How to replace the elements of a matrix using the conditions if,else?
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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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采纳的回答
Matt Fig
2011-6-5
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 个评论
Nick
2019-10-30
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?
更多回答(6 个)
Rain
2013-12-11
编辑:madhan ravi
2019-1-15
Hi,
Actually, there is a very simple way to do it:
X=rand(10,10);
X(X>0.5) = [-1];
Hope it is helpful.
4 个评论
Majid Mahmoodabadi
2022-4-16
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.
Ivan van der Kroon
2011-6-5
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.
Ivan van der Kroon
2011-6-5
Just implement it for multiple matrices using element multplication:
Logical=(L>0.5).*(M==0).*(N<0.5);
T=-Logical+(M+1).*Logical;
3 个评论
Ivan van der Kroon
2011-6-6
Sorry, typo: it should be T=-Logical+(M+1).*(1-Logical); Ones and zeros.
Kelly Kearney
2013-12-11
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
2015-5-3
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
2016-10-19
This is very important message for me, spend hours try to debug this and thank you all
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Kinga Gyeltshen
2021-3-16
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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