How can I count the occurrences of a list item

Question

Given an item  how can I count its occurrences in a list in Python

User · Answer

I had this problem today and rolled my own solution before I thought to check SO. This:

dict((i,a.count(i)) for i in a)

is really, really slow for large lists. My solution

def occurDict(items):
    d = {}
    for i in items:
        if i in d:
            d[i] = d[i]+1
        else:
            d[i] = 1
return d

is actually a bit faster than the Counter solution, at least for Python 2.7.

User · Answer

Below are the three solutions   Fastest is using a for loop and storing it in a Dict   import time from collections import Counter   def countElement a       g          for i in a          if i in g               g i    1         else               g i   1     return g   z    1 1 1 1 2 2 2 2 3 3 4 5 5 234 23 3 12 3 123 12 31 23 13 2 4 23 42 42 34 234 23 42 34 23 423 42 34 23 423 4 234 23 42 34 23 4 23 423 4 23 4     Solution 1 - Faster st   time monotonic   for i in range 1000000       b   countElement z  et   time monotonic   print b  print  Simple for loop and storing it in dict - Duration      format et - st     Solution 2 - Fast st   time monotonic   for i in range 1000000       a   Counter z  et   time monotonic   print  a  print  Using collections Counter - Duration      format et - st     Solution 3 - Slow st   time monotonic   for i in range 1000000       g   dict   i  z count i   for i in set z    et   time monotonic   print g  print  Using list comprehension - Duration      format et - st     Result    Solution 1 - Faster     1  4  2  5  3  4  4  6  5  2  234  3  23  10  12  2  123  1  31  1  13  1  42  5  34  4  423  3  Simple for loop and storing it in dict - Duration  12 032000000000153     Solution 2 - Fast    Counter  23  10  4  6  2  5  42  5  1  4  3  4  34  4  234  3  423  3  5  2  12  2  123  1  31  1  13  1   Using collections Counter - Duration  15 889999999999418     Solution 3 - Slow     1  4  2  5  3  4  4  6  5  2  34  4  423  3  234  3  42  5  12  2  13  1  23  10  123  1  31  1  Using list comprehension - Duration  33 0

User · Answer

list count x  returns the number of times x appears in a list  see  http   docs python org tutorial datastructures html more-on-lists

User · Answer

If you want to count all values at once you can do it very fast using numpy arrays and bincount as follows  import numpy as np a   np array  1  2  3  4  1  4  1   np bincount a    which gives   gt  gt  gt  array  0  3  1  1  2

User · Answer

Although it is very old question  but as i didn t find a one liner  i made one       original numbers in list l    1  2  2  3  3  3  4     empty dictionary to hold pair of number and its count d         loop through all elements and store count   d update   i d get i  0  1    for i in l    print d

User · Answer

sum  1 for elem in  lt yourlist gt  if elem   lt your value gt      This will return the amount of occurences of your value

User · Answer

Why not using Pandas   import pandas as pd  l     a    b    c    d    a    d    a      converting the list to a Series and counting the values my count   pd Series l  value counts   my count   Output   a    3 d    2 b    1 c    1 dtype  int64   If you are looking for a count of a particular element  say a  try   my count  a     Output   3

User · Answer

Another way to get the number of occurrences of each item  in a dictionary   dict  i  a count i   for i in a

User · Answer

Given a list X  import numpy as np  X    1  -1  1  -1  1   The dictionary which shows i  frequency i  for elements of this list is   i X count i  for i in np unique X    Output   -1  2  1  3

User · Answer

Given an item  how can I count its occurrences in a list in Python   Here s an example list   gt  gt  gt  l   list  aaaaabbbbcccdde    gt  gt  gt  l   a    a    a    a    a    b    b    b    b    c    c    c    d    d    e    list count There s the list count method  gt  gt  gt  l count  b   4  This works fine for any list  Tuples have this method as well   gt  gt  gt  t   tuple  aabbbffffff    gt  gt  gt  t   a    a    b    b    b    f    f    f    f    f    f    gt  gt  gt  t count  f   6  collections Counter And then there s collections Counter  You can dump any iterable into a Counter  not just a list  and the Counter will retain a data structure of the counts of the elements  Usage   gt  gt  gt  from collections import Counter  gt  gt  gt  c   Counter l   gt  gt  gt  c  b   4  Counters are based on Python dictionaries  their keys are the elements  so the keys need to be hashable  They are basically like sets that allow redundant elements into them  Further usage of collections Counter You can add or subtract with iterables from your counter   gt  gt  gt  c update list  bbb     gt  gt  gt  c  b   7  gt  gt  gt  c subtract list  bbb     gt  gt  gt  c  b   4  And you can do multi-set operations with the counter as well   gt  gt  gt  c2   Counter list  aabbxyz     gt  gt  gt  c - c2                     set difference Counter   a   3   c   3   b   2   d   2   e   1    gt  gt  gt  c   c2                     addition of all elements Counter   a   7   b   6   c   3   d   2   e   1   y   1   x   1   z   1    gt  gt  gt  c   c2                     set union Counter   a   5   b   4   c   3   d   2   e   1   y   1   x   1   z   1    gt  gt  gt  c  amp  c2                     set intersection Counter   a   2   b   2    Why not pandas  Another answer suggests   Why not use pandas   Pandas is a common library  but it s not in the standard library  Adding it as a requirement is non-trivial  There are builtin solutions for this use-case in the list object itself as well as in the standard library  If your project does not already require pandas  it would be foolish to make it a requirement just for this functionality

User · Answer

I would use filter    take Lukasz s example    gt  gt  gt  lst    1  2  3  4  1  4  1   gt  gt  gt  len filter lambda x  x  1  lst   3

User · Answer

if you want a number of occurrences for the particular element    gt  gt  gt  from collections import Counter  gt  gt  gt  z     blue    red    blue    yellow    blue    red    gt  gt  gt  single occurrences   Counter z   gt  gt  gt  print single occurrences get  blue    3  gt  gt  gt  print single occurrences values    dict values  3  2  1

User · Answer

If you only want one item s count  use the count method    gt  gt  gt   1  2  3  4  1  4  1  count 1  3   Don t use this if you want to count multiple items  Calling count in a loop requires a separate pass over the list for every count call  which can be catastrophic for performance  If you want to count all items  or even just multiple items  use Counter  as explained in the other answers

User · Answer

I ve compared all suggested solutions  and a few new ones  with perfplot  a small project of mine    Counting one item  For large enough arrays  it turns out that    numpy sum numpy array a     1     is slightly faster than the other solutions     Counting all items  As established before     numpy bincount a    is what you want       Code to reproduce the plots     from collections import Counter from collections import defaultdict import numpy import operator import pandas import perfplot   def counter a       return Counter a    def count a       return dict  i  a count i   for i in set a     def bincount a       return numpy bincount a    def pandas value counts a       return pandas Series a  value counts     def occur dict a       d          for i in a          if i in d              d i    d i  1         else              d i    1     return d   def count unsorted list items items       counts   defaultdict int      for item in items          counts item     1     return dict counts    def operator countof a       return dict  i  operator countOf a  i   for i in set a     perfplot show      setup lambda n  list numpy random randint 0  100  n        n range  2  k for k in range 20        kernels           counter  count  bincount  pandas value counts  occur dict          count unsorted list items  operator countof                equality check None      logx True      logy True          2     from collections import Counter from collections import defaultdict import numpy import operator import pandas import perfplot   def counter a       return Counter a    def count a       return dict  i  a count i   for i in set a     def bincount a       return numpy bincount a    def pandas value counts a       return pandas Series a  value counts     def occur dict a       d          for i in a          if i in d              d i    d i  1         else              d i    1     return d   def count unsorted list items items       counts   defaultdict int      for item in items          counts item     1     return dict counts    def operator countof a       return dict  i  operator countOf a  i   for i in set a     perfplot show      setup lambda n  list numpy random randint 0  100  n        n range  2  k for k in range 20        kernels           counter  count  bincount  pandas value counts  occur dict          count unsorted list items  operator countof                equality check None      logx True      logy True

User · Answer

Python  gt   2 6  defaultdict   amp  amp   lt  2 7  Counter  OrderedDict  from collections import defaultdict def count unsorted list items items                param items  iterable of hashable items to count      type items  iterable       returns  dict of counts like Py2 7 Counter      rtype  dict             counts   defaultdict int      for item in items          counts item     1     return dict counts      Python  gt   2 2  generators  def count sorted list items items                param items  sorted iterable of items to count      type items  sorted iterable       returns  generator of  item  count  tuples      rtype  generator             if not items          return     elif len items     1          yield  items 0   1          return     prev item   items 0      count   1     for item in items 1            if prev item    item              count    1         else              yield  prev item  count              count   1             prev item   item     yield  item  count      return   import unittest class TestListCounters unittest TestCase       def test count unsorted list items self           D                                         2     2 1                   2 2     2 2                   2 2 2 2 3 3 5 5     2 4    3 2    5 2                           for inp  exp outp in D              counts   count unsorted list items inp               print inp  exp outp  counts             self assertEqual counts  dict  exp outp             inp  exp outp   UNSORTED WIN     2 2 4 2     2 3    4 1            self assertEqual dict  exp outp    count unsorted list items inp          def test count sorted list items self           D                                         2     2 1                   2 2     2 2                   2 2 2 2 3 3 5 5     2 4    3 2    5 2                           for inp  exp outp in D              counts   list  count sorted list items inp                print inp  exp outp  counts             self assertEqual counts  exp outp           inp  exp outp   UNSORTED FAIL     2 2 4 2     2 3    4 1            self assertEqual exp outp  list  count sorted list items inp                     2 2    4 1    2 1

User · Answer

Counting the occurrences of one item in a list  For counting the occurrences of just one list item you can use count     gt  gt  gt  l     a   b   b    gt  gt  gt  l count  a   1  gt  gt  gt  l count  b   2   Counting the occurrences of all items in a list is also known as  tallying  a list  or creating a tally counter   Counting all items with count    To count the occurrences of items in l one can simply use a list comprehension and the count   method    x l count x   for x in set l      or similarly with a dictionary dict  x l count x   for x in set l     Example     gt  gt  gt  l     a   b   b    gt  gt  gt    x l count x   for x in set l      a   1     b   2    gt  gt  gt  dict  x l count x   for x in set l     a   1   b   2    Counting all items with Counter    Alternatively  there s the faster Counter class from the collections library  Counter l    Example    gt  gt  gt  l     a   b   b    gt  gt  gt  from collections import Counter  gt  gt  gt  Counter l  Counter   b   2   a   1     How much faster is Counter   I checked how much faster Counter is for tallying lists  I tried both methods out with a few values of n and it appears that Counter is faster by a constant factor of approximately 2   Here is the script I used   from   future   import print function import timeit  t1 timeit Timer  Counter l                       import random import string from collections import Counter n 1000 l  random choice string ascii letters  for x in range n                       t2 timeit Timer    x l count x   for x in set l                      import random import string n 1000 l  random choice string ascii letters  for x in range n                       print  Counter       t1 repeat repeat 3 number 10000   print  count         t2 repeat repeat 3 number 10000    And the output   Counter      0 46062711701961234  0 4022796869976446  0 3974247490405105  count        7 779430688009597  7 962715800967999  8 420845870045014

User · Answer

Count of all elements with itertools groupby    Antoher possiblity for getting the count of all elements in the list could be by means of itertools groupby     With  duplicate  counts  from itertools import groupby  L     a    a    a    t    q    a    d    a    d    c      Input list  counts     i  len list c    for i c in groupby L          Create value-count pairs as list of tuples  print counts    Returns     a   3     t   1     q   1     a   1     d   1     a   1     d   1     c   1     Notice how it combined the first three a s as the first group  while other groups of a are present further down the list  This happens because the input list L was not sorted  This can be a benefit sometimes if the groups should in fact be separate   With unique counts  If unique group counts are desired  just sort the input list   counts     i  len list c    for i c in groupby sorted L    print counts    Returns     a   5     c   1     d   2     q   1     t   1     Note  For creating unique counts  many of the other answers provide easier and more readable code compared to the groupby solution  But it is shown here to draw a parallel to the duplicate count example

User · Answer

May not be the most efficient  requires an extra pass to remove duplicates    Functional implementation     arr   np array   a   a   b   b   b   c    print set map lambda x     x   list arr  count x     arr      returns        c   1     b   3     a   2     or return as dict    print dict map lambda x     x   list arr  count x     arr      returns       b   3   c   1   a   2

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To count the number of diverse elements having a common type   li     A0   c5   A8   A2   A5   c2   A3   A9    print sum 1 for el in li if el 0    A  and el 1  in  01234     gives  3    not 6

User · Answer

Use Counter if you are using Python 2 7 or 3 x and you want the number of occurrences for each element   gt  gt  gt  from collections import Counter  gt  gt  gt  z     blue    red    blue    yellow    blue    red    gt  gt  gt  Counter z  Counter   blue   3   red   2   yellow   1

User · Answer

l2  1  feto    feto  1   feto      feto   1 2 3   feto      count 0  def Test l              global count          if len l   0               return count         count l count  feto           for i in l               if type i  is list                  count  Test i          return count        print Test l2     this will recursive count or search for the item in the list even if it in list of lists

User · Answer

If you can use pandas  then value counts is there for rescue     gt  gt  gt  import pandas as pd  gt  gt  gt  a    1  2  3  4  1  4  1   gt  gt  gt  pd Series a  value counts   1    3 4    2 3    1 2    1 dtype  int64   It automatically sorts the result based on frequency as well    If you want the result to be in a list of list  do as below   gt  gt  gt  pd Series a  value counts   reset index   values tolist     1  3    4  2    3  1    2  1

User · Answer

def countfrequncyinarray arr1       r len arr1      return  i arr1 count i  for i in range 1 r 1   arr1  4 4 4 4  a countfrequncyinarray arr1  print a

User · Answer

use  timeit to see which operation is more efficient    np array counting operations should be faster   from collections import Counter  mylist    1 7 7 7 3 9 9 9 7 9 10 0    types counts Counter mylist   print types counts

User · Answer

It was suggested to use numpy s bincount  however it works only for 1d arrays with non-negative integers  Also  the resulting array might be confusing  it contains the occurrences of the integers from min to max of the original list  and sets to 0 the missing integers    A better way to do it with numpy is to use the unique function with the attribute return counts set to True  It returns a tuple with an array of the unique values and an array of the occurrences of each unique value     a    1  1  0  2  1  0  3  3  a uniq  counts   np unique a  return counts True     array  0  1  2  3    array  2  3  1  2    and then we can pair them as  dict zip a uniq  counts       0  2  1  3  2  1  3  2    It also works with other data types and  2d lists   e g    gt  gt  gt  a      a    b    b    b      a    c    c    a     gt  gt  gt  dict zip  np unique a  return counts True      a   3   b   3   c   2

User · Answer

You can also use countOf method of a built-in module operator    gt  gt  gt  import operator  gt  gt  gt  operator countOf  1  2  3  4  1  4  1   1  3

[python] How can I count the occurrences of a list item?

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