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

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Emerson De Souza on 5 Jun 2011
Commented: Nicholas Nurre on 30 Oct 2019
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

Matt Fig on 5 Jun 2011
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

Show 1 older comment
Matt Fig on 5 Jun 2011
LENGTH should only be used with vectors, as in my example. For a general array, use NUMEL instead of length.
for ii = 1:numel(x)
...
Emerson De Souza on 5 Jun 2011
Thank you Mattfig,
that fixed the problem.
Wish you a nice day
Emerson
Nicholas Nurre on 30 Oct 2019
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?

Rain on 11 Dec 2013
Edited: madhan ravi on 15 Jan 2019
Hi,
Actually, there is a very simple way to do it:
X=rand(10,10);
X(X>0.5) = [-1];

Bineet_Mehra on 20 Aug 2016
Hello Rain,
Thanks for the solution. lets consider different situation. For example if i have a 100*100 matrix of angles. I was to replace all the angle > 90 deg with 180-that angle. How can i acheive this ?
Thanks a lot
Björn Kok on 14 Mar 2018
X(X>90)=180
Mohammad Safayet Hossain on 16 Feb 2019
Great command

Ivan van der Kroon on 5 Jun 2011
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 Comment

Emerson De Souza on 5 Jun 2011
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

Ivan van der Kroon on 5 Jun 2011
Just implement it for multiple matrices using element multplication:
Logical=(L>0.5).*(M==0).*(N<0.5);
T=-Logical+(M+1).*Logical;

Emerson De Souza on 5 Jun 2011
Hi Ivan,
I tried the suggestion above, using 5x5 matrices L,M,N,but all elements of T equals zero. I guess we are making some mistakes.
Run the lines below and verify by yourself
clear all
close all
L=rand(5,5);
M=rand(5,5);
N=rand(5,5);
Logical=(L>0.5).*(M==0).*(N<0.5);
T=-Logical+(M+1).*Logical;
Thank you again for trying to help to solve this question
Emerson
Ivan van der Kroon on 6 Jun 2011
Sorry, typo: it should be T=-Logical+(M+1).*(1-Logical); Ones and zeros.
Kelly Kearney on 11 Dec 2013
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;

yashar khatib shahidi on 3 May 2015
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

Jyahway Dong on 19 Oct 2016
This is very important message for me, spend hours try to debug this and thank you all