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How to Calculate Rolling Correlation in Python?

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  • Last Updated : 29 Jan, 2022

Correlation generally determines the relationship between two variables. The rolling correlation measure the correlation between two-time series data on a rolling window Rolling correlation can be applied to a specific window width to determine short-term correlations. 

Calculating Rolling Correlation in Python

Let’s use sales data of two products A and B in the last 60 months to calculate the rolling correlation. Pandas package provides a function called rolling.corr() to calculate the rolling correlation.

Syntax:

data1.rolling(width).corr(data2)

Where, 

  • data1, data2 – data/column of interest (type series)
  • width – Rolling window width (int)

Note: The width of the rolling window should be 3 or greater in order to calculate correlations.

Data Used:  

Python3




# import pandas module
import pandas as pd
 
# read the data
data = pd.read_csv('product_sales.csv')
 
# display top 10 rows
print(data.head(10))
 
# display column names
print(data.columns)

Output:

Example 2:

Here, we used the window width of 6, which shows the successive 6 months rolling correlation. We could see a significant correlation between two products sales any sudden dip or rise in correlation signals an unusual event, that caused the dip.

Python3




data['Product A'].rolling(6).corr(data['Product B'])
 
# formatting the output
k = 1
for i, j in enumerate(data['Product A'].rolling(6).corr(data['Product B'])):
    if (i >= 5 and i < 12):
        print(f'The correlation in sales during months\
        {k} through {i+1} is {j}')
        i = 0
        k += 1

Output:

Now’s let us try the same for 3-month correlation as shown below,

Example 3:

Python3




data['Product A'].rolling(3).corr(data['Product B'])
 
# formatting the output
k = 1
for i, j in enumerate(data['Product A'].rolling(3).corr(data['Product B'])):
    if (i >= 3 and i < 12):
        print(
            f'The correlation in sales during months {k} \
            through {i+1} is {j}')
        i = 0
        k += 1

Output:


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