictionary)
filename = 'pickled_tuple_of_objects.p'
with open(filename, 'wb') as filehandler:
    pickle.dump(items_to_save, filehandler)

Reload pickled tuple, unpack the objects, and print them.

filename = 'pickled_tuple_of_objects.p'
with open(filename, 'rb') as filehandler:
    reloaded_tuple = pickle.load(filehandler)

reloaded_array = reloaded_tuple[0]
reloaded_list = reloaded_tuple[1]
reloaded_dict = reloaded_tuple[2]

print ('Reloaded array:')
print (reloaded_array)
print ('\nReloaded list:')
print (reloaded_list)
print ('\n Reloaded dictionary')
print (reloaded_dict)


Reloaded array:
[[0.40193978 0.55173167 0.89411291 0.84625061]
 [0.86540981 0.27835353 0.43359222 0.31579122]]

Reloaded list:
['A', 'B', 'C']

 Reloaded dictionary
{'name': 'Bob', 'Age': 42}

117. Genetic Algorithms 2 – a multiple objective genetic algorithm (NSGA-II)

Note: As there is quite a substantial amount of code in this post, you may also copy the code as a single block from here.

If you have not looked at our description of a more simple genetic algorithm, with a single objective, then we advise you to look at that first (here). The example here will assume some basic understand of genetic algorithms.

In our previous example of a genetic algorithm, we looked at a genetic algorithm that optimised a single parameter. But what if we have two or more objectives to optimise?

An example in healthcare modelling is in deciding which hospitals should provide a specialist service in order to 1) minimise travel time for patients who need an emergency hospital admission, while 2) ensuring the hospital has sufficient admissions to maintain expertise and a 24/7 service, and 3) and ensuring no hospital is over-loaded with too many admissions.

One option when considering multiple objectives is to combine different objectives into a single number. This requires standardising the values in some way and giving them weights for their importance. This can simplify the optimisation problem but may be sensitive to how individual objectives are weighted (if following this approach it will probably be sensible to re-run the algorithm using alternative weighting schemes).

Pareto fronts

The approach we will describe here takes an alternative approach and seeks to explicitly investigate the trade-off between different objectives. When one objective cannot be improved without the worsening of another objective we are on what is known as the ‘Pareto front’. That is easy to visualise with two objectives, but Pareto fronts exist across any number of objectives – we are on the Pareto front when we cannot improve one objective without necessarily worsening at least one other objective (as shown by the red line in the figure below).

More detail on Pareto fronts here: https://pythonhealthcare.org/2018/09/27/93-exploring-the-best-possible-trade-off-between-competing-objectives-identifying-the-pareto-front/


The algorithm implemented here is based on an algorithm called ‘NSGA-II’. This was published by Deb et al.

Deb et al. (2002) A Fast and Elitist Multiobjective Genetic Algorithm: NSGA-II. IEEE Transactions on Evolutionary Computation. 6, 182-197.

With NSGA-II we maintain a population of a given size (in the original paper that is fixed; in our implementation we define a range – the population must be larger than a minimum size, but not exceed a given maximum size.

The key process steps of NSGA-II are:

1) Start with a random population of solutions (P), encoded in binary form (‘chromosomes’) 2) Create a child population (Q) through crossover and mutation 3) Combine P & Q and score for all objectives 4) Identify the first Pareto front (F1); that is all solutions where there are no other solutions that are at least equally good in all objectives and better in at least one objective 5) If F1 is larger than the maximum permitted solution then reduce the size of F1 by ‘crowding selection’ (see below). 6) If F1 is smaller than the required population size then repeat Pareto selection (after removal of selected already selected). This new set of solutions is F2. If the total number selected solutions is greater than the permitted maximum population size then reduce the size of just the latest selection (in this case F2) so that the number of all selected solutions is equal to the maximum permitted population size. 7) Repeat Pareto selection until the required population size is reached (and then reduce the last selected Pareto front by ‘crowding selection’ as required to avoid exceeding the maximum permitted population size). 8) The selected Pareto fronts for the new population, P. 9) Repeat from (2) for the required number of generations or until some other ‘stop’ criterion is reached. 10) Perform a final Pareto selection so that the final reported population are just those on the first Pareto front.

A minimum population size is required to maintain diversity of solutions. If we only selected the first Pareto front we may have a very small population which would lack the diversity required to reach good final solutions.

This process is shown diagrammatically below (from Deb et al., 2002):

Crowding distances/selection

When we are using a Pareto front to select solutions (as described here), all solutions are on the optimal front, that is in each solution there is no other solution that is at least as good in all scores, and better in at least one score. We therefore cannot rank solutions by performance. In order to select solutions, if we need to control the number of solutions we are generating we can use ‘crowding distances’. Crowding distances give a measure of closeness in performance to other solutions. The crowding distance is the average distance to its two neighbouring solutions. we will bias the selection of solutions to those with greater distances to neighbouring solutions.

We will use a ‘tournament’ method to select between two individual solutions. The crowding distances are calculated for all solutions. Two solutions are picked at random and the one with the greater crowding distance will be selected (with the unselected solution being returned back). This process is repeated until the require number of solutions have been selected.

An example is shown below – by using crowding distances we have reduced the number of points in densely packed regions by more than in the sparsely populated areas.

The Code

We will break the code down into sections and functions.

To enale easier visualisation we will look to identify a Pareto population based on two competing objectives. But all the code will work on as many objectives as you wish (though it is not recommended to try to optimise more than five objectives).

Import required libraries

import random as rn
import numpy as np
import matplotlib.pyplot as plt
# For use in Jupyter notebooks only:
% matplotlib inline

Create reference solutions

In this example we will pick the easiest example possible. We will create two reference chromosomes as ‘ideal solutions’. This function may be used to create more references to mimic optimising against more than two objectives.

We will pick a chromosome length of 25 (this mimics solutions that would be coded as 25 binary ‘genes’).

In real life this binary representation would lead to better or worse solutions. We might, for example, use the binary notation to denote open/closed hospitals in a location problem. Or any number of binary digits might be used to encode the value of a variable.

In real life this binary representation would lead to better or worse solutions. We might, for example, use the binary notation to denote open/closed hospitals in a location problem. Or any number of binary digits might be used to encode the value of a variable.

def create_reference_solutions(chromosome_length, solutions):
    Function to create reference chromosomes that will mimic an ideal solution 
    references = np.zeros((solutions, chromosome_length))
    number_of_ones = int(chromosome_length / 2)
    for solution in range(solutions):
        # Build an array with an equal mix of zero and ones
        reference = np.zeros(chromosome_length)
        reference[0: number_of_ones] = 1
        # Shuffle the array to mix the zeros and ones
        references[solution,:] = reference
    return references

Show an example set of reference solutions:

print (create_reference_solutions(25, 2))

print (create_reference_solutions(25, 2))

[[0. 1. 0. 1. 0. 0. 1. 1. 0. 1. 1. 1. 0. 0. 0. 1. 0. 0. 0. 0. 1. 1. 1. 1.
 [0. 1. 0. 1. 0. 0. 0. 1. 0. 0. 1. 0. 1. 1. 0. 1. 1. 1. 0. 0. 1. 0. 1. 1.

Evaluating solutions

Normally this section of the code would be more complex – it decodes the chromosome and applies it to a problem, and assesses the performance. For example we might use the chromosome to store whether hospitals are open/closed and then a piece of code will test travel distances for all patients to the open hospitals, and will also calculate the number of admissions to each hospital.

In our ‘toy’ example here we will simply calculate how many binary digits in each solution are the same as our two reference solutions (so we will have two scores for each solution).

We have two functions. The first will calculate the score (‘fitness’) against a single reference. The second will loop through all references (equivalent to all objectives) and call for the score/fitness to be calculated.

def calculate_fitness(reference, population):
    Calculate how many binary digits in each solution are the same as our 
    reference solution.
    # Create an array of True/False compared to reference
    identical_to_reference = population == reference
    # Sum number of genes that are identical to the reference
    fitness_scores = identical_to_reference.sum(axis=1)
    return fitness_scores
def score_population(population, references):
    Loop through all reference solutions and request score/fitness of
    population against that reference solution.
    scores = np.zeros((population.shape[0], references.shape[0]))
    for i, reference in enumerate(references):
        scores[:,i] = calculate_fitness(reference, population)
    return scores 

Calculate crowding and select a population based on crowding scores

We have two functions here. The first to calculate the crowding of a population, based on similarity of scores. The second is a Tournament selection method that uses those crowding scores to pick a given number of solutions from a population.

def calculate_crowding(scores):
    Crowding is based on a vector for each individual
    All scores are normalised between low and high. For any one score, all
    solutions are sorted in order low to high. Crowding for chromsome x
    for that score is the difference between the next highest and next
    lowest score. Total crowding value sums all crowding for all scores
    population_size = len(scores[:, 0])
    number_of_scores = len(scores[0, :])

    # create crowding matrix of population (row) and score (column)
    crowding_matrix = np.zeros((population_size, number_of_scores))

    # normalise scores (ptp is max-min)
    normed_scores = (scores - scores.min(0)) / scores.ptp(0)

    # calculate crowding distance for each score in turn
    for col in range(number_of_scores):
        crowding = np.zeros(population_size)

        # end points have maximum crowding
        crowding[0] = 1
        crowding[population_size - 1] = 1

        # Sort each score (to calculate crowding between adjacent scores)
        sorted_scores = np.sort(normed_scores[:, col])

        sorted_scores_index = np.argsort(
            normed_scores[:, col])

        # Calculate crowding distance for each individual
        crowding[1:population_size - 1] = \
            (sorted_scores[2:population_size] -
             sorted_scores[0:population_size - 2])

        # resort to orginal order (two steps)
        re_sort_order = np.argsort(sorted_scores_index)
        sorted_crowding = crowding[re_sort_order]

        # Record crowding distances
        crowding_matrix[:, col] = sorted_crowding

    # Sum crowding distances of each score
    crowding_distances = np.sum(crowding_matrix, axis=1)

    return crowding_distances
def reduce_by_crowding(scores, number_to_select):
    This function selects a number of solutions based on tournament of
    crowding distances. Two members of the population are picked at
    random. The one with the higher croding dostance is always picked
    population_ids = np.arange(scores.shape[0])

    crowding_distances = calculate_crowding(scores)

    picked_population_ids = np.zeros((number_to_select))

    picked_scores = np.zeros((number_to_select, len(scores[0, :])))

    for i in range(number_to_select):

        population_size = population_ids.shape[0]

        fighter1ID = rn.randint(0, population_size - 1)

        fighter2ID = rn.randint(0, population_size - 1)

        # If fighter # 1 is better
        if crowding_distances[fighter1ID] >= crowding_distances[

            # add solution to picked solutions array
            picked_population_ids[i] = population_ids[

            # Add score to picked scores array
            picked_scores[i, :] = scores[fighter1ID, :]

            # remove selected solution from available solutions
            population_ids = np.delete(population_ids, (fighter1ID),

            scores = np.delete(scores, (fighter1ID), axis=0)

            crowding_distances = np.delete(crowding_distances, (fighter1ID),
            picked_population_ids[i] = population_ids[fighter2ID]

            picked_scores[i, :] = scores[fighter2ID, :]

            population_ids = np.delete(population_ids, (fighter2ID), axis=0)

            scores = np.delete(scores, (fighter2ID), axis=0)

            crowding_distances = np.delete(
                crowding_distances, (fighter2ID), axis=0)

    # Convert to integer 
    picked_population_ids = np.asarray(picked_population_ids, dtype=int)
    return (picked_population_ids)

Pareto selection

Pareto selection involves two functions. The first will select a single Pareto front. The second will, as necessary, repeat Pareto front selection to build a population within defined size limits, and, as necessary, will reduce a Pareto front by applying crowding selection.

def identify_pareto(scores, population_ids):
    Identifies a single Pareto front, and returns the population IDs of
    the selected solutions.
    population_size = scores.shape[0]
    # Create a starting list of items on the Pareto front
    # All items start off as being labelled as on the Parteo front
    pareto_front = np.ones(population_size, dtype=bool)
    # Loop through each item. This will then be compared with all other items
    for i in range(population_size):
        # Loop through all other items
        for j in range(population_size):
            # Check if our 'i' pint is dominated by out 'j' point
            if all(scores[j] >= scores[i]) and any(scores[j] > scores[i]):
                # j dominates i. Label 'i' point as not on Pareto front
                pareto_front[i] = 0
                # Stop further comparisons with 'i' (no more comparisons needed)
    # Return ids of scenarios on pareto front
    return population_ids[pareto_front]
def build_pareto_population(
        population, scores, minimum_population_size, maximum_population_size):
    As necessary repeats Pareto front selection to build a population within
    defined size limits. Will reduce a Pareto front by applying crowding 
    selection as necessary.    
    unselected_population_ids = np.arange(population.shape[0])
    all_population_ids = np.arange(population.shape[0])
    pareto_front = []
    while len(pareto_front) < minimum_population_size:
        temp_pareto_front = identify_pareto(
                scores[unselected_population_ids, :], unselected_population_ids)
        # Check size of total parteo front. 
        # If larger than maximum size reduce new pareto front by crowding
        combined_pareto_size = len(pareto_front) + len(temp_pareto_front)
        if combined_pareto_size > maximum_population_size:
            number_to_select = combined_pareto_size - maximum_population_size
            selected_individuals = (reduce_by_crowding(
                    scores[temp_pareto_front], number_to_select))
            temp_pareto_front = temp_pareto_front[selected_individuals]
        # Add latest pareto front to full Pareto front
        pareto_front = np.hstack((pareto_front, temp_pareto_front))
        # Update unselected population ID by using sets to find IDs in all
        # ids that are not in the selected front
        unselected_set = set(all_population_ids) - set(pareto_front)
        unselected_population_ids = np.array(list(unselected_set))

    population = population[pareto_front.astype(int)]
    return population

Population functions

There are four functions concerning the population

1) Create a random population

2) Breed by crossover – two children produced from two parents

3) Randomly mutate population (small probability that any given gene in population will switch between 1/0 – applied to the child population)

4) Breed population -Create child population by repeatedly calling breeding function (two parents producing two children), applying genetic mutation to the child population, combining parent and child population, and removing duplicate chromosomes.

def create_population(individuals, chromosome_length):
    Create random population with given number of individuals and chroosome
    # Set up an initial array of all zeros
    population = np.zeros((individuals, chromosome_length))
    # Loop through each row (individual)
    for i in range(individuals):
        # Choose a random number of ones to create
        ones = rn.randint(0, chromosome_length)
        # Change the required number of zeros to ones
        population[i, 0:ones] = 1
        # Sfuffle row
    return population
def breed_by_crossover(parent_1, parent_2):
    Combine two parent chromsomes by crossover to produce two children.
    # Get length of chromosome
    chromosome_length = len(parent_1)
    # Pick crossover point, avoding ends of chromsome
    crossover_point = rn.randint(1,chromosome_length-1)
    # Create children. np.hstack joins two arrays
    child_1 = np.hstack((parent_1[0:crossover_point],
    child_2 = np.hstack((parent_2[0:crossover_point],
    # Return children
    return child_1, child_2
def randomly_mutate_population(population, mutation_probability):
    Randomly mutate population with a given individual gene mutation
    probability. Individual gene may switch between 0/1.
    # Apply random mutation
    random_mutation_array = np.random.random(size=(population.shape))
    random_mutation_boolean = \
        random_mutation_array <= mutation_probability

    population[random_mutation_boolean] = \
    # Return mutation population
    return population
def breed_population(population):
    Create child population by repetedly calling breeding function (two parents
    producing two children), applying genetic mutation to the child population,
    combining parent and child population, and removing duplicatee chromosomes.
    # Create an empty list for new population
    new_population = []
    population_size = population.shape[0]
    # Create new popualtion generating two children at a time
    for i in range(int(population_size/2)):
        parent_1 = population[rn.randint(0, population_size-1)]
        parent_2 = population[rn.randint(0, population_size-1)]
        child_1, child_2 = breed_by_crossover(parent_1, parent_2)
    # Add the child population to the parent population
    # In this method we allow parents and children to compete to be kept
    population = np.vstack((population, np.array(new_population)))
    population = np.unique(population, axis=0)
    return population

Main algorithm code

Now let’s put it all together!

# Set general parameters
chromosome_length = 50
starting_population_size = 5000
maximum_generation = 250
minimum_population_size = 500
maximum_population_size = 1000

# Create two reference solutions 
# (this is used just to illustrate GAs)
references = create_reference_solutions(chromosome_length, 2)

# Create starting population
population = create_population(
        starting_population_size, chromosome_length)

# Loop through the generations of genetic algorithm

for generation in range(maximum_generation):
    if generation %10 ==0:
        print ('Generation (out of %i): %i '%(maximum_generation, generation))
    # Breed
    population = breed_population(population)
    # Score population
    scores = score_population(population, references)
    # Build pareto front
    population = build_pareto_population(
            population, scores, minimum_population_size, maximum_population_size)

# Get final pareto front
scores = score_population(population, references)
population_ids = np.arange(population.shape[0]).astype(int)
pareto_front = identify_pareto(scores, population_ids)
population = population[pareto_front, :]
scores = scores[pareto_front]

Plot final Pareto front

This is the only part of the code that works only for two objectives.

You will see in this example we have a well-described trade-off between achieving the two objectives. With the population sizes used and the number of generations we achieved 96% maximum score on each objective (a larger population and/or more generations would get us to 100% on this quite simple example, though Pareto algorithms should not be expected to find the perfect solution for all objectives, but rather they should identify good solutions).

# Plot Pareto front (for two scores only)  
x = scores[:, 0]/chromosome_length*100
y = scores[:, 1]/chromosome_length*100
plt.xlabel('Objective A - % maximum obtainable')
plt.ylabel('Objective B - % maximum obtainable')