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90% Percentile of Matrix

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Jonathan Moorman
Jonathan Moorman 2020 年 8 月 17 日
コメント済み: Star Strider 2020 年 9 月 3 日
Hi! I am trying to find a way to calculate the 90% percentile of a matrix (1,n) . I provide an example below, and the value I would need to find is 0.292. (Note the matrix will changes in length). How could this be done?
[0.289, 0.254, 0.287, 0.292, 0.289, 0.267, 0.289, 0.304, 0.284, 0.282]

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Star Strider
Star Strider 2020 年 8 月 17 日
Here are two possibilities, the second of which gives you the exact value in your Question:
v = [0.289, 0.254, 0.287, 0.292, 0.289, 0.267, 0.289, 0.304, 0.284, 0.282];
Out1 = prctile(v, 90)
Out2 = interp1(linspace(1/numel(v),1,numel(v)), sort(v), 0.9)
producing:
Out1 =
0.2980
Out2 =
0.2920
.

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Star Strider
Star Strider 2020 年 9 月 1 日
I do not have the vector to work with. (I can barely read the images of it.)
The 90th percentile may not actually be an element of the vector, simply what that element would be if it were there, since that by default uses linear interpolation.
See if this version does what you want, with the interpolation method now 'nearest':
Out2 = interp1(linspace(1/numel(v),1,numel(v)), sort(v), 0.9, 'nearest')
An anonymous function version of it would be:
prctlv = @(v) interp1(linspace(1/numel(v),1,numel(v)), sort(v), 0.9, 'nearest');
That will make it easier to work with. Call it as:
a90Per = prctlv(a90Export);
I tested that with the original ‘v’ and it again gave the desired result.
Jonathan Moorman
Jonathan Moorman 2020 年 9 月 3 日
Working great again. Thanks!
Star Strider
Star Strider 2020 年 9 月 3 日
As always, my pleasure!

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その他の回答 (1 件)

jonas
jonas 2020 年 8 月 17 日
There is probably a one-liner for this, but I guess you could use
A = [0.289, 0.254, 0.287, 0.292, 0.289, 0.267, 0.289, 0.304, 0.284, 0.282];
B = sort(A);
id = round(numel(A).*0.9)
B(id)

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