Do not use the for loop to calculate the first number greater than 0 in each line

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slevin Lee
slevin Lee 2022-11-19
编辑： Jan 2022-11-19
x=randn(10,10)
...
  2 个评论
Jan
Jan 2022-11-19
This sounds like a homework question. So please show, what you have tried so far and ask a specific question about Matlab. The forum will not solve your homework, but assist you in case of questions concerning Matlab.
slevin Lee
slevin Lee 2022-11-19
编辑：Jan 2022-11-19
x=randn(10,10)
x1=x>0
x2=cumsum(x1,2)
x3=x2==0
x4=sum(x3,2)+1
x5=x(:,x4).*eye(10,10)
answer=sum(x5,2)
sorry!
I just want to see if there's a better way

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采纳的回答

Jan
Jan 2022-11-19
编辑：Jan 2022-11-19
x = randn(10, 10)
y = x.'; % Required for the last step y(y3==1)
y1 = y > 0
y2 = cumsum(y1)
% Now a row can contain multiple 1's. Another CUMSUM helps:
y3 = cumsum(y2)
y(y3 == 1)
Another way:
x = randn(10, 10)
m = x > 0
y = cumsum(x .* m, 2)
y(~m) = Inf % Mask leading zeros
min(y, [], 2)
Or with beeing nitpicking for the formulation "no for loop":
result = zeros(1, 10);
m = x > 0;
k = 1;
while k <= 10
row = x(k, :);
result(k) = row(find(row(m(k, :)), 1));
k = k + 1;
end
And another method:
[row, col] = find(x > 0);
first = splitapply(@min, col, row);
x(sub2ind(size(x), 1:10, first.'))

更多回答（1 个）

DGM
DGM 2022-11-19
编辑：DGM 2022-11-19
Ran in:
Here's one way.
x = randn(10,10)
x = 10×10
0.5355 -1.4904 -0.9662 -0.7693 0.6789 0.6031 0.2375 -1.0743 0.6916 -0.8738 -0.0884 0.5313 -0.2913 1.0506 -1.1930 0.1070 0.2047 -0.2512 2.0512 0.1435 -0.1235 0.5840 1.3542 -0.3692 1.1048 -0.4211 -2.1762 -0.3553 0.8296 -2.1670 -0.6151 0.7344 -1.1325 -1.9521 -1.5595 0.8968 -0.8484 -2.5253 -0.1668 1.8909 1.0192 -0.4541 0.9363 -0.1635 -1.7753 -0.2266 1.1185 2.0062 -1.3746 -0.3283 -0.2484 -1.0466 1.2285 0.4551 -1.1110 -1.1021 0.3378 0.2945 -0.2886 1.0569 1.4228 -1.4178 0.5278 0.1037 0.7001 -0.3782 0.3081 1.4122 0.6484 -1.0982 -0.0277 -1.6063 -0.8334 -0.2916 -1.0505 -0.6081 -0.4252 -1.0511 0.4218 0.2551 -0.2180 0.5246 0.2145 1.4641 2.2411 -1.4396 -0.2848 -0.4727 2.5402 2.0645 0.5302 0.6045 -0.7767 1.2057 -0.9240 0.0234 1.1590 -0.0401 1.3279 0.5901
[~,idx] = min(x<=0,[],2) % note the logical inversion
idx = 10×1
1 2 2 2 1 3 1 9 2 1
firstnz = x(sub2ind(size(x),(1:size(x,1))',idx)) % get the values from the array
firstnz = 10×1
0.5355 0.5313 0.5840 0.7344 1.0192 1.2285 1.4228 0.4218 0.5246 0.5302

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