How do you find the IQR in Numpy?

Question

Is there a baked-in Numpy/Scipy function to find the interquartile range? I can do it pretty easily myself, but mean() exists which is basically sum/len<

Answer 1

np.percentile takes multiple percentile arguments, and you are slightly better off doing:

q75, q25 = np.percentile(x, [75 ,25])
iqr = q75 - q25

or

iqr = np.subtract(*np.percentile(x, [75, 25]))

than making two calls to percentile:

In [8]: x = np.random.rand(1e6)

In [9]: %timeit q75, q25 = np.percentile(x, [75 ,25]); iqr = q75 - q25
10 loops, best of 3: 24.2 ms per loop

In [10]: %timeit iqr = np.subtract(*np.percentile(x, [75, 25]))
10 loops, best of 3: 24.2 ms per loop

In [11]: %timeit iqr = np.percentile(x, 75) - np.percentile(x, 25)
10 loops, best of 3: 33.7 ms per loop

Answer 2

Ignore this if Jaime's answer works for your case. But if not, according to this answer, to find the exact values of 1st and 3rd quartiles, you should consider doing something like:

samples = sorted([28, 12, 8, 27, 16, 31, 14, 13, 19, 1, 1, 22, 13])

def find_median(sorted_list):
    indices = []

    list_size = len(sorted_list)
    median = 0

    if list_size % 2 == 0:
        indices.append(int(list_size / 2) - 1)  # -1 because index starts from 0
        indices.append(int(list_size / 2))

        median = (sorted_list[indices[0]] + sorted_list[indices[1]]) / 2
        pass
    else:
        indices.append(int(list_size / 2))

        median = sorted_list[indices[0]]
        pass

    return median, indices
    pass

median, median_indices = find_median(samples)
Q1, Q1_indices = find_median(samples[:median_indices[0]])
Q2, Q2_indices = find_median(samples[median_indices[-1] + 1:])

IQR = Q3 - Q1

quartiles = [Q1, median, Q2]

Code taken from the referenced answer.

Answer 3

There is now an iqr function in scipy.stats. It is available as of scipy 0.18.0. My original intent was to add it to numpy, but it was considered too domain-specific.

You may be better off just using Jaime's answer, since the scipy code is just an over-complicated version of the same.