Filter löschen
Filter löschen

Zero a cell matrix with varying number of zeros

3 Ansichten (letzte 30 Tage)
Mohammed Kagalwala
Mohammed Kagalwala am 30 Okt. 2019
Kommentiert: ME am 31 Okt. 2019
Hi,
I have a cell matrix, C, that's n x n, and an array A of size n x 1. The entries of array A dictate the no. of column of zeros in the matrix M.
i.e. M1 has a dimension 6 x A(1), M2 has a dimension 6 x A(2)
Given an index i, what I wish to do is the following,
Column i of C is filled with matrix Mi of dimension 6xA(i). So all n entries in column i are replaced with a matrix of zeros.
I know the following code
C(:,:) = {zeros(3,3)}
will give me zeros eveyrwhere, I want to leverage similar syntax to solve my problem. Avoiding the use of for loops.
Thank you.

Antworten (1)

ME
ME am 30 Okt. 2019
I'm not 100% sure I follow, but as I understand you want to take some list of columns and turn all elements in that column to zeros?
If so then you could use something like:
C(:,cols) = 0
where cols is an array containing the indices of the columns to be zeroed. For example setting cols = [1 3] would put zeros in the first and thrid columns.
If that's not what you meant then please could you explain some more?
  6 Kommentare
Mohammed Kagalwala
Mohammed Kagalwala am 30 Okt. 2019
C is a cell of a fixed size nxn. Each cell column will have a unique size based on A. i.e. something like this,
cell.PNG
In this case A = [5 1 5 10 8 8 8 8 9 8 10 2].
This was however created using a for loop. I want to make this same cell structure without the use of for loops.
ME
ME am 31 Okt. 2019
Oh, I see! Sorry, I completely missed the fact you wanted to do this inside a cell array. I don’t have a good answer for you right now but I’ll have a think on it.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by