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Скачать или смотреть Fast and Efficient Ways to Iterate Over Pandas DataFrame Rows

  • vlogize
  • 2025-05-28
  • 0
Fast and Efficient Ways to Iterate Over Pandas DataFrame Rows
Fast way to iterate over rows and return column names where cells meet thresholdpythonpandas
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Описание к видео Fast and Efficient Ways to Iterate Over Pandas DataFrame Rows

Learn how to quickly iterate over rows in a Pandas DataFrame and return column names where values meet a specified threshold, all while boosting performance and efficiency!
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This video is based on the question https://stackoverflow.com/q/66974948/ asked by the user 'Tyler Rinker' ( https://stackoverflow.com/u/1000343/ ) and on the answer https://stackoverflow.com/a/66975270/ provided by the user 'Ch3steR' ( https://stackoverflow.com/u/12416453/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions.

Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Fast way to iterate over rows and return column names where cells meet threshold

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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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Fast Ways to Iterate Over Pandas DataFrame Rows

When handling large datasets in Python, particularly with Pandas DataFrames, you may often need to analyze your data based on certain thresholds. A common task is to identify which columns in a DataFrame have values exceeding a set threshold for each row. This guide will guide you through an effective method to achieve this, using a sample DataFrame filled with numeric values.

The Problem

Suppose you have a Pandas DataFrame with 10,000 rows and 700 columns. The intention is to find out which column names contain values that meet or exceed a specific threshold, say 0.05. You want to return this information in a dictionary format, where each key is an index label (like animal names) and the corresponding value is a list of columns that meet the condition.

Sample DataFrame

Here’s how a small portion of such a DataFrame might look:

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

The DataFrame resembles this:

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

We desire the output to look something like this:

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

The Solution

To tackle this task efficiently, we can leverage dictionary comprehension combined with boolean indexing. Below is a succinct method to achieve the goal with minimal execution time.

Code Implementation

Here’s the code that accomplishes this:

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

Breakdown of the Code

Boolean Indexing: df.ge(0.05) creates a boolean DataFrame where True indicates that the value meets or exceeds the threshold of 0.05.

Extract Values: The values method extracts the boolean values so that we can use them easily.

Dictionary Comprehension: We use a dictionary comprehension to map each index (animal names) to a list of column names that fulfill our condition.

Performance Test

To assess the performance of this solution, we can use the timeit library in Python:

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

On average, you should find that this method completes in about 167 milliseconds for the entire DataFrame.

Conclusion

In summary, iterating over rows and identifying column names that meet a specified threshold can be executed efficiently in Pandas using boolean indexing and dictionary comprehension. By structuring your DataFrame this way, you save time and resources, particularly when handling larger datasets. This method is not only fast but also easy to implement, making it a powerful strategy for data analysis tasks.

Keep this technique in mind as you work with data in Python – it’s bound to challenge your efficiency and provide answers swiftly!

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