Question about array alternating between positive and negative

Question

William Fantin 2019-5-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/462150-question-about-array-alternating-between-positive-and-negative

评论： James Tursa 2019-5-15

Hello,

I have solved a differential equation symbolically and then evaulated it based on user input. The result is different every time. Each time I obtain an array from the graph. For example, lets say I get an array, a=[-1,1,2,-2,1,5,-4,1]. Is there a way to find out how many times the sign of an array changes from positive to negative? The reason I am trying to do this is I need to know how many times the curve crosses the origin.

Thanks!

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

James Tursa 2019-5-15

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/462150-question-about-array-alternating-between-positive-and-negative#answer_375048

编辑：James Tursa 2019-5-15

在 MATLAB Online 中打开

Seems like there should be a shorter way, but assuming you are looking for all 0 crossings and not just the positive to negative ones:

sum(logical(diff(sign(a))))

This will double book the exact 0's, however. So if you expect you might get those, you will need something more sophisticated.

2 个评论
显示无隐藏无

William Fantin 2019-5-15

IT WORKED!!!! THANK YOU!!! i had sunusoidal motion about the x-axis and it crossed 6 times and your method gave me 6! what did you mean it wont exactly count zeros correctly?

THanks again

James Tursa 2019-5-15

在 MATLAB Online 中打开

E.g.,

>> a = [-2 -3 0 1 2 -3 5]
a =
    -2    -3     0     1     2    -3     5
>> sum(logical(diff(sign(a))))
ans =
     4

There's really only three 0 crossings, but the calculation above gave a result of 4. The 0 crossing from -3 --> 0 --> 1 got double booked because the sign changed twice even though it was only one crossing. So if you expect you might get exact 0's you will need a more sophisticated method to detect them and check for 0 crossings vs 0 peaks etc and adjust the answer accordingly.

请先登录，再进行评论。