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]]

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

40. Removing duplicate data in NumPy and Pandas

Both NumPy and Pandas offer easy ways of removing duplicate rows. Pandas offers a more powerful approach if you wish to remove rows that are partly duplicated.


With numpy we use np.unique() to remove duplicate rows or columns (use the argument axis=0 for unique rows or axis=1 for unique columns). Continue reading “40. Removing duplicate data in NumPy and Pandas”