MATLAB Answers

90% Percentile of Matrix

10 views (last 30 days)
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]

  0 Comments

Sign in to comment.

Accepted Answer

Star Strider
Star Strider on 17 Aug 2020
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
.

  6 Comments

Show 3 older comments
Star Strider
Star Strider on 1 Sep 2020
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 on 3 Sep 2020
Working great again. Thanks!
Star Strider
Star Strider on 3 Sep 2020
As always, my pleasure!

Sign in to comment.

More Answers (1)

jonas
jonas on 17 Aug 2020
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)

  0 Comments

Sign in to comment.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by