Looping matrix differences for every iterations

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Anusha
Anusha 2013-10-18
评论： Jan 2013-10-18
I have this scenario
A=[6;7;8] W=[3;9;1] d=|Ai-W| for each iterations take one value from A subtract with every value in W
1 iteration
d=
6-3
6-9
6-1
Like this for another iteration

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Andrei Bobrov
Andrei Bobrov 2013-10-18
编辑：Andrei Bobrov 2013-10-18
d = bsxfun(@minus,A(:)',W(:));
  2 个评论
Anusha
Anusha 2013-10-18
not like this sir..
for first iteration d=
3
-3
5
for second iteration d1=
4
-2
6
d2=
5
-1
7
each differences A take one by one value
Jan
Jan 2013-10-18
@Anusha: Please read Andrei's suggestion again and think twice, why he prefers this method. It is a bda programming practice to insert an index in the name of a variable. His approach uses and index as index (!):
d(:, 1), d(:, 2), d(:, 3)
This is much more useful and efficient. You can e.g. expand the problem easily to millions of elements.

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