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

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'开发者_开发问答], and I want [('a', 2), ('b', 3)].

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

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)]

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)]

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

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)]
>>>

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}

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.

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)]

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()

With help of pandas you can do like:

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