Fastest way to count number of occurrences in a Python list

Question

I have a Python list and I want to know what\'s the quickest way to count the number of occurrences of the item, \'1\' in this list. In my actual case, the item

Answer 1

a = ['1', '1', '1', '1', '1', '1', '2', '2', '2', '2', '7', '7', '7', '10', '10']
print a.count("1")

It's probably optimized heavily at the C level.

Edit: I randomly generated a large list.

In [8]: len(a)
Out[8]: 6339347

In [9]: %timeit a.count("1")
10 loops, best of 3: 86.4 ms per loop

Edit edit: This could be done with collections.Counter

a = Counter(your_list)
print a['1']

Using the same list in my last timing example

In [17]: %timeit Counter(a)['1']
1 loops, best of 3: 1.52 s per loop

My timing is simplistic and conditional on many different factors, but it gives you a good clue as to performance.

Here is some profiling

In [24]: profile.run("a.count('1')")
         3 function calls in 0.091 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    0.091    0.091 <string>:1(<module>)
        1    0.091    0.091    0.091    0.091 {method 'count' of 'list' objects}

        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Prof
iler' objects}



In [25]: profile.run("b = Counter(a); b['1']")
         6339356 function calls in 2.143 seconds

   Ordered by: standard name

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
        1    0.000    0.000    2.143    2.143 <string>:1(<module>)
        2    0.000    0.000    0.000    0.000 _weakrefset.py:68(__contains__)
        1    0.000    0.000    0.000    0.000 abc.py:128(__instancecheck__)
        1    0.000    0.000    2.143    2.143 collections.py:407(__init__)
        1    1.788    1.788    2.143    2.143 collections.py:470(update)
        1    0.000    0.000    0.000    0.000 {getattr}
        1    0.000    0.000    0.000    0.000 {isinstance}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Prof
iler' objects}
  6339347    0.356    0.000    0.356    0.000 {method 'get' of 'dict' objects}

Answer 2

By the use of Counter dictionary counting the occurrences of all element as well as most common element in python list with its occurrence value in most efficient way.

If our python list is:-

l=['1', '1', '1', '1', '1', '1', '2', '2', '2', '2', '7', '7', '7', '10', '10']

To find occurrence of every items in the python list use following:-

\>>from collections import Counter

\>>c=Counter(l)

\>>print c

Counter({'1': 6, '2': 4, '7': 3, '10': 2})

To find most/highest occurrence of items in the python list:-

\>>k=c.most_common()

\>>k

[('1', 6), ('2', 4), ('7', 3), ('10', 2)]

For Highest one:-

\>>k[0][1]

6

For the item just use k[0][0]

\>>k[0][0]

'1'

For nth highest item and its no of occurrence in the list use follow:-

**for n=2 **

\>>print k[n-1][0] # For item

2

\>>print k[n-1][1] # For value

4

Answer 3

You can convert list in string with elements seperated by space and split it based on number/char to be searched..

Will be clean and fast for large list..

>>>L = [2,1,1,2,1,3]
>>>strL = " ".join(str(x) for x in L)
>>>strL
2 1 1 2 1 3
>>>count=len(strL.split(" 1"))-1
>>>count
3

Answer 4

You can use pandas, by transforming the list to a pd.Series then simply use .value_counts()

import pandas as pd
a = ['1', '1', '1', '1', '1', '1', '2', '2', '2', '2', '7', '7', '7', '10', '10']
a_cnts = pd.Series(a).value_counts().to_dict()

Input  >> a_cnts["1"], a_cnts["10"]
Output >> (6, 2)

Answer 5

Combination of lambda and map function can also do the job:

list_ = ['a', 'b', 'b', 'c']
sum(map(lambda x: x=="b", list_))
:2