Given an item, how do I count its occurrences in a list in Python?
31 Answers
If you only want a single item's count, use the count method:
>>> [1, 2, 3, 4, 1, 4, 1].count(1) 3 Important: this is very slow if you are counting multiple different items
Each count call goes over the entire list of n elements. Calling count in a loop n times means n * n total checks, which can be catastrophic for performance.
If you want to count multiple items, use Counter, which only does n total checks.
Use Counter if you are using Python 2.7 or 3.x and you want the 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}) 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] Another way to get the number of occurrences of each item, in a dictionary:
dict((i, a.count(i)) for i in a) 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.
2list.count(x) returns the number of times x appears in a list
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
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, ) 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)) b = perfplot.bench( 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, ) b.save("out.png") b.show() 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]) 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]] Why not using Pandas?
import pandas as pd my_list = ['a', 'b', 'c', 'd', 'a', 'd', 'a'] # converting the list to a Series and counting the values my_count = pd.Series(my_list).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 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.
2Count 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.
# 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)] Although it is very old question, since 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) # {1: 1, 2: 2, 3: 3, 4: 1} 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 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} 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
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 I would use filter(), take Lukasz's example:
>>> lst = [1, 2, 3, 4, 1, 4, 1] >>> len(filter(lambda x: x==1, lst)) 3 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} 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) 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} Alternatively, you can also implement the counter by yourself. This is the way I do:
item_list = ['me', 'me', 'you', 'you', 'you', 'they'] occ_dict = {} for item in item_list: if item not in occ_dict: occ_dict[item] = 1 else: occ_dict[item] +=1 print(occ_dict) Output: {'me': 2, 'you': 3, 'they': 1}
mot = ["compte", "france", "zied"] lst = ["compte", "france", "france", "france", "france"] dict((x, lst.count(x)) for x in set(mot)) this gives
{'compte': 1, 'france': 4, 'zied': 0} sum([1 for elem in <yourlist> if elem==<your_value>]) This will return the amount of occurences of your_value
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
1if 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]) test = [409.1, 479.0, 340.0, 282.4, 406.0, 300.0, 374.0, 253.3, 195.1, 269.0, 329.3, 250.7, 250.7, 345.3, 379.3, 275.0, 215.2, 300.0] for i in test: print('{} numbers {}'.format(i, test.count(i))) import pandas as pd test = [409.1, 479.0, 340.0, 282.4, 406.0, 300.0, 374.0, 253.3, 195.1, 269.0, 329.3, 250.7, 250.7, 345.3, 379.3, 275.0, 215.2, 300.0] #turning the list into a temporary dataframe test = pd.DataFrame(test) #using the very convenient value_counts() function df_counts = test.value_counts() df_counts then you can use df_counts.index and df_counts.values to get the data.
x = ['Jess', 'Jack', 'Mary', 'Sophia', 'Karen', 'Addison', 'Joseph','Jack', 'Jack', 'Eric', 'Ilona', 'Jason'] the_item = input('Enter the item that you wish to find : ') how_many_times = 0 for occurrence in x: if occurrence == the_item : how_many_times += 1 print('The occurrence of', the_item, 'in', x,'is',how_many_times) Created a list of names wherein the name 'Jack' is repeated. In order to check its Occurrence, I ran a for loop in the list named x. Upon each iteration, if the loop variable attains the value same that of received from the user and stored in the variable the_item, the variable how_many_times gets incremented by 1. After attaining some value...We print how_many_times which stores the value of the occurance of the word 'jack'
