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Hi everyone,

I have two arrays:

y = [1 1 1 1 1 1.5 1.7 1.9 2 2.7]

x = [25:35]

How can I find the corresponding x value for the first number that breaks the chain in y (in this case 1.5)?

y values are different for each loop but the x is almost stable.

Furthermore, how can I store the x y values of the first value that breaks the chain in a loop. Thank you so much.

Star Strider
on 21 Feb 2021

Edited: Star Strider
on 21 Feb 2021

Try this:

y = [1 1 1 1 1 1.5 1.7 1.9 2 2.7];

x = [25:35];

TF = ischange(y,'variance');

Idx = find(TF,1,'first')

Out = x(Idx)

producing:

Out =

30

I’m not certain what you want, or how robust this would be to your other data, so you may need to experiment with it.

Star Strider
on 22 Feb 2021

As always, my pleasure!

‘Now, how can I calculate the slope of the line between the first and last?’

Please define ‘first and last’.

Also, for ths slope calculation, do you want to consider only those two points, or all the points in between as well? The first involves a simple calculation of only those two points, and the second involves a least-squares linear fit. (Both of these are straightforward in MATLAB, so that is not the issue. I just need to know which of these you want.) What data do you want the slope to be calculated with respect to? I assume this is essentially with ‘x’ and ‘y’ defined as previously.

‘... and how can I store the original files of them in two different folders when the slope is greater than A?’

Do you want to store one set in a folder when the slope is less than or equal to ‘A’ and in another folder when it is greater than ‘A’, or something else? Do the folders already exist or will it be necessary to create them?

the cyclist
on 21 Feb 2021

Because x and y are not the same length, I'm not sure how to make the "correspondence" between them. However, this code snippet will give the index of the first element that is different from the preceding ones:

idx = find(diff(y)~=0,1) + 1;

Maybe you can figure out the rest from there, or explain more about what you mean.

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