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Скачать или смотреть How to Use If-Else Logic to Compute a New Column in a Pandas DataFrame

  • vlogize
  • 2025-03-31
  • 0
How to Use If-Else Logic to Compute a New Column in a Pandas DataFrame
If-else logic to set value of dataframe columnpythonpandasdataframelambda
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Описание к видео How to Use If-Else Logic to Compute a New Column in a Pandas DataFrame

Learn how to utilize `if-else` statements to create a new column in your Pandas DataFrame, ensuring that zero values are not included in the ranking. Perfect for data analysis tasks!
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This video is based on the question https://stackoverflow.com/q/70217280/ asked by the user 'hansolo' ( https://stackoverflow.com/u/4668284/ ) and on the answer https://stackoverflow.com/a/70217969/ provided by the user 'eshirvana' ( https://stackoverflow.com/u/1367454/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

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The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license.

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How to Use If-Else Logic to Compute a New Column in a Pandas DataFrame

When working with data in a Pandas DataFrame, you often encounter scenarios where you need to manipulate existing data to derive new insights. A common requirement is to create a new column based on the conditions of existing ones. In this post, we’ll explore how to use if-else logic to create a new column in a DataFrame, ensuring that certain values (like zeros) are handled appropriately.

The Problem

Imagine you have a DataFrame that contains user sessions, and you want to create a new column that ranks the sessions while ignoring any rows with zero sessions. The DataFrame looks like this:

IDSessions123440056782009101112199131415160Your goal is to add a new column, new_col, that ranks the sessions, but it should give a rank of 0 for any ID with 0 sessions. The desired output would be:

IDSessionsnew_col12344001.00000056782000.99998791011121990.9999741314151600The Solution

To achieve this, you can use an approach that incorporates conditional logic to replace zero values with NaN, followed by using the ranking function. Let’s break down two efficient methods of achieving this.

Method 1: Using df.loc

Use loc to Filter: We can directly rank only those sessions greater than zero using the loc method.

[[See Video to Reveal this Text or Code Snippet]]

Understanding the Code:

df.loc[df.Sessions > 0, 'Sessions']: This filters the session values that are greater than zero.

.rank(ascending=True, pct=True): This ranks those values in ascending order and returns the percentage rank.

Method 2: Using replace with rank

Replace Zeros with NaN: Another approach is to replace zeros with NaN which will be ignored during the ranking process.

[[See Video to Reveal this Text or Code Snippet]]

Understanding the Code:

df['Sessions'].replace(0, np.NaN): This replaces any session counts of zero with NaN.

.rank(pct=True): This ranks the remaining values, treating NaN as missing.

.fillna(0): Finally, it fills the NaNs created from the zeros back to zero in the new column.

Conclusion

Using if-else logic in Pandas can be straightforward with the right methods. By incorporating either of the above strategies, you can easily create a new column that meets your needs while handling zeros properly. This not only makes your data more insightful but also enables more effective analysis.

Give these methods a try on your own DataFrames and see how they can enhance your data manipulation tasks. Happy coding!

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