Evaluation of a for loop
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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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Walter Roberson
le 13 Août 2012
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.
Plus de réponses (2)
Wayne King
le 13 Août 2012
Modifié(e) : Wayne King
le 13 Août 2012
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
Azzi Abdelmalek
le 13 Août 2012
Modifié(e) : Azzi Abdelmalek
le 13 Août 2012
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)
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