# Detect sign change in Matlab

조회 수: 149 (최근 30일)
K 2012년 12월 11일
답변: Sindar 2019년 10월 29일
Hi everyone, I have one question. I have two values : prior_value and current_value. Is there any easy way or any function that can be used to detect sign change between this two values? positive to negative or the other way. Thank you.

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### 채택된 답변

Muruganandham Subramanian 2012년 12월 11일
>>help sign

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### 추가 답변 (2개)

Jan 2012년 12월 11일
편집: Jan 2012년 12월 11일
Either:
if ~isequal(sign(previous), sign(current))
Or the trivial:
if previos * current < 0
In both cases the 0 must be considered also, e.g. by "<= 0".
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Derek O'Connor 2012년 12월 18일
편집: Derek O'Connor 2012년 12월 18일
@Jan, the statement ``if previos * current < 0 is prone to overflow and underflow which will give the wrong result'', contains an implicit question: do you recommend a statement that is prone to overflow and underflow which will give the wrong result? Sorry for the idiomatic English.
Your argument ``But how often does this happen in a real problem?'' reminds me of Intel's argument that the FDIV BUG didn't matter because it happened very rarely. IBM's analysis found that it happened many thousands of times per day, contrary to Intel's rarely. Remember, this error-prone sign test could be buried deep in a large program, called millions of times, and you have no idea what values are thrown at it.
For a nice history of the Pentium FDIV bug see here:
and for the sign test see here:
In a previous comment you say ``Nothing is wrong with the comparison of the SIGNs [ if sign(previous) == sign(current) ], see my first suggestion. It makes it only harder to handle the cases of zero values.''
How about if sign(1/previous) == sign(1/current) ?
Mathematically sign(x) = sign(1/x), x ~= 0, so this test handles all non-zero values along with signed zeros.
The difficulties in defining machine precision are discussed by Nick Higham and Cleve Moler here:
Jan 2012년 12월 18일
@Derek: I do not think that this dicussion matchs the complexity level of the OP's question. If a user asks for different signs of two values, the topic "a*b<0" should be mentioned. I hope, that the OP feels encouraged to find such trivial solutions by his own in the future.
There is absolutely no doubt, that a multiplication can lead to numerical difficulties. Neither the famous FDIV story nor a discussion about common errors is required to prove this.
sign(1/x) will suffer from 1/realmin ~= realmax.

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Sindar 2019년 10월 29일
A method safe against overflow.
counting zeros as not a sign change
Sign changes:
if sign(prior_value)*sign(current_value) == -1
Sign stays the same:
if sign(prior_value)*sign(current_value) ~= -1
counting zeros as a sign change (even 0,0)
Sign changes:
if sign(prior_value)*sign(current_value) ~= 1
Sign stays the same:
if sign(prior_value)*sign(current_value) == 1
In all expressions, replace '*' with '.*' to allow prior_value and current_value to be vectors (or matrices), and return a vector of locations where the sign changes. If you want to check whether all the signs change, wrap with all(). Wrap with any() to check if any of the signs change.

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