How to sum up row values in a matrix?
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Dear All
I have matrix A
A = [ 1 2 3 5;
3 4 5 4;
];
I want to add row values like that using a loop ( without manual input)
A(1,1) + A(1,2) = B1
A(1,3) + A(1,4) = B2
A(2,1) + A(2,2) = B3
A(2,3) + A(2,4) = B4
B= [ B1 B2;
B3 B4
];
How can I do that any tips
Many Thanks in advance
6 个评论
Image Analyst
2012-9-14
编辑：Image Analyst
2012-9-14
Looks like you've done it (almost) except that you need to flip your B's to the other side:
B1 = A(1,1) + A(1,2);
B2 = A(1,3) + A(1,4);
B3 = A(2,1) + A(2,2);
B4 = A(2,3) + A(2,4);
B= [ B1 B2;
B3 B4
];
No loop needed (for such a small array). Why do you want to use a loop?
Yuli Hartini
2017-1-2
Can you tell how to use a loop? Sorry for the late comment
Image Analyst
2017-1-2
m = randi(4, 3, 5); % Sample data in a 3 by 5 matrix
[rows, columns] = size(m) % Get dimensions of the matrix.
% Preallocate space for the sums of the rows.
rowSums = zeros(rows, 1);
for row = 1 : rows
% Get the sum for this row across all columns in this row.
for col = 1 : columns
rowSums(row) = rowSums(row) + m(row, col);
end
% Print the sums to the command window
fprintf('For row #%d the sum over the columns = %f.\n', row, rowSums(row));
end
Yuli Hartini
2017-1-2
编辑：Yuli Hartini
2017-1-2
What if I have matrix M, M = [1 2 0.2; 2 3 0.1; 3 4 0.4] And I want values like this.. Values= [0.2; (0.2+01); (0.2+0.1+0.4)]
Yuli Hartini
2017-1-2
Help me please
Image Analyst
2017-1-2
I'm not sure of your rule, but it looks like you might be doing
Values = cumsum(M(:, end))
采纳的回答
Renda Mohammedjuhar
2019-4-30
0 个投票
I have a matrix like [1 2 3 4] I want an output [1 3 6 10]
更多回答（4 个）
Azzi Abdelmalek
2012-9-14
编辑：Azzi Abdelmalek
2012-9-14
A = [ 1 2 3 5;3 4 5 4]
res=reshape(sum(reshape(A',1,2,[])),2,2)'
%or
res=A(:,[1 3])+A(:,[2 4])
%or
n=size(A,2)/2
res=[sum(A(1,1:n)) sum(A(1,n+1:end)); sum(A(2,1:n)) sum(A(2,n+1:end))]
Image Analyst
2012-9-14
编辑：Image Analyst
2012-9-14
Here's one way:
A = [ 1 2 3 5;
3 4 5 4]
% Get the sliding sum.
a2 = conv2(A, [1 1], 'valid');
% Extract just the first and last column.
output = [a2(:,1) a2(:,3)]
Sayanta
2012-9-14
4 个评论
Azzi Abdelmalek
2012-9-14
编辑：Azzi Abdelmalek
2012-9-14
B=A(1:2,:)
n=size(B,2)/2
res=reshape(sum(reshape(B',1,n,[])),2,2)'
Image Analyst
2012-9-14
See my other answer. I was wondering - but usually when it needs to be general for some variable number of rows or columns, people will say that in advance so they get the general answer the first time.
Azzi Abdelmalek
2012-9-14
or simpler
n=size(A,2)/2
res=[sum(A(1,1:n)) sum(A(1,n+1:end)); sum(A(2,1:n)) sum(A(2,n+1:end))]
Image Analyst
2012-9-14
Yeah, that's probably better - more direct - as long as he has a 2 row array. In his example here (which he incorrectly posted as an answer), he has a 7 row by 8 column array. See my build on your solution for when it has any number of rows.
Image Analyst
2012-9-14
编辑：Image Analyst
2012-9-14
A=[...
0.0018 0.0008 0.0000 0.0000 0.2304 0.7345 0.0159 0.0166
0.0024 0.0016 0.0001 0.0000 0.2161 0.7441 0.0165 0.0192
0.0029 0.0027 0.0002 0.0000 0.2084 0.7475 0.0169 0.0214
0.0034 0.0040 0.0003 0.0000 0.2041 0.7479 0.0172 0.0230
0.0038 0.0055 0.0005 0.0001 0.2016 0.7468 0.0175 0.0243
0.0041 0.0072 0.0007 0.0001 0.1999 0.7450 0.0177 0.0253
0.0044 0.0090 0.0009 0.0001 0.1988 0.7429 0.0178 0.0261]
[rows columns] = size(A)
% Get the sliding sum
a2 = conv2(A, ones(1, columns/2), 'valid')
% Extract just the first and last column.
B = [a2(:,1) a2(:,end)]
Or, building off Azzi's solution and making it work for a 2D array of any number of rows:
B = [sum(A(:,1:columns/2), 2) sum(A(:,(columns/2)+1:end), 2)]
This is probably the most direct way. And it's only 1 line of code instead of 2.
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