How do I count the occurrences of a list item?
PythonListCountPython Problem Overview
Given an item, how do I count its occurrences in a list in Python?
Python Solutions
Solution 1 - Python
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.
Solution 2 - Python
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})
Solution 3 - Python
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]
Solution 4 - Python
Another way to get the number of occurrences of each item, in a dictionary:
dict((i, a.count(i)) for i in a)
Solution 5 - Python
># 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.
Solution 6 - Python
list.count(x) returns the number of times x appears in a list
see: http://docs.python.org/tutorial/datastructures.html#more-on-lists
Solution 7 - Python
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,
)
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))
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()
Solution 8 - Python
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])
Solution 9 - Python
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]]
Solution 10 - Python
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
Solution 11 - Python
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.
Solution 12 - Python
# 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)]
Solution 13 - Python
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.
Solution 14 - Python
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}
Solution 15 - Python
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
Solution 16 - Python
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}
Solution 17 - Python
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
Solution 18 - Python
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
Solution 19 - Python
I would use filter(), take Lukasz's example:
>>> lst = [1, 2, 3, 4, 1, 4, 1]
>>> len(filter(lambda x: x==1, lst))
3
Solution 20 - Python
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}
Solution 21 - Python
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)
Solution 22 - Python
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}
Solution 23 - Python
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}
Solution 24 - Python
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}
Solution 25 - Python
sum([1 for elem in <yourlist> if elem==<your_value>])
This will return the amount of occurences of your_value
Solution 26 - Python
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
Solution 27 - Python
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])
Solution 28 - Python
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)))
Solution 29 - Python
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.
Solution 30 - Python
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'
Solution 31 - Python
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)