How to get unique values with respective occurrence count from a list in Python?

If your items are grouped (i.e. similar items come together in a bunch), the most efficient method to use is itertools.groupby:

>>> [(g[0], len(list(g[1]))) for g in itertools.groupby(['a', 'a', 'b', 'b', 'b'])]
[('a', 2), ('b', 3)]

If you are willing to use a 3rd party library, NumPy offers a convenient solution. This is particularly efficient if your list contains only numeric data.

import numpy as np

L = ['a', 'a', 'b', 'b', 'b']

res = list(zip(*np.unique(L, return_counts=True)))

# [('a', 2), ('b', 3)]

To understand the syntax, note np.unique here returns a tuple of unique values and counts:

uniq, counts = np.unique(L, return_counts=True)

print(uniq)    # ['a' 'b']
print(counts)  # [2 3]

See also: What are the advantages of NumPy over regular Python lists?

With Python 2.7+, you can use collections.Counter.

Otherwise, see this counter receipe.

Under Python 2.7+:

from collections import Counter
input =  ['a', 'a', 'b', 'b', 'b']
c = Counter( input )

print( c.items() )

Output is:

[('a', 2), ('b', 3)]

>>> mylist=['a', 'a', 'b', 'b', 'b']
>>> [ (i,mylist.count(i)) for i in set(mylist) ]
[('a', 2), ('b', 3)]