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

Another way to do this would be

mylist = [1, 1, 2, 3, 3, 3, 4, 4, 4, 4]
mydict = {}
for i in mylist:
    if i in mydict: mydict[i] += 1
    else: mydict[i] = 1

then to get the list of tuples,

mytups = [(i, mydict[i]) for i in mydict]

This only goes over the list once, but it does have to traverse the dictionary once as well. However, given that there are a lot of duplicates in the list, then the dictionary should be a lot smaller, hence faster to traverse.

Nevertheless, not a very pretty or concise bit of code, I'll admit.

Answer 2

Convert any data structure into a pandas series s:

CODE:

for i in sort(s.value_counts().unique()):
  print i, (s.value_counts()==i).sum()

Answer 3

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)]

Answer 4

I know this isn't a one-liner... but to me I like it because it's clear to me that we pass over the initial list of values once (instead of calling count on it):

>>> from collections import defaultdict
>>> l = ['a', 'a', 'b', 'b', 'b']
>>> d = defaultdict(int)
>>> for i in l:
...  d[i] += 1
... 
>>> d
defaultdict(<type 'int'>, {'a': 2, 'b': 3})
>>> list(d.iteritems())
[('a', 2), ('b', 3)]
>>>

Answer 5

With help of pandas you can do like:

import pandas as pd
dict(pd.value_counts(my_list))