Cumulative sum with conditions
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I am counting a number of 1's in an array.
But I want to count the cumulative sums only when the number of 1's are followed by a sequence of zeros.
% For instance,
h1=[1 0 0]
h2=[1 1 0]
h3=[1 1 1]
h4=[0 0 1]
h5=[0 1 0].
I want to count the cumulative sums (S) of those arrays as follows.
% The ideal outputs are:
S1= 1
S2= 2
S3= 0
S4= 0
S5= 0
since h2: h4 do not meet the condition of 1's followed by 0's.
I would truly appreciate if any one could advise me an effective way of programming the conditional sums.
댓글 수: 1
Rik
2018년 10월 3일
So you have these as numbered variables? Then you should really have them in a single array. Also, I don't understand the rules here. Are we meant to look row by row? If so, doesn't the 5th row count as well (as there is a 0 after the 1)?
채택된 답변
Bruno Luong
2018년 10월 4일
편집: Bruno Luong
2018년 10월 4일
Change the last line to make it can handle with single row:
m = size(h,1);
d = diff(h,1,2);
[r,c] = find(d==-1);
b = sum(abs(d),2)==1;
s = accumarray(r(:),c(:),[m 1]).*double(b)
This because find() rotates the result for row input array.
추가 답변 (4개)
Sean de Wolski
2018년 10월 3일
h = [1 0 0;
1 1 0;
0 0 1;
1 1 1;
0 1 0];
s = sum(h, 2).*~h(:,end)
requiring the last element to be zero. This wouldn't work for [1 0 1] but I don't know what you'd want for that.
댓글 수: 0
yp78
2018년 10월 3일
편집: yp78
2018년 10월 3일
댓글 수: 2
Sean de Wolski
2018년 10월 3일
In that case what I have above should work. It returns the sum and requires that the last element be a zero. If the last element is non-zero, it returns a zero.
With [1 0 1] it would say zero because the last element is not a zero.
Bruno Luong
2018년 10월 4일
편집: Bruno Luong
2018년 10월 4일
h = [1 0 0;
1 1 0;
0 0 1;
1 1 1;
0 1 0];
[a,c] = max(1-h,[],2);
s = (c-1).*a.*h(:,1)
댓글 수: 3
Bruno Luong
2018년 10월 4일
편집: Bruno Luong
2018년 10월 4일
Can you explain why it returns 0? Still not clear the rule...
Do you only count only if the row is composing a sequence of 1s followed by a sequence of 0s (of at least one 0)?
Bruno Luong
2018년 10월 4일
편집: Bruno Luong
2018년 10월 4일
m = size(h,1);
d = diff(h,1,2);
[r,c] = find(d==-1);
b = sum(abs(d),2)==1;
s = accumarray(r(:),c(:),[m 1]).*double(b)
댓글 수: 4
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