# How to replace the elements of a matrix using the conditions if,else?

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Emerson De Souza 2011 年 6 月 5 日

I want to replace the elements of a matrix
using different conditions. For instance, let all
elements larger than 0.5 be replaced by -1, else
keep the way it is.
So I thought I should simply write the command below:
X=rand(10,10);
if X(:,:)>0.5;
T(:,:)=-1;
else T=X(:,:);
end;
but it does not work because T==X.
Could someone tell me how to correct these command lines?
Thank you
Emerson

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### 採用された回答

Matt Fig 2011 年 6 月 5 日
IF statements do not pick out elements of an array like you are imagining that they do. When you write this:
if conditional
% Do something
end
for non-scalar conditional, the IF statement will pass if and only if all of the elements in conditional are true or non-zero. For example:
x = [1 2 3]; % All are non-zero, passes conditional...
if x
disp('In if')
end
but now change it to:
x = [1 2 0]; % All are not non-zero, fails conditional...
if x
disp('In if')
end
So the way you have to do what you want is either through logical indexing, or by single value through iteration.
% iteration approach - look at one value at a time!
x = [.1 .2 .3 .7 .8 .9 .1 .2 .3]; % Work with this array.
T = zeros(size(x)); % Make another array to fill up...
for ii = 1:length(x)
if x(ii)>.5
T(ii) = 99;
else
T(ii) = -100;
end
end
% Logical indexing approach
x = [.1 .2 .3 .7 .8 .9 .1 .2 .3]; % Work with this array.
T(x>.5) = 99;
T(x<=.5) = -100
##### 4 件のコメント表示非表示 3 件の古いコメント
Nicholas Nurre 2019 年 10 月 30 日
Mattfig,
What should be done if each individual element must be sorted through but there are thousands of elements. My script takes way too long to run. Is there any optimiziation?

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### その他の回答 (6 件)

Rain 2013 年 12 月 11 日

Hi,
Actually, there is a very simple way to do it:
X=rand(10,10);
X(X>0.5) = [-1];
##### 3 件のコメント表示非表示 2 件の古いコメント
Mohammad Safayet Hossain 2019 年 2 月 16 日
Great command

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Ivan van der Kroon 2011 年 6 月 5 日
You don;t need the if-statment here but only the logicals. This gives you a matrix with ones where X is larger than 0.5 and zeros other wise
(X>0.5)
T = X.* (X<=0.5)-(X>0.5);
Alls values smaller than or equal to 0.5 are kept while the others are set to zero and then a matrix is added that has entries of -1 for the entries of X larger than 0.5.
##### 1 件のコメント表示非表示 なし
Emerson De Souza 2011 年 6 月 5 日
Hi Ivan
For this particular case you are right.
But how do we write, if the matrix T is build up
based on conditions of several matrices?
For example: The elements of T equals -1 if:
1) the elements of L>0.5 AND
2) the elements of M=0 AND
3) the elements of N<0.5
if any of these conditions is not satisfied, T equals the elements of M+1
Hope there is a way to do it
Emerson

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Ivan van der Kroon 2011 年 6 月 5 日
Just implement it for multiple matrices using element multplication:
Logical=(L>0.5).*(M==0).*(N<0.5);
T=-Logical+(M+1).*Logical;
##### 3 件のコメント表示非表示 2 件の古いコメント
Kelly Kearney 2013 年 12 月 11 日
That seems a little convoluted (I see how you're combining both assignments into one, but for a beginner the syntax might not be clear). This might be better:
T = M + 1;
T(L > 0.5 & M == 0 & N < 0.5) = -1;

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yashar khatib shahidi 2015 年 5 月 3 日
I have a row vector like a = [3 4 6 8 9] then I want to replace third element which is 6 with 5 and 8 so the new vector becomes b = [3 4 5 8 8 9]. What is the function to convert the vector. Thanks
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Jyahway Dong 2016 年 10 月 19 日
This is very important message for me, spend hours try to debug this and thank you all
##### 0 件のコメント表示非表示 -1 件の古いコメント

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Kinga Gyeltshen 2021 年 3 月 16 日
I have a 9x9 matrix and in every iteration i want to retain the 3x3 matrix and on to it I want to add the 4,5,6,7,8,9 (row, column) element to the 3x3 matrix to form 4x4 matrix. The fourth element (row,column) should get replaced with the remaining element of the 9x9 matrix. Can anyone help me with a simple and condensed algorithm to get it done please.
Thank you.

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