Can anyone explain me StandardScaler?

Question

I am unable to understand the page of the StandardScaler in the documentation of sklearn.

Can anyone explain this to me in simple terms?

Answer 1

The answers above are great, but I needed a simple example to alleviate some concerns that I have had in the past. I wanted to make sure it was indeed treating each column separately. I am now reassured and can't find what example had caused me concern. All columns ARE scaled separately as described by those above.

CODE

import pandas as pd
import scipy.stats as ss
from sklearn.preprocessing import StandardScaler


data= [[1, 1, 1, 1, 1],[2, 5, 10, 50, 100],[3, 10, 20, 150, 200],[4, 15, 40, 200, 300]]

df = pd.DataFrame(data, columns=['N0', 'N1', 'N2', 'N3', 'N4']).astype('float64')

sc_X = StandardScaler()
df = sc_X.fit_transform(df)

num_cols = len(df[0,:])
for i in range(num_cols):
    col = df[:,i]
    col_stats = ss.describe(col)
    print(col_stats)

OUTPUT

DescribeResult(nobs=4, minmax=(-1.3416407864998738, 1.3416407864998738), mean=0.0, variance=1.3333333333333333, skewness=0.0, kurtosis=-1.3599999999999999)
DescribeResult(nobs=4, minmax=(-1.2828087129930659, 1.3778315806221817), mean=-5.551115123125783e-17, variance=1.3333333333333337, skewness=0.11003776770595125, kurtosis=-1.394993095506219)
DescribeResult(nobs=4, minmax=(-1.155344148338584, 1.53471088361394), mean=0.0, variance=1.3333333333333333, skewness=0.48089217736510326, kurtosis=-1.1471008824318165)
DescribeResult(nobs=4, minmax=(-1.2604572012883055, 1.2668071116222517), mean=-5.551115123125783e-17, variance=1.3333333333333333, skewness=0.0056842140599118185, kurtosis=-1.6438177182479734)
DescribeResult(nobs=4, minmax=(-1.338945389819976, 1.3434309690153527), mean=5.551115123125783e-17, variance=1.3333333333333333, skewness=0.005374558840039456, kurtosis=-1.3619131970819205)

NOTE:

The scipy.stats module is correctly reporting the "sample" variance, which uses (n - 1) in the denominator. The "population" variance would use n in the denominator for the calculation of variance. To understand better, please see the code below that uses scaled data from the first column of the data set above:

Code

import scipy.stats as ss

sc_Data = [[-1.34164079], [-0.4472136], [0.4472136], [1.34164079]]
col_stats = ss.describe([-1.34164079, -0.4472136, 0.4472136, 1.34164079])
print(col_stats)
print()

mean_by_hand = 0
for row in sc_Data:
    for element in row:
        mean_by_hand += element
mean_by_hand /= 4

variance_by_hand = 0
for row in sc_Data:
    for element in row:
        variance_by_hand += (mean_by_hand - element)**2
sample_variance_by_hand = variance_by_hand / 3
sample_std_dev_by_hand = sample_variance_by_hand ** 0.5

pop_variance_by_hand = variance_by_hand / 4
pop_std_dev_by_hand = pop_variance_by_hand ** 0.5

print("Sample of Population Calcs:")
print(mean_by_hand, sample_variance_by_hand, sample_std_dev_by_hand, '\n')
print("Population Calcs:")
print(mean_by_hand, pop_variance_by_hand, pop_std_dev_by_hand)

Output

DescribeResult(nobs=4, minmax=(-1.34164079, 1.34164079), mean=0.0, variance=1.3333333422778562, skewness=0.0, kurtosis=-1.36000000429325)

Sample of Population Calcs:
0.0 1.3333333422778562 1.1547005422523435

Population Calcs:
0.0 1.000000006708392 1.000000003354196

Answer 2

Following is a simple working example to explain how standarization calculation works. The theory part is already well explained in other answers.

>>>import numpy as np
>>>data = [[6, 2], [4, 2], [6, 4], [8, 2]]
>>>a = np.array(data)

>>>np.std(a, axis=0)
array([1.41421356, 0.8660254 ])

>>>np.mean(a, axis=0)
array([6. , 2.5])

>>>from sklearn.preprocessing import StandardScaler
>>>scaler = StandardScaler()
>>>scaler.fit(data)
>>>print(scaler.mean_)

#Xchanged = (X−μ)/σ  WHERE σ is Standard Deviation and μ is mean
>>>z=scaler.transform(data)
>>>z

Calculation

As you can see in the output, mean is [6. , 2.5] and std deviation is [1.41421356, 0.8660254 ]

Data is (0,1) position is 2 Standardization = (2 - 2.5)/0.8660254 = -0.57735027

Data in (1,0) position is 4 Standardization = (4-6)/1.41421356 = -1.414

Result After Standardization

Check Mean and Std Deviation After Standardization

Note: -2.77555756e-17 is very close to 0.

References

1. Compare the effect of different scalers on data with outliers

2. What's the difference between Normalization and Standardization?

3. Mean of data scaled with sklearn StandardScaler is not zero

Answer 3

Intro: I assume that you have a matrix X where each row/line is a sample/observation and each column is a variable/feature (this is the expected input for any sklearn ML function by the way -- X.shape should be [number_of_samples, number_of_features]).

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Core of method: The main idea is to normalize/standardize i.e. μ = 0 and σ = 1 your features/variables/columns of X, individually, before applying any machine learning model.

StandardScaler() will normalize the features i.e. each column of X, INDIVIDUALLY, so that each column/feature/variable will have μ = 0 and σ = 1.

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P.S: I find the most upvoted answer on this page, wrong. I am quoting "each value in the dataset will have the sample mean value subtracted" -- This is neither true nor correct.

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See also: How and why to Standardize your data: A python tutorial

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Example:

from sklearn.preprocessing import StandardScaler
import numpy as np

# 4 samples/observations and 2 variables/features
data = np.array([[0, 0], [1, 0], [0, 1], [1, 1]])
scaler = StandardScaler()
scaled_data = scaler.fit_transform(data)

print(data)
[[0, 0],
 [1, 0],
 [0, 1],
 [1, 1]])

print(scaled_data)
[[-1. -1.]
 [ 1. -1.]
 [-1.  1.]
 [ 1.  1.]]

Verify that the mean of each feature (column) is 0:

scaled_data.mean(axis = 0)
array([0., 0.])

Verify that the std of each feature (column) is 1:

scaled_data.std(axis = 0)
array([1., 1.])

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The maths:

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UPDATE 08/2020: Concerning the input parameters with_mean and with_std to False/True, I have provided an answer here: StandardScaler difference between “with_std=False or True” and “with_mean=False or True”