# How can I count the occurrences of a list item?

## How can I count the occurrences of a list item?

### Accepted Answer

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.

### Popular Answer

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})
```

Read more... Read less...

**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)
```

`list.count(x)`

returns the number of times `x`

appears in a list

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

## 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.

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))
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,
)
```