Finding the most common subarray within a numpy array

Question

Example data:

array(
  [[ 1.,  1.],
   [ 2.,  1.],
   [ 0.,  1.],
   [ 0.,  0.],
   [ 0.,  0.]])

with a desired result of

>>> [0.,0.]

ie) The most common pair.

Approaches that don't seem to work:

Using statistics as numpy arrays are unhashable.

Using scipy.stats.mode as this returns the mode over each axis, eg) for our example it gives

mode=array([[ 0.,  1.]])

piman314 · Accepted Answer · 2018-04-06 14:19:59Z

9

You can do this efficiently with numpy using the unique function:

pairs, counts = np.unique(a, axis=0, return_counts=True)
print(pairs[counts.argmax()])

Returns: [ 0. 0.]

answered Apr 6, 2018 at 14:19

piman314

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2 Comments

draco_alpine Over a year ago

jic anyone needs to know, for .argmax() ~ "In case of multiple occurrences of the maximum values, the indices corresponding to the first occurrence are returned."

jpp Over a year ago

Note the axis argument is only available in (fairly recent) numpy v.1.13.

jpp · Accepted Answer · 2018-04-06 14:18:26Z

2

One way via the standard library is to use collections.Counter.

This gives you both the most common pair and the count. Use [0] index on Counter.most_common() to retrieve the highest count.

import numpy as np
from collections import Counter

A = np.array(
  [[ 1.,  1.],
   [ 2.,  1.],
   [ 0.,  1.],
   [ 0.,  0.],
   [ 0.,  0.]])

c = Counter(map(tuple, A)).most_common()[0]

# ((0.0, 0.0), 2)

The only complication is you need to convert to tuple as Counter only accepts hashable objects.

answered Apr 6, 2018 at 14:18

jpp

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