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

Question

I have a list which has repeating items and I want a list of the unique items with their frequency.

For example, I have [\'a\', \'a\', \'b\', \'b\', \'b\']

Answer 1

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?

Answer 2

A solution without hashing:

def lcount(lst):
   return reduce(lambda a, b: a[0:-1] + [(a[-1][0], a[-1][1]+1)] if a and b == a[-1][0] else a + [(b, 1)], lst, [])

>>> lcount([])
[]
>>> lcount(['a'])
[('a', 1)]
>>> lcount(['a', 'a', 'a', 'b', 'b'])
[('a', 3), ('b', 2)]

Answer 3

the "old school way".

>>> alist=['a', 'a', 'b', 'b', 'b']
>>> d={}
>>> for i in alist:
...    if not d.has_key(i): d[i]=1  #also: if not i in d
...    else: d[i]+=1
...
>>> d
{'a': 2, 'b': 3}

Answer 4

Here's one way:

your_list = ['a', 'a', 'b', 'b', 'b']

count_dictionary = {}

for letter in your_list:

    if letter in count_dictionary:

        count_dictionary[letter] +=1 

    else:

        count_dictionary[letter] = 1

Answer 5

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)]

Answer 6

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