How to sum up row values in a matrix?
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Sayanta
il 14 Set 2012
Risposto: Renda Mohammedjuhar
il 30 Apr 2019
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 Commenti
Image Analyst
il 2 Gen 2017
I'm not sure of your rule, but it looks like you might be doing
Values = cumsum(M(:, end))
Risposta accettata
Renda Mohammedjuhar
il 30 Apr 2019
I have a matrix like [1 2 3 4] I want an output [1 3 6 10]
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Più risposte (4)
Azzi Abdelmalek
il 14 Set 2012
Modificato: Azzi Abdelmalek
il 14 Set 2012
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))]
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Image Analyst
il 14 Set 2012
Modificato: Image Analyst
il 14 Set 2012
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)]
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Sayanta
il 14 Set 2012
4 Commenti
Azzi Abdelmalek
il 14 Set 2012
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
il 14 Set 2012
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
il 14 Set 2012
Modificato: Image Analyst
il 14 Set 2012
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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