Evaluation of a for loop

Question

Jürgen 2012-8-13

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hello

is this pseudo code:

for Counter=1:size(M,2)
  if someting
     delete row from M
  end
end

If would expect that if the size of M changing the range of Counter would change do but that does not happen, instead Counter will rang from 1 to the begin size of M

so size(M,2) is not evaluated each time that the loop runs?

I notices with while ...end you do not have the problem or am I completely mistaken?

thanks for any insights

regards,Jürgen

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Answer 1

Walter Roberson 2012-8-13

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The bounds and increments of "for" loops are recorded at the time the "for" statement is executed, and changes to any of the bounds or increment in the body of the loop do not have any effect on the execution of the loop. If you need to have the bounds vary, use a "while" loop.

In the particular case of deleting elements of the array, a "forward" for loop is not appropriate. Suppose Counter is 3 and you delete element 3, then what was in element 4 "falls down" to occupy element 3, so in order to test that element to see if it should be deleted, you would have to re-test at location 3 rather than proceeding on to location 4.

One of the tricks for avoiding this problem is to run the "for" loop backwards:

for Counter = size(M,1) : -1 : 1
  if someting
     delete row from M
  end
end

Note here I adjusted to size(M,1) which is the row count (as per your "delete row"); your pseudo-code was using the count of columns to decide whether to delete rows.

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Jürgen 2012-8-13

ok ( yes pseudo code was wrong sorry for that ) nice idea a backwards loop

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Answer 2

Wayne King 2012-8-13

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编辑：Wayne King 2012-8-13

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Yes, the for loop termination condition is set the first time you enter the loop

m = 3;
for jj = 1:m
 disp('Hi');
 m = m-1;
end

but the value of m is zero upon exiting the loop

If you wanted to use the value of size(M,2) to know when to execute the loop, you can use a break()

    m = 3;
    for jj = 1:m
     disp('Hi');
     m = m-1;
     if (m<2)
       break;
    end
    end

Or a while loop instead of a for

m = 3;
while m>1
 disp('Hi');
 m= m-1;
end

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Jürgen 2012-8-13

yes , I choose the while loop to solve the problem, I don't like to use break, long ago , I learned that breaking out a loop with 'break' is not an ideal solutions

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Answer 3

Azzi Abdelmalek 2012-8-13

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编辑：Azzi Abdelmalek 2012-8-13

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the idea is to:

replace the rows by "nan"
at the end remove the rows containing "nan"

 A=magic(10) % example
 for k=1:size(A,2)
    if rand(1)>0.5   % it's just an example 
        A(:,k)=nan   % replace the row by "nan"
    end
 end
 A(:,any(isnan(A)))=[] %remove the rows containing nan

% this code suppose that your initial matrix don't contain nan. if your matrix contain "nan" use instead this code

A=magic(10) % example
A=num2cell(A) % convert to cell
for k=1:size(A,2)
  if rand(1)>0.5   % it's just an example 
      A(:,k)={'remove'}   % replace the row by "remove"
  end
end
ind=cellfun(@(x) ~isnumeric(x),A)
A(:,find(mean(ind)==1))=[]
A=cell2mat(A)