how to find the sum of multiple non zero elements in an array
显示 更早的评论
hi, i have an array that has like zero and non zero elements like eg: 0 0 0 1 1 1 0 0 0 3 3 3 0 0 0
so i want output i=3 and j=9 that is the sum of first set on nonzero elements seporate and the second seporate
1 个评论
Jan
2013-6-4
Is "i" and "j" really required? Note that this might get extremely complicated, when you have 5000 such blocks and have to avoid collisions with other existing variables.
采纳的回答
a = [0 0 0 1 1 1 0 0 0 3 3 3 0 0 0];
c = cumsum(a);
index = strfind([a,0] ~= 0, [true, false]);
result = [c(index(1)), diff(c(index))];
Not tested yet!
2 个评论
Mitson Monteiro
2013-6-5
Jan
2013-6-5
@Mitson: Do you mean something like this:
index = strfind([0, a] ~= 0, [false, true]) + 1;
更多回答（3 个）
Roger Stafford
2013-6-4
There is an ambiguity in the posing of this problem. Are the sets of nonzero numbers distinguished from one another by their values or by their contiguity? For example, if your vector were [1 3 0 3 1] do you want [2 6] (distinguished by values) as a result or [4 4] (distinguished by contiguity)? The following assumes the latter.
Let x be the row vector of your numbers.
t = [0,x,0];
c = cumsum(t);
f = find(diff(t~=0)~=0);
r = c(f(2:2:end))-c(f(1:2:end));
My apologies to you, Jan, if this seems too similar to your solution. I couldn't resist giving it.
3 个评论
Matt J
2013-6-4
For example, if your vector were [1 3 0 3 1] do you want [2 6] (distinguished by values) as a result or [4 4] (distinguished by contiguity)? The following assumes the latter.
Yes, and my solution assumes the latter.
Roger Stafford
2013-6-4
编辑：Roger Stafford
2013-6-4
Don't you mean the former, Matt?
Matt J
2013-6-4
Um. Yes, the former.
Matt J
2013-6-4
[u,i,j]=unique([0 0 0 1 1 1 0 0 0 3 3 3 0 0 0])
result = histc(j,1:max(j)).*u
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2013-6-4
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