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Скачать или смотреть Deleting Unwanted Rows in a 3D Matrix in MATLAB

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  • 2025-05-25
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Deleting Unwanted Rows in a 3D Matrix in MATLAB
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Описание к видео Deleting Unwanted Rows in a 3D Matrix in MATLAB

Learn how to effectively delete rows in a `3D matrix` based on column values in MATLAB, solving common indexing issues.
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This video is based on the question https://stackoverflow.com/q/71064859/ asked by the user 'Laura' ( https://stackoverflow.com/u/17744565/ ) and on the answer https://stackoverflow.com/a/71065133/ provided by the user 'Adriaan' ( https://stackoverflow.com/u/5211833/ ) 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: How to delete rows based on value of first column in matrix

Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l...
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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Deleting Unwanted Rows in a 3D Matrix in MATLAB: A Step-by-Step Guide

When working with matrices in MATLAB, you may encounter situations where you need to filter out certain rows based on the values in a specific column. One common scenario is deleting rows where the values in the first column fall below a threshold or exceed a certain maximum. In this guide, we will explore how to achieve this task efficiently and correctly using MATLAB code.

The Problem

Let's say you have a 3D matrix named xyposframe, and you want to delete all rows where the first column contains numbers either below 150 or above 326. Initially, you might try to implement this using two separate conditions, but this can lead to issues with matrix dimensions, resulting in errors.

The initial attempt may look something like this:

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

However, you may find that this does not yield the expected result.

The Cause of the Problem

The main issue here is that after executing the first line to delete rows, the size of xyposframe is reduced. Consequently, when you try to access indexabove, which references the original matrix size, an error occurs because xyposframe now has fewer rows than before.

Solution Overview

To delete the unwanted rows safely:

Split the operation into two distinct steps, handling each condition separately.

Combine conditions into a single logical statement that can filter rows in one go.

Let's break down these solutions.

Method 1: Splitting the Operation

This approach involves deleting rows according to one condition before addressing the next condition. Here’s how to implement this:

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

Method 2: Combining Conditions

A more efficient way would be to combine both conditions and execute in a single line. This avoids the need for multiple indexing operations on the matrix:

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

Explanation of the Code

The expressions xyposframe(:, 1) < 150 and xyposframe(:, 1) > 326 return logical arrays that indicate which rows fall under these conditions.

By using | (logical OR), we can filter xyposframe to only include rows where the first condition is true or the second condition is true.

Finalizing the statements with , : preserves all columns in the remaining rows.

Conclusion

In summary, when you need to delete rows from a 3D matrix in MATLAB based on the values of the first column, it is crucial to ensure that your indexing logic accounts for changes in matrix dimensions. By either splitting the operations or combining the conditions into one logical statement, you can effectively manage your data without running into errors.

Feel free to try out the provided MATLAB code snippets in your own projects, and you’ll find this method to be both robust and efficient!

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