Sum every i-th column in matrix seperately

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

Answer 1

[Zikobrelli]

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

[Anya]

Please look at my code above and example. Your answer is not clear to me

[Zikobrelli]

 Sum=[]
for i=1:size(C,2)
if (sum(C(:,i))~=0 & sum(C(:,i))~=4)
Sum=[Sum sum(C(:,i))]
end
end

Sum is the answer. As i wanted, i does not include the sums of columns whose values are either all 1 or 0.

for your example, Sum=[ 1 3 2 1 1 2 3 2] Hope this helps

[Anya]

I tried to put your code into mine but it's not working as I expected. Sorry I'm new with Matlab but with the code above columns with all ones or zeros are not skipped

[Zikobrelli]

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];

v=[];

[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

This isnt working for you?

[Zikobrelli]

enter your matrix instead of C. i coied it as a vector, sorry

[Anya]

No, the code is not working, it never enters the if not print 'I-th column is zero or ones , move on'

C = [randi(2,4,10)-1; 1:10]
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

[Zikobrelli]

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]

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))

Answer 2

[Roger Stafford]

   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".

[Anya]

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

Answer 3

[Jos (10584)]

sumColsC = sum(C,1) 
NotInteresting = sumColsC == 0 | sumColsC == size(C,2)
sumColsC(NotInteresting) = []

[Anya]

I done it on this way but it doesn't work , it never enters in the if statements , where is my mistake or how can I apply your code to mine?

C = [randi(2,4,10)-1; 1:10]

[a,b] = size(C);

for i = 1:b

    if (sum(C(:,i)) == 0)
        'I-th column is zero , move on'
         i=i+1;

    end

    if (sum(C(:,i)) == a)
        'I-th column is ones, move on'
        i=i+1;

    end

[Zikobrelli]

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)]

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"?