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How to Find the Most Frequent Value in Each Group of a Pandas DataFrame?

Linda Hamilton
Release: 2024-12-01 08:22:10
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How to Find the Most Frequent Value in Each Group of a Pandas DataFrame?

Select Most Common Value for Each Group in a DataFrame

To clean data that contains multiple string columns, it's necessary to group the rows by certain columns and select the most common value for a specific column within each group. This article demonstrates how to accomplish this task using the powerful Pandas library.

Code Correction for Specific Error Messages

The code provided in the initial query contains some errors, which have been corrected below:

import pandas as pd

source = pd.DataFrame({
    'Country': ['USA', 'USA', 'Russia', 'USA'], 
    'City': ['New York', 'New York', 'Saint Petersburg', 'New York'],
    'Short Name': ['NY', 'New', 'Spb', 'NY']})

# Group by 'Country' and 'City' and calculate the most frequent 'Short Name' in each group
result = source.groupby(['Country', 'City'])['Short Name'].apply(lambda x: pd.Series.mode(x)[0][0])
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Explanation

  1. Use the latest Series.mode: The original code attempts to apply statistics.mode to each group, which doesn't handle multiple modes well and can raise an error. Instead, the more recent pd.Series.mode function is used, which explicitly returns a Series of all the modes, solving the issue.
  2. Handle multiple modes: To ensure that only a single most common value is selected, the code extracts the first element from the Series returned by Series.mode. This is achieved by using the 0 syntax.

Additional Options

If a DataFrame is preferred as the result:

result = source.groupby(['Country', 'City'])['Short Name'].agg(pd.Series.mode).to_frame()
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If you want separate rows for each mode:

result = source.groupby(['Country', 'City'])['Short Name'].apply(pd.Series.mode)
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Note: If you're willing to accept any mode value as the selection, you can use a lambda function that extracts the first mode from the Series:

result = source.groupby(['Country', 'City'])['Short Name'].agg(lambda x: pd.Series.mode(x)[0])
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