How can I sum every nth row?

If there's a matrix, for example:

A = [1 2 3 4 5 6 ; 1 3 5 7 9 11 ; 2 4 6 8 10 12]'

A =

     1     1     2
     2     3     4
     3     5     6
     4     7     8
     5     9    10
     6    11    12
     7    13    14

I'm trying to sum 3 rows at a time (like an attached jpg file).

So I want to get:

 B = 
     6     9     12
     15    27    30
     7     13    14

Thanks a lot!

[madhan ravi]

Can you elaborate which row you want to sum in this example?

[AJ]

for example, in column 1 in A,

1+2+3= 6, 4+5+6= 15, and 7 doesn't have pairs so it's just 7.

Then column 1 in B goes 6, 15, 7.

Well.. I'm not sure this is enough.. Sorry for the late reply!

Answer 1

[Paolo]

 A = [1 2 3 4 5 6 7; 1 3 5 7 9 11 13; 2 4 6 8 10 12 14]';
 [n,col] = size(A);
 index = 1:n;
 elem = [repmat(3,1,floor(n/3))];
 endv = n-sum(elem);
 if(~endv)
     endv = [];
 end
 index = mat2cell(index,1,[elem,endv])';
 B = cell2mat(cellfun(@(x) sum(A(x,:),1),index,'un',0));

 B =

     6     9    12
    15    27    30
     7    13    14

[AJ]

Thanks!

but should I assign index every time when the size of the data varies?

If the data is big, then i need to make that index with another code, right?

[Paolo]

How are indexes chosen?

[AJ]

um, i mean,

in your answer above, there's

index = {1:3;4:6;7}

if data i want to use has 7xn double form it always works,

but if data has 10x2 form, then I need to assign index as :

index = {1:3;4:6;7:9;10}

Thus, I have to change 'index' when the size of the data changes.

Am I getting right?

[Paolo]

I've updated my answer.

[AJ]

Ohhhhh Thanks a lot!!!

[Paolo]

You can say thank you to me and Stephen by accepting and voting for either one of the questions :)

Answer 2

[Peng Li]

Just a comment that this could be done without involving a cell, which is the least type that I'd like to use among others.

A = [1 2 3 4 5 6 7; 1 3 5 7 9 11 13; 2 4 6 8 10 12 14]';
sumEveryRow = 3;
intBlock    = floor(size(A, 1) / sumEveryRow)*sumEveryRow;

temp        = reshape(A(1:intBlock, :)', size(A, 2), sumEveryRow, []);

sumA        = [squeeze(sum(temp, 2))'; sum(A(intBlock+1:end, :), 1)];

This should be quick than the two above answers, both involving a cell type.

a test below

A = [1 2 3 4 5 6 7; 1 3 5 7 9 11 13; 2 4 6 8 10 12 14]';
A = repmat(A, 1000, 1);

tic
sumEveryRow = 3;
intBlock    = floor(size(A, 1) / sumEveryRow)*sumEveryRow;

temp        = reshape(A(1:intBlock, :)', size(A, 2), sumEveryRow, []);

sumA        = [squeeze(sum(temp, 2))'; sum(A(intBlock+1:end, :), 1)];
toc

tic
[n,col] = size(A);
index = 1:n;
elem = [repmat(3,1,floor(n/3))];
endv = n-sum(elem);
if(~endv)
    endv = [];
end
index = mat2cell(index,1,[elem,endv])';
B = cell2mat(cellfun(@(x) sum(A(x,:),1),index,'un',0));
toc

tic
N = 3;
S = size(A);
V = N*ones(1,ceil(S(1)/N));
V(end) = 1+mod(size(A,1)-1,3);
C = mat2cell(A,V,S(2));
Z = cellfun(@(m)sum(m,1),C,'uni',0);
toc

Elapsed time is 0.008036 seconds.
Elapsed time is 0.039083 seconds.
Elapsed time is 0.022779 seconds.

I repeated A 100 times to amphasize the effect of computational time.

[Yongqi Shi]

excellent!

Answer 3

[Stephen23]

A = [1 2 3 4 5 6 7; 1 3 5 7 9 11 13; 2 4 6 8 10 12 14]'
N = 3;
S = size(A);
V = N*ones(1,ceil(S(1)/N));
V(end) = 1+mod(size(A,1)-1,3);
C = mat2cell(A,V,S(2));
Z = cellfun(@(m)sum(m,1),C,'uni',0);

Giving

>> Z{:}
ans =
     6     9    12
ans =
    15    27    30
ans =
     7    13    14

[AJ]

wow here's an another idea! Thank you!!