Find the most frequent number in a NumPy array

You may use

values, counts = np.unique(a, return_counts=True)

ind = np.argmax(counts)
print(values[ind])  # prints the most frequent element

ind = np.argpartition(-counts, kth=10)[:10]
print(values[ind])  # prints the 10 most frequent elements

If some element is as frequent as another one, this code will return only the first element.

If you're willing to use SciPy:

>>> from scipy.stats import mode
>>> mode([1,2,3,1,2,1,1,1,3,2,2,1])
(array([ 1.]), array([ 6.]))
>>> most_frequent = mode([1,2,3,1,2,1,1,1,3,2,2,1])[0][0]
>>> most_frequent
1.0

If your list contains all non-negative ints, you should take a look at numpy.bincounts:

http://docs.scipy.org/doc/numpy/reference/generated/numpy.bincount.html

and then probably use np.argmax:

a = np.array([1,2,3,1,2,1,1,1,3,2,2,1])
counts = np.bincount(a)
print(np.argmax(counts))

For a more complicated list (that perhaps contains negative numbers or non-integer values), you can use np.histogram in a similar way. Alternatively, if you just want to work in python without using numpy, collections.Counter is a good way of handling this sort of data.

from collections import Counter
a = [1,2,3,1,2,1,1,1,3,2,2,1]
b = Counter(a)
print(b.most_common(1))