Aggregation with Lambda Function for Last 30 Days with Python

Introduction

In this article, we will explore how to use a lambda function in pandas to perform aggregation on a specific date range. We’ll also dive into the issue of NaN values that can occur when merging the aggregated data back into the original DataFrame.

Aggregation Basics

Before we begin, let’s review some basic concepts of aggregation in pandas.

• Grouping: When you group DataFrames by one or more columns, you’re creating a set of subgroups to operate on.
• Aggregation Functions: Pandas provides various built-in functions for performing different types of aggregations, such as sum, mean, max, and so on.

Using Lambda Function for Aggregation

Here’s an example of how we can use a lambda function for aggregation:

import pandas as pd

# Create sample data
data = {
    'user_ID': [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
    'Date': ['2012-09-01 10:00:00', '2012-09-02 11:00:00', '2012-09-03 12:00:00',
             '2012-10-01 13:00:00', '2012-10-02 14:00:00', '2012-10-03 15:00:00',
             '2012-10-04 16:00:00', '2012-09-01 18:00:00', '2012-09-02 19:00:00',
             '2012-09-03 20:00:00', '2012-09-04 21:00:00', '2012-09-05 22:00:00'],
    'sales': [10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 20.0, 20.0, 20.0,
              20.0, 20.0]
}

df = pd.DataFrame(data)

# Group by user_ID and calculate the sum of sales for each day
grouped_df = df.groupby('user_ID')['sales'].agg(lambda x: x[(x > (x.max() - pd.to_timedelta(30, unit='d')))]).sum().reset_index()

Merging the Aggregated Data

To merge the aggregated data back into the original DataFrame, you can use the following methods:

• Reset Index and Merge: Reset the index of the aggregated DataFrame and then merge it with the original DataFrame on the ‘user_ID’ column.

merged_df = df.merge(grouped_df.reset_index(), left_on=‘user_ID’, right_on=‘user_ID’)

*   **Use a Common Column for Merging**: If you don't want to reset the index of the aggregated DataFrame, you can use a common column (in this case, 'user_ID') for merging.

    ```markdown
merged_df = df.merge(grouped_df, left_on='user_ID', right_on='user_ID')

Handling NaN Values in Aggregated Data

When using lambda functions for aggregation, it’s possible to encounter NaN values in the resulting DataFrame. In this case, we can use various methods to handle these missing values.

• Drop Missing Values: Drop any rows with missing values from the aggregated DataFrame.

grouped_df = grouped_df.dropna()

*   **Fill Missing Values with a Specific Value**: Fill missing values in the aggregated DataFrame with a specific value (e.g., 0 or the mean of the column).

    ```markdown
grouped_df['sales_30d_lag'] = grouped_df['sales_30d_lag'].fillna(0)

• Interpolate Missing Values: Interpolate missing values in the aggregated DataFrame.

import pandas as pd

Create sample data

data = { ‘user_ID’: [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2], ‘Date’: [‘2012-09-01 10:00:00’, ‘2012-09-02 11:00:00’, ‘2012-09-03 12:00:00’, ‘2012-10-01 13:00:00’, ‘2012-10-02 14:00:00’, ‘2012-10-03 15:00:00’, ‘2012-10-04 16:00:00’, ‘2012-09-01 18:00:00’, ‘2012-09-02 19:00:00’, ‘2012-09-03 20:00:00’, ‘2012-09-04 21:00:00’, ‘2012-09-05 22:00:00’], ‘sales’: [10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 20.0, 20.0, 20.0, 20.0, 20.0] }

df = pd.DataFrame(data)

Group by user_ID and calculate the sum of sales for each day

grouped_df = df.groupby(‘user_ID’)[‘sales’].agg(lambda x: x[(x > (x.max() - pd.to_timedelta(30, unit=’d’)))]).sum().reset_index()

Interpolate missing values

import pandas as pd

Create sample data

data = { ‘user_ID’: [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2], ‘Date’: [‘2012-09-01 10:00:00’, ‘2012-09-02 11:00:00’, ‘2012-09-03 12:00:00’, ‘2012-10-01 13:00:00’, ‘2012-10-02 14:00:00’, ‘2012-10-03 15:00:00’, ‘2012-10-04 16:00:00’, ‘2012-09-01 18:00:00’, ‘2012-09-02 19:00:00’, ‘2012-09-03 20:00:00’, ‘2012-09-04 21:00:00’, ‘2012-09-05 22:00:00’], ‘sales’: [10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 20.0, 20.0, 20.0, 20.0, 20.0] }

df = pd.DataFrame(data)

Group by user_ID and calculate the sum of sales for each day

grouped_df = df.groupby(‘user_ID’)[‘sales’].agg(lambda x: x[(x > (x.max() - pd.to_timedelta(30, unit=’d’)))]).sum().reset_index()

Interpolate missing values with linear interpolation

interpolated_df = grouped_df.interpolate(method=‘linear’)

Conclusion
----------

In this article, we explored how to use lambda functions for aggregation in pandas. We discussed the importance of handling NaN values that can occur when merging aggregated data back into the original DataFrame. By understanding these concepts and using various methods for handling missing values, you'll be able to perform efficient and effective aggregations in your DataFrames.

Example Use Case: Merging Aggregated Data with the Original DataFrame
-----------------------------------------------------------------

Here's an example of how we can merge the aggregated data with the original DataFrame:

```markdown
import pandas as pd

# Create sample data
data = {
    'user_ID': [1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
    'Date': ['2012-09-01 10:00:00', '2012-09-02 11:00:00', '2012-09-03 12:00:00',
             '2012-10-01 13:00:00', '2012-10-02 14:00:00', '2012-10-03 15:00:00',
             '2012-10-04 16:00:00', '2012-09-01 18:00:00', '2012-09-02 19:00:00',
             '2012-09-03 20:00:00', '2012-09-04 21:00:00', '2012-09-05 22:00:00'],
    'sales': [10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 10.0, 20.0, 20.0, 20.0,
              20.0, 20.0]
}

df = pd.DataFrame(data)

# Group by user_ID and calculate the sum of sales for each day
grouped_df = df.groupby('user_ID')['sales'].agg(lambda x: x[(x > (x.max() - pd.to_timedelta(30, unit='d')))]).sum().reset_index()

# Interpolate missing values with linear interpolation
interpolated_df = grouped_df.interpolate(method='linear')

# Merge the aggregated data with the original DataFrame
merged_df = df.merge(interpolated_df.reset_index(), left_on='user_ID', right_on='user_ID')

print(merged_df)

This will produce a DataFrame that includes the original sales values and the aggregated sales for each day:

Last modified on 2023-07-07