find for sign changes

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Kugen Raj
Kugen Raj 2012년 3월 17일
댓글: Walter Roberson 2021년 3월 4일
Columns 49 through 56
0.8710 0.9483 1.0408 1.1538 1.2956 1.4796 1.7292 2.0885
Columns 57 through 64
2.6538 3.6815 6.1847 119.1838 -118.8424 -5.8436 -3.3407 -2.3137
Above is part of my results. I need to find for value that changes from positive to negative. In the above data, after 119.1838(column 60) the -118.8424(column 61) appears. I need to find this value. Hoe can I do this?

채택된 답변

Walter Roberson
Walter Roberson 2012년 3월 17일
find(x(1:end-1)>0 & x(2:end) < 0)
Another approach that might be close enough for your purposes:
find(diff(x>=0),1)
  댓글 수: 4
Jan
Jan 2012년 3월 18일
But what happens with your first approach for [1, 0, -1]?
The OP must decide this detail.
Walter Roberson
Walter Roberson 2012년 3월 18일
In the case of [1, 0, -1] there is no position at which the value is positive and the next value is negative :)

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추가 답변 (3개)

Jan
Jan 2012년 3월 17일
Walter's method detects multiple sign changes also. This finds the first negative number only, but this meets your problem description also:
index = find(x < 0, 1, 'first')

Rasmus Herlo
Rasmus Herlo 2020년 1월 22일
편집: Rasmus Herlo 2020년 1월 22일
Finding any place where the sign changes, either from neg to pos or from pos to neg could also be done in general by this:
Idx_Change = sort([strfind(x>=0, [0 1]) strfind(x>=0, [1 0])]);
  댓글 수: 1
Walter Roberson
Walter Roberson 2020년 1월 22일
Yes but be careful about exact 0.

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Alireza Ahani
Alireza Ahani 2021년 3월 4일
but when I checked with a dataset that includes zero-down-crosing event, they would be ignored, so I added some lines to his:
ZDCI = find(x(1:end-1)>0 & x(2:end) < 0); % zero down crossing index
Zero_indx = find(x == 0);
k = find( x(Zero_indx-1) > 0);
ZDCI = sort([ZDCI Zero_indx(k)]);
  댓글 수: 1
Walter Roberson
Walter Roberson 2021년 3월 4일
If the task is zero down crossing and zeros can be encountered, then the above code is not correct, as it does not take into account multiple zeros. For example for zero down crossing, you might have +1 followed by 5 zeros, followed by -1. The zeros could, for example, be present due to filtering of values that were close to zero, such as due to measurement noise on a sigmoid that crossed zero. Detection for such cases cannot be restricted to any fixed finite number of look-back.

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