By weakish


2010-04-08 13:30:00 8 Comments

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

22 comments

@Akash Swain 2018-11-02 08:43:50

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

@Tim Skov Jacobsen 2018-10-30 12:07:43

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)]

@AndyK 2018-10-21 21:59:07

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.

>>> a = [['a', 'b', 'b', 'b'], ['a', 'c', 'c', 'a']]
>>> dict(zip(*np.unique(a, return_counts=True)))
{'a': 3, 'b': 3, 'c': 2}

@Aaron Hall 2016-04-13 12:50:27

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

Here's an example list:

>>> l = list('aaaaabbbbcccdde')
>>> l
['a', 'a', 'a', 'a', 'a', 'b', 'b', 'b', 'b', 'c', 'c', 'c', 'd', 'd', 'e']

list.count

There's the list.count method

>>> l.count('b')
4

This works fine for any list. Tuples have this method as well:

>>> t = tuple('aabbbffffff')
>>> t
('a', 'a', 'b', 'b', 'b', 'f', 'f', 'f', 'f', 'f', 'f')
>>> 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:

>>> from collections import Counter
>>> c = Counter(l)
>>> 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:

>>> c.update(list('bbb'))
>>> c['b']
7
>>> c.subtract(list('bbb'))
>>> c['b']
4

And you can do multi-set operations with the counter as well:

>>> c2 = Counter(list('aabbxyz'))
>>> c - c2                   # set difference
Counter({'a': 3, 'c': 3, 'b': 2, 'd': 2, 'e': 1})
>>> c + c2                   # addition of all elements
Counter({'a': 7, 'b': 6, 'c': 3, 'd': 2, 'e': 1, 'y': 1, 'x': 1, 'z': 1})
>>> c | c2                   # set union
Counter({'a': 5, 'b': 4, 'c': 3, 'd': 2, 'e': 1, 'y': 1, 'x': 1, 'z': 1})
>>> c & 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.

@jpp 2019-01-03 01:36:32

While "why not Pandas" is appropriate, it should probably be accompanied by "when to use NumPy", i.e. for large numeric arrays. The deciding factor isn't just project limitations, there are memory efficiencies with NumPy which become apparent with big data.

@ravi tanwar 2018-07-07 08:29:21

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)

@Alex Riabov 2018-07-07 08:53:11

While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value.

@DataBender 2018-06-29 14:05:39

if you want a number of occurrences for the particular element:

>>> from collections import Counter
>>> z = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']
>>> single_occurrences = Counter(z)
>>> print(single_occurrences.get("blue"))
3
>>> print(single_occurrences.values())
dict_values([3, 2, 1])

@blue-sky 2018-03-01 10:38:08

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}

@Thirupathi Thangavel 2018-01-17 07:56:22

If you can use pandas, then value_counts is there for rescue.

>>> import pandas as pd
>>> a = [1, 2, 3, 4, 1, 4, 1]
>>> 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

>>> pd.Series(a).value_counts().reset_index().values.tolist()
[[1, 3], [4, 2], [3, 1], [2, 1]]

@Ɓukasz 2010-04-08 13:31:52

If you only want one item's count, use the count method:

>>> [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.

@cpp-coder 2017-09-09 19:15:33

mylist = [1,7,7,7,3,9,9,9,7,9,10,0] print sorted(set([i for i in mylist if mylist.count(i)>2]))

@Nico Schlömer 2017-09-13 10:32:48

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.

enter image description here

Counting all items

As established before,

numpy.bincount(a)

is what you want.

enter image description here


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,
    )

@Mukarram Pasha 2018-03-03 09:20:46

numpy.bincount() will work only for lists with int items.

@whackamadoodle3000 2017-08-23 01:20:52

sum([1 for elem in <yourlist> if elem==<your_value>])

This will return the amount of occurences of your_value

@user2314737 2014-05-28 10:58:37

Counting the occurrences of one item in a list

For counting the occurrences of just one list item you can use count()

>>> l = ["a","b","b"]
>>> l.count("a")
1
>>> 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:

>>> l = ["a","b","b"]
>>> [[x,l.count(x)] for x in set(l)]
[['a', 1], ['b', 2]]
>>> 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:

>>> l = ["a","b","b"]
>>> from collections import Counter
>>> 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]

@fhucho 2015-11-11 22:34:43

Counter is way faster for bigger lists. The list comprehension method is O(n^2), Counter should be O(n).

@Martijn Pieters 2017-05-23 10:13:51

Counter is not faster by a factor of 2, Counter is faster by a factor of n (O(n^2) vs O(n)).

@Jacob Schneider 2018-04-23 04:02:46

count() also works on strings, just saying

@Bram Vanroy 2018-06-19 21:27:58

I have found that when using this a lot (talking about millions of strings) that it is very slow because of its calls to isinstance. So if you are certain about the data that you're working with, it might be better to write a custom function without type and instance checking.

@Aakash Goel 2017-04-28 10:36:56

from collections import Counter
country=['Uruguay', 'Mexico', 'Uruguay', 'France', 'Mexico']
count_country = Counter(country)
output_list= [] 

for i in count_country:
    output_list.append([i,count_country[i]])
print output_list

Output list:

[['Mexico', 2], ['France', 1], ['Uruguay', 2]]

@Chris_Rands 2017-05-23 09:42:52

Code only answers and not encouraged, and moreover the use of Counter() has been suggested many times in previous answers

@Shoresh 2016-10-17 17:15:38

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

@vishes_shell 2016-09-18 20:08:11

You can also use countOf method of a built-in module operator.

>>> import operator
>>> operator.countOf([1, 2, 3, 4, 1, 4, 1], 1)
3

@Chris_Rands 2017-05-23 09:41:21

How is countOf is implemented? How does it compare to the more obvious list.count (which benefits from C implementation)? Are there any advantages?

@Silfverstrom 2010-04-08 13:34:15

list.count(x) returns the number of times x appears in a list

see: http://docs.python.org/tutorial/datastructures.html#more-on-lists

@tj80 2011-10-20 22:38:08

Another way to get the number of occurrences of each item, in a dictionary:

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

@Nicolas78 2012-10-10 00:30:31

this looks like one of the constructs I often come up with in the heat of the battle, but it will run through a len(a) times which means quadratic runtime complexity (as each run depends on len(a) again).

@hugo24 2013-08-23 09:20:20

would dict((i,a.count(i)) for i in set(a)) be more correct and faster?

@Clément 2013-10-07 09:46:31

@hugo24: A bit, but it won't be asymptotically faster in the worst case; it will take n * (number of different items) operations, not counting the time it takes to build the set. Using collections.Counter is really much better.

@Wes Turner 2011-08-14 08:47:39

# Python >= 2.6 (defaultdict) && < 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 >= 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)]

@plaes 2011-08-14 08:58:15

This is a bit "enterprisey"...

@eyquem 2011-08-14 15:40:12

@Wes Turner Happily, you use Python. Imagine the same in Java or C....

@Wes Turner 2011-08-21 12:32:36

@plaes : How so? If by 'enterprisey', you mean "documented" in preparation for Py3k annotations, I agree.

@Adam Lewis 2013-02-27 21:06:18

This is a great example, as I am developing mainly in 2.7, but have to have migration paths to 2.4.

@flonk 2013-11-19 10:53:07

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

>>> array([0, 3, 1, 1, 2])

@D Blanc 2011-11-07 19:18:54

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.

@chaosflaws 2015-06-08 21:29:35

Counter sorts the entries while yours does not, hence the speed difference (True at the time of writing, not sure if it was when you wrote the answer. Still, it might be relevant for someone scrolling down.)

@Martijn Pieters 2017-04-22 18:05:30

Counter in Python 2 was a little on the slow side, yes. It uses C-optimised code to do the counting in Python 3 however, and now beats your loop with ease.

@eyquem 2011-08-14 15:51:59

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

@user52028778 2011-04-29 07:44:22

If you are using Python 2.7 or 3 and you want number of occurrences for each element:

>>> from collections import Counter
>>> z = ['blue', 'red', 'blue', 'yellow', 'blue', 'red']
>>> Counter(z)
Counter({'blue': 3, 'red': 2, 'yellow': 1})

@Bram Vanroy 2018-06-19 21:26:32

I have found that when using this a lot (talking about millions of strings) that it is very slow because of its calls to isinstance. So if you are certain about the data that you're working with, it might be better to write a custom function without type and instance checking.

@user2357112 2018-11-14 04:08:00

@BramVanroy: What isinstance calls? Even with millions of strings, calling Counter only involves one isinstance call, to check whether its argument is a mapping. You most likely misjudged what's eating all your time.

@Bram Vanroy 2018-11-14 08:45:57

You misinterpreted what I meant: Counter checks the types of your data before it creates the Counter. This takes relatively much time and if you know the type of your data in advance. If you look at Counter's update method, you'll see it has to go through three if-statements before doing something. If you call update frequently, this adds up quickly. When you have control over your data and you know that the input will be indeed an iterable, then you can skip the first two checks. As I said, I only noticed this when working with millions of updates so it's an edge case.

@user2357112 2018-11-14 18:20:32

@BramVanroy: If you're performing millions of updates rather than just counting millions of strings, that's a different story. The optimization effort in Counter has gone into counting large iterables, rather than counting many iterables. Counting a million-string iterable will go faster with Counter than with a manual implementation. If you want to call update with many iterables, you may be able to speed things up by joining them into one iterable with itertools.chain.

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