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While loop question in row addition sums. Want row to be less then 2.

1 Ansicht (letzte 30 Tage)
tban
tban am 9 Apr. 2017
Geschlossen: MATLAB Answer Bot am 20 Aug. 2021
I am generating a matrix such as this,
1.0739 0.9517 0.1078 0.4077 0.7066 0.7983
0.6003 1.1885 0.1747 0.3946 0.3541 0.0705
0.7753 0.0929 0.0649 0.0188 0.0298 0.0954
0.4378 0.1140 2.8074 0.5060 0.7248 0.2897
0.1666 0.4086 0.3344 0.0417 0.5798 2.7909
0.0513 1.0805 0.4291 1.3061 0.1078 0.6336
0.5828 0.4903 2.0129 0.5368 0.2691 0.6476
0.1921 0.0626 0.1121 0.1333 0.0121 0.0682
0.2640 0.1722 0.3182 0.2676 1.0384 0.5087
0.5586 0.1019 0.1335 0.4800 0.5046 1.1357
0.2461 0.0407 0.2132 0.7918 0.1956 1.2577
1.4788 0.0586 0.3212 1.1473 0.3263 0.2079
0.2761 2.1830 0.6020 0.5124 0.0489 0.0253
0.0572 0.1084 0.3595 0.1271 0.6155 0.0778
0.2312 0.1831 0.0423 0.1079 0.6713 0.1707
0.2744 0.6776 0.3923 1.0708 0.1322 0.1165
I also have a matrix the same size full of zeros. I want to loop trough every row and sum every value until the next value yields a value above 2.
So a row such as [1.0739 0.9517 0.1078 0.4077 0.7066 0.7983 ...] would loop trough every value and if the sum is not 2 yet, it would rewrite it in the other matrix that is now zeros.
initiate_matrix_z = zeros(16,6);
initiate_matrix_p = %the numbered matrix
for r=1:16 %rows
while (sum <= 2) %note that all sums must be below 2
for c=1:6 %for all values in row
sum = initiate_matrix_p(r,c) + sum %take the value and add it to previous sum
end
end
At the end of this I want the following:
1.0739 0 0 0 0 0 %second value makes sum of row exceed 2
0.6003 1.1885 0.1747 0 0 0 %fourth value makes sum of row exceed 2
0.7753 0.0929 0.0649 0.0188 0.0298 0.0954 %no value makes sum(row)>2
0.4378 0.1140 0 0 0 0 %third value makes sum of row exceed 2
  2 Kommentare
Image Analyst
Image Analyst am 9 Apr. 2017
DON'T USE sum AS THE NAME OF YOUR VARIABLE BECAUSE IT'S THE NAME OF A VERY IMPORTANT BUILT IN VARIABLE. Now, do you NEED to use a loop, or can you use cumsum() like the vast majority of good MATLAB programmers would use?
tban
tban am 9 Apr. 2017
Thanks for the feedback, but too much salt man.

Antworten (2)

Image Analyst
Image Analyst am 9 Apr. 2017
Try this:
m = [...
1.0739 0.9517 0.1078 0.4077 0.7066 0.7983
0.6003 1.1885 0.1747 0.3946 0.3541 0.0705
0.7753 0.0929 0.0649 0.0188 0.0298 0.0954
0.4378 0.1140 2.8074 0.5060 0.7248 0.2897
0.1666 0.4086 0.3344 0.0417 0.5798 2.7909
0.0513 1.0805 0.4291 1.3061 0.1078 0.6336
0.5828 0.4903 2.0129 0.5368 0.2691 0.6476
0.1921 0.0626 0.1121 0.1333 0.0121 0.0682
0.2640 0.1722 0.3182 0.2676 1.0384 0.5087
0.5586 0.1019 0.1335 0.4800 0.5046 1.1357
0.2461 0.0407 0.2132 0.7918 0.1956 1.2577
1.4788 0.0586 0.3212 1.1473 0.3263 0.2079
0.2761 2.1830 0.6020 0.5124 0.0489 0.0253
0.0572 0.1084 0.3595 0.1271 0.6155 0.0778
0.2312 0.1831 0.0423 0.1079 0.6713 0.1707
0.2744 0.6776 0.3923 1.0708 0.1322 0.1165]
[rows, columns] = size(m);
output = zeros(size(m));
for row = 1 : rows
cdf = cumsum(m(row, :))
index = find(cdf < 2, 1, 'last');
cdf(index+1:end) = 0;
output(row, :) = cdf;
end
output % Print to command window

Walter Roberson
Walter Roberson am 9 Apr. 2017
temp = cumsum( initiate_matrix_p, 2);
mask = temp < 2;
initiate_matrix_z(mask) = temp(mask);
No loop.

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