Sum every i-th column in matrix seperately

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Anya
Anya 2014-6-12
评论： Anya 2014-6-12
Hi all,
I have randomly generated matrix and I want to go through all columns and to sum every i-th column separately. I don't want to sum all the columns but to sum depending on the counter in for loop for example:
[a,b] = size(C);
for i = 1:b
S = sum(C(:,i))
S = 0 %but this doesn't work, result is sum of elements of all columns

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Zikobrelli
Zikobrelli 2014-6-12
Try sum(A(:,[1 4]))
where A is your random matrix.The line above will give you the sums of column 1 and 4.
ex: A=spiral(4)
sum(A(:,[1 4]))
ans = 34 46
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Zikobrelli
Zikobrelli 2014-6-12
You're using randi(generates random integer numbers) to build your matrix. then you add a last line (1:10) which means that you can NEVER have an all zeros column ,because the last line does not contain 0 Which means that the ONLY way to enter the if statement is to get an all ones column, which indeed does not happen often. The code works, you just need to run the program enough or choose a more suited matrix :)
Anya
Anya 2014-6-12
I can't believe that I missed this last control row . I don't have a lot of experience so I didn't noticed. I change the code now by changing the sum function and it works ! Thanks
sum(C(1:end-1,i))

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更多回答（2 个）

Roger Stafford
Roger Stafford 2014-6-12
S = sum(C(:,i:i:end),1);
This results in a row vector, S, consisting of the sum of every i-th column of C, as requested. It is not clear where you want the i-spaced columns to start. This starts at the i-th column. If you want them to start at the first column, change "i:i:end" to "1:i:end".
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Anya
Anya 2014-6-12
I think that I was not clear enough. I will try to explain on the example matrix C. Using for loop I want to iterate through the columns and count number of non-zero elements. They are written after the last row. If I found the column with all ones or all zeros I want to skip that column and move on.
C =
1 1 1 1 0 1 1 1 1 0
0 1 1 0 0 0 1 1 1 0
0 1 0 1 0 0 0 1 0 0
0 1 1 0 1 0 0 0 0 0
1 4 3 2 1 1 1 3 2 0

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Jos (10584)
Jos (10584) 2014-6-12
sumColsC = sum(C,1)
NotInteresting = sumColsC == 0 | sumColsC == size(C,2)
sumColsC(NotInteresting) = []
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Zikobrelli
Zikobrelli 2014-6-12
you're generating a random matrix.So yes, sometimes, you will not enter the if condition :)
try this
v=[]; C = [randi(2,4,10)-1]
[a,b] = size(C);
for i = 1:b
if (sum(C(:,i)) == 0) || (sum(C(:,i)) == a)
'I-th column is zero or ones , move on'
else v=[v sum(C(:,i))]
end
end
Jos (10584)
Jos (10584) 2014-6-12
When you change the iterator in a for-loop, it will reset at the end
for k=1:10
disp(k) ;
k = 1 ;
disp(k) ;
end
You should be clearer about your goals. What do you mean with "move on"?

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