CALCULATION OF MAXIMUM VALUE IN EVERY ROW
显示 更早的评论
suppose there is a matrix of 3x3 size i want to calculate the maximum value in every row.
ex: 3 4 5
6 9 0
1 2 3
the result i should get at the end should be like this
0 0 1
0 1 0
0 0 1
[EDITED, code formatted, Jan]
2 个评论
Brahima DRABO
2016-1-16
suppose your matrix is A: result = A==max(A,[],2)
Stephen23
2016-1-16
@Brahima DRABO: you should put this as an answer too: you would get votes.
Note to others: ignore the accepted answer, it is very poor code.
采纳的回答
Bart
2012-3-6
try this:
suppose the input matrix is A, the output result matrix is B;
max = 0;
k = 0;
l = 0;
for i=1:3
for j=1:5
place(i,j) = 0;
end
end
for i=1:3
for j=1:3
if A(i,j)>=max
max = A(i,j);
place(i,j) = 1
end
end
end
for k=1:3
for l=1:3
if place(k,l) = 1
if place(k,l+1) = 1
B(k,l) = 0;
end
if place[k,l+2) = 1
B(k,l) = 0;
else B(k,l) = 1;
end
else B(k,l) = 0;
end
end
end
1 个评论
Jan
2012-3-7
Shadowing the built-in function "max" by a variable is a frequent source of porblems and bugs.
The creation of the variable "place" element by element is very inefficient and should be avoided. "place = zeros(3, 5);" is nicer, faster and consumes less memory. Although this is a tiny example here and runtime does not matter, it is a good idea to follow a clean programming style.
Using nested loops to create B element by element is slow also, again due to the missing pre-allocation.
It is not useful to declare k=0 and l=0 at the beginning.
The algorithm wastes a lot of time. Try it with a 1000x1000 array to feel the difference.
更多回答（3 个）
Jonathan Sullivan
2012-3-7
m = [ 3 4 5
6 9 0
1 2 3];
mx = max(m,[],2); % EDIT: was max(m,2); Oops
out = bsxfun(@eq,m,mx);
Jan
2012-3-6
[v, sub] = max(a, [], 2); % [EDITED], was "max(a, 2)"
sa = size(a);
b = zeros(sa);
index = sub2ind(sa, 1:sa(1), sub);
b(index) = 1;
I cannot try it currently and I find the documentation of sub2ind confusing in this point: Perhaps the 2nd and 3rd argument must be swapped: sub2ind(sa, sub, 1:sa(1)).
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