Is there a "freq" function in numpy/python?

Question

Suppose you have:

arr = np.array([1,2,1,3,3,4])

Is there a built in function that returns the most frequent element?

Answer 1

Yes, Python's collections.Counter has direct support for finding the most frequent elements:

>>> from collections import Counter

>>> Counter('abracadbra').most_common(2)
[('a', 4), ('r', 2)]

>>> Counter([1,2,1,3,3,4]).most_common(2)
[(1, 2), (3, 2)]

With numpy, you might want to start with the histogram() function or the bincount() function.

With scipy, you can search for the modal element with mstats.mode.

Answer 2

the pandas module might also be of help here. pandas is a neat data analysis package for python and also has support for this problem.

import pandas as pd 
arr = np.array([1,2,1,3,3,4])
arr_df = pd.Series(arr) 
value_counts = arr_df.value_counts()
most_frequent = value_counts.max()

this returns

> most_frequent 
2 

Answer 3

This will work for any type, integer or not, and the return is always a numpy array:

def most_common(a, n=1) :
    if a.dtype.kind not in 'bui':
        items, _ = np.unique(a, return_inverse=True)
    else:
        items, _ = None, a
    counts = np.bincount(_)
    idx = np.argsort(counts)[::-1][:n]
    return idx.astype(a.dtype) if items is None else items[idx]

>>> a = np.fromiter('abracadabra', dtype='S1')
>>> most_common(a, 2)
array(['a', 'r'], 
      dtype='|S1')
>>> a = np.random.randint(10, size=100)
>>> a
array([0, 0, 0, 9, 3, 9, 1, 2, 6, 3, 0, 4, 3, 2, 4, 7, 2, 8, 8, 2, 9, 7, 0,
       3, 5, 2, 5, 0, 4, 2, 4, 7, 8, 5, 4, 0, 1, 6, 1, 0, 2, 0, 5, 1, 3, 8,
       8, 6, 3, 5, 4, 3, 3, 5, 0, 7, 3, 0, 2, 5, 4, 2, 4, 2, 8, 1, 4, 4, 7,
       4, 4, 3, 7, 4, 0, 1, 0, 8, 8, 1, 1, 2, 1, 4, 2, 5, 1, 0, 7, 2, 0, 0,
       0, 8, 9, 9, 8, 1, 3, 8])
>>> most_common(a, 5)
array([0, 4, 2, 8, 3])