How to get unique values with respective occurance 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'], and I want [('a', 2), ('b', 3)]

Looking for a simple way to do this without looping twice.

Answer 1

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 2

When Python 2.7 comes out you can use its collections.Counter class

otherwise see counter receipe

Under Python 2.7a3

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

print( c.items() )

output is

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

Answer 3

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 4

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 5

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 6

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

Answer 7

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)]