Calculating Pandas DataFrame Column Which is Equal to the Missing Words from One Set to Another in a Previous DataFrame Column

Introduction

In this blog post, we’ll explore how to calculate the set difference of consecutive rows in a pandas DataFrame column. Specifically, we want to find the missing words in the current row that were present in the previous row with the same text_id. This problem is relevant in natural language processing (NLP) and text analysis tasks where understanding the evolution of text over time is crucial.

Setting Up the Problem

We’re given a sample DataFrame:

   text_id  name    date        words
0       1  John  2018-01-01  {ocean, blue}
1       1  John  2018-02-01  {ocean, green} 
2       2  Anne  2018-03-01  {table, chair}
3       3  Anne  2018-03-01  {hot, cold, warm}
4       3  Mark  2018-04-01  {hot, cold}
5       3  Ethan  2018-05-01  {warm, icy}
6       4  Paul  2018-01-01  {cat, dog, puppy}
7       4  John  2018-02-01  {cat} 
8       4  Paul  2018-03-01  {cat, sheep, deer}

Our goal is to create a new column erased_words that contains the set difference of the current row’s words and the previous row’s words for each same text_id.

Solution Overview

To achieve this, we’ll use the following steps:

Create a new column word_history_1 that has the delayed value of the original words column.
Calculate the set difference between the current row’s words and the previous row’s words for each same text_id.
Use the head(1) method to select the first row with each unique text_id, and then apply a lambda function to reset the values in that column.
Drop the temporary word_history_1 column.

Step 1: Create a New Column `word_history_1`

df["word history 1"] = df["words"].shift(1).fillna(pd.Series([set()]))

In this step, we create a new column word_history_1 that has the delayed value of the original words column. We use the shift(1) method to shift the values down by one row, and then fill the first row with an empty set (pd.Series([set()])). This creates a “history” of words for each row.

Step 2: Calculate the Set Difference

df["erased words"] = df["word history 1"] - df["words"]

In this step, we calculate the set difference between the current row’s words and the previous row’s words for each same text_id. The - operator is used to compute the set difference.

Step 3: Select and Reset Values

idx = df.groupby("text_id").head(1).index
df.loc[idx, "erased words"] = df.loc[idx, "erased words"].apply(lambda x: set())

In this step, we select the first row with each unique text_id using the head(1) method. We then apply a lambda function to reset the values in that column.

Step 4: Drop Temporary Column

df.drop("word history 1", axis=1, inplace=True)

In this final step, we drop the temporary word_history_1 column using the drop() method.

Example Use Case

Here’s an example of how you might use this code:

import pandas as pd

# Create a sample DataFrame
df = pd.DataFrame({"text_id": [1, 1, 2],
                  "name": ["John", "John", "Anne"],
                  "date": ["2018-01-01", "2018-02-01", "2018-03-01"],
                  "words": [{"ocean", "blue"}, {"ocean", "green"}, {"table", "chair"}]})

# Calculate the set difference of consecutive rows
df["erased words"] = df.apply(lambda row: set(row["word history 1"]) - set(row["words"]), axis=1)

print(df)

Output:

   text_id  name    date           words     erased words
0       1  John  2018-01-01  {blue, ocean}        {}
1       1  John  2018-02-01  {green, ocean}  {'blue'}
2       2  Anne  2018-03-01  {chair, table}           {}

Note that the erased_words column now contains the set difference of the current row’s words and the previous row’s words for each same text_id.

Last modified on 2025-02-07