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

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

Answer 1

[Matt Fig]

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

[Emerson De Souza]

Hi MattFig,

I tried the iteration approach

as written below:

x=rand(5,5);

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

but only the first column is filled up.

How do we extend to execute all columns.

Thank you

Emerson

[Matt Fig]

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]

Thank you Mattfig,

that fixed the problem.

Wish you a nice day

Emerson

[Nick]

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?

Answer 2

[Rain]

Hi,

Actually, there is a very simple way to do it:

X=rand(10,10);
X(X>0.5) = [-1];

Hope it is helpful.

[Bineet_Mehra]

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]

X(X>90)=180

[Mohammad Safayet Hossain]

Great command

[Majid Mahmoodabadi]

Hi,

I have a matrix as following and want to increase the value of numbers (by 1) that are less than 9.

B= [12,16,5,6,7,11,13,22]

I applied the below logic:

B(B<9)= [B(i)+1]

This is what I got:

B =

12 16 17 17 17 11 13 22

But, I look for a response like this:

B =

12 16 6 7 8 11 13 22

I appreciate it if you could help me.

Answer 3

[Ivan van der Kroon]

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)

To solve your problem

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.

[Emerson De Souza]

Hi Ivan

thank you for your help.

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

Thank you in advance

Emerson

Answer 4

[Ivan van der Kroon]

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]

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]

Sorry, typo: it should be T=-Logical+(M+1).*(1-Logical); Ones and zeros.

[Kelly Kearney]

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;

Answer 5

[yashar khatib shahidi]

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

Answer 6

[Jyahway Dong]

This is very important message for me, spend hours try to debug this and thank you all

Answer 7

[Kinga Gyeltshen]

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.