Given a value, find percentile % with Numpy

Question

There are probably better words to describe this question, however what I am trying to do is the opposite of np.percentile(). I have a list of n numbers, and I want to see what percentile of them are smaller than a given value. Right now the way I get this value is by continuously trying different decimals. What I want Numpy to tell me is this:

Given threshold = 0.20 (input), about 99.847781% (output) of the items in list d are below this percentile.

What I do right now to get this number is pretty sketchy:

>>> np.percentile(np.absolute(d), 99.847781) 0.19999962082827874 >>> np.percentile(np.absolute(d), 99.8477816) 0.19999989822334402 >>> np.percentile(np.absolute(d), 99.8477817) 0.19999994445584851 >>> np.percentile(np.absolute(d), 99.8477818) 0.19999999068835939 ...

Are you looking for sum(d < given_value) / len(d)? If you're using python2 you'd have to cast one of the operands to float — pault
– pault, Commented Jul 30, 2018 at 14:51

cookiedough · Accepted Answer · 2018-07-30 15:21:26Z

18

If I'm understanding your question correctly, something like

sum(d < threshold) / len(d)

should do it.

Edit: I missed the absolute value in the question -

sum(np.abs(d) < threshold) / float(len(d))

edited Jul 30, 2018 at 15:21

cookiedough

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answered Jul 30, 2018 at 14:52

Tim

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3 Comments

cookiedough Over a year ago

Edited your code to be python functional, but it was the right idea! I guess I was forgetting the basics! Thanks.

pault Over a year ago

@cookiedough your edit is not equivalent to what is posted here. If you wanted the absolute value, do sum(np.absolute(d) < 0.2) / float(len(d))

Tim Over a year ago

If d is a numpy array, the code I gave is valid python code (assuming you're using python3).

Martin Marek · Accepted Answer · 2022-03-30 14:36:52Z

Assuming d is a NumPy array, in general, you can do:

(d < threshold).mean()

And for absolute values specifically:

(np.abs(d) < threshold).mean()

Robert Robison · Accepted Answer · 2023-10-18 14:21:05Z

The other answers are great. But if there's a chance some values in the array could be identical to the threshold (e.g., array of integers), this trick will handle that:

( (d < threshold).mean() + (d <= threshold).mean() ) / 2

Just averaging using less than and less than or equal to.

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