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How to efficiently count each element in a list in Python? [duplicate]

This question already has answers here:
Using a dictionary to count the items in a list
(8 answers)
Closed 7 months ago.
Given an unordered list of values like
a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
How can I get the frequency of each value that appears in the list, like so?
# `a` has 4 instances of `1`, 4 of `2`, 2 of `3`, 1 of `4,` 2 of `5`
b = [4, 4, 2, 1, 2] # expected output

In Python 2.7 (or newer), you can use collections.Counter:
>>> import collections
>>> a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
>>> counter = collections.Counter(a)
>>> counter
Counter({1: 4, 2: 4, 5: 2, 3: 2, 4: 1})
>>> counter.values()
dict_values([2, 4, 4, 1, 2])
>>> counter.keys()
dict_keys([5, 1, 2, 4, 3])
>>> counter.most_common(3)
[(1, 4), (2, 4), (5, 2)]
>>> dict(counter)
{5: 2, 1: 4, 2: 4, 4: 1, 3: 2}
>>> # Get the counts in order matching the original specification,
>>> # by iterating over keys in sorted order
>>> [counter[x] for x in sorted(counter.keys())]
[4, 4, 2, 1, 2]
If you are using Python 2.6 or older, you can download an implementation here.

If the list is sorted, you can use groupby from the itertools standard library (if it isn't, you can just sort it first, although this takes O(n lg n) time):
from itertools import groupby
a = [5, 1, 2, 2, 4, 3, 1, 2, 3, 1, 1, 5, 2]
[len(list(group)) for key, group in groupby(sorted(a))]
Output:
[4, 4, 2, 1, 2]

Python 2.7+ introduces Dictionary Comprehension. Building the dictionary from the list will get you the count as well as get rid of duplicates.
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> d = {x:a.count(x) for x in a}
>>> d
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}
>>> a, b = d.keys(), d.values()
>>> a
[1, 2, 3, 4, 5]
>>> b
[4, 4, 2, 1, 2]

Count the number of appearances manually by iterating through the list and counting them up, using a collections.defaultdict to track what has been seen so far:
from collections import defaultdict
appearances = defaultdict(int)
for curr in a:
appearances[curr] += 1

In Python 2.7+, you could use collections.Counter to count items
>>> a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>>
>>> from collections import Counter
>>> c=Counter(a)
>>>
>>> c.values()
[4, 4, 2, 1, 2]
>>>
>>> c.keys()
[1, 2, 3, 4, 5]

Counting the frequency of elements is probably best done with a dictionary:
b = {}
for item in a:
b[item] = b.get(item, 0) + 1
To remove the duplicates, use a set:
a = list(set(a))

You can do this:
import numpy as np
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
np.unique(a, return_counts=True)
Output:
(array([1, 2, 3, 4, 5]), array([4, 4, 2, 1, 2], dtype=int64))
The first array is values, and the second array is the number of elements with these values.
So If you want to get just array with the numbers you should use this:
np.unique(a, return_counts=True)[1]

Here's another succint alternative using itertools.groupby which also works for unordered input:
from itertools import groupby
items = [5, 1, 1, 2, 2, 1, 1, 2, 2, 3, 4, 3, 5]
results = {value: len(list(freq)) for value, freq in groupby(sorted(items))}
results
format: {value: num_of_occurencies}
{1: 4, 2: 4, 3: 2, 4: 1, 5: 2}

I would simply use scipy.stats.itemfreq in the following manner:
from scipy.stats import itemfreq
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq = itemfreq(a)
a = freq[:,0]
b = freq[:,1]
you may check the documentation here: http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.stats.itemfreq.html

from collections import Counter
a=["E","D","C","G","B","A","B","F","D","D","C","A","G","A","C","B","F","C","B"]
counter=Counter(a)
kk=[list(counter.keys()),list(counter.values())]
pd.DataFrame(np.array(kk).T, columns=['Letter','Count'])

seta = set(a)
b = [a.count(el) for el in seta]
a = list(seta) #Only if you really want it.

Suppose we have a list:
fruits = ['banana', 'banana', 'apple', 'banana']
We can find out how many of each fruit we have in the list like so:
import numpy as np
(unique, counts) = np.unique(fruits, return_counts=True)
{x:y for x,y in zip(unique, counts)}
Result:
{'banana': 3, 'apple': 1}

This answer is more explicit
a = [1,1,1,1,2,2,2,2,3,3,3,4,4]
d = {}
for item in a:
if item in d:
d[item] = d.get(item)+1
else:
d[item] = 1
for k,v in d.items():
print(str(k)+':'+str(v))
# output
#1:4
#2:4
#3:3
#4:2
#remove dups
d = set(a)
print(d)
#{1, 2, 3, 4}

For your first question, iterate the list and use a dictionary to keep track of an elements existsence.
For your second question, just use the set operator.

def frequencyDistribution(data):
return {i: data.count(i) for i in data}
print frequencyDistribution([1,2,3,4])
...
{1: 1, 2: 1, 3: 1, 4: 1} # originalNumber: count

I am quite late, but this will also work, and will help others:
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
freq_list = []
a_l = list(set(a))
for x in a_l:
freq_list.append(a.count(x))
print 'Freq',freq_list
print 'number',a_l
will produce this..
Freq [4, 4, 2, 1, 2]
number[1, 2, 3, 4, 5]

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counts = dict.fromkeys(a, 0)
for el in a: counts[el] += 1
print(counts)
# {1: 4, 2: 4, 3: 2, 4: 1, 5: 2}

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
# 1. Get counts and store in another list
output = []
for i in set(a):
output.append(a.count(i))
print(output)
# 2. Remove duplicates using set constructor
a = list(set(a))
print(a)
Set collection does not allow duplicates, passing a list to the set() constructor will give an iterable of totally unique objects. count() function returns an integer count when an object that is in a list is passed. With that the unique objects are counted and each count value is stored by appending to an empty list output
list() constructor is used to convert the set(a) into list and referred by the same variable a
Output
D:\MLrec\venv\Scripts\python.exe D:/MLrec/listgroup.py
[4, 4, 2, 1, 2]
[1, 2, 3, 4, 5]

Simple solution using a dictionary.
def frequency(l):
d = {}
for i in l:
if i in d.keys():
d[i] += 1
else:
d[i] = 1
for k, v in d.iteritems():
if v ==max (d.values()):
return k,d.keys()
print(frequency([10,10,10,10,20,20,20,20,40,40,50,50,30]))

#!usr/bin/python
def frq(words):
freq = {}
for w in words:
if w in freq:
freq[w] = freq.get(w)+1
else:
freq[w] =1
return freq
fp = open("poem","r")
list = fp.read()
fp.close()
input = list.split()
print input
d = frq(input)
print "frequency of input\n: "
print d
fp1 = open("output.txt","w+")
for k,v in d.items():
fp1.write(str(k)+':'+str(v)+"\n")
fp1.close()

from collections import OrderedDict
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
def get_count(lists):
dictionary = OrderedDict()
for val in lists:
dictionary.setdefault(val,[]).append(1)
return [sum(val) for val in dictionary.values()]
print(get_count(a))
>>>[4, 4, 2, 1, 2]
To remove duplicates and Maintain order:
list(dict.fromkeys(get_count(a)))
>>>[4, 2, 1]

i'm using Counter to generate a freq. dict from text file words in 1 line of code
def _fileIndex(fh):
''' create a dict using Counter of a
flat list of words (re.findall(re.compile(r"[a-zA-Z]+"), lines)) in (lines in file->for lines in fh)
'''
return Counter(
[wrd.lower() for wrdList in
[words for words in
[re.findall(re.compile(r'[a-zA-Z]+'), lines) for lines in fh]]
for wrd in wrdList])

For the record, a functional answer:
>>> L = [1,1,1,1,2,2,2,2,3,3,4,5,5]
>>> import functools
>>> >>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc,1)] if e<=len(acc) else acc+[0 for _ in range(e-len(acc)-1)]+[1], L, [])
[4, 4, 2, 1, 2]
It's cleaner if you count zeroes too:
>>> functools.reduce(lambda acc, e: [v+(i==e) for i, v in enumerate(acc)] if e<len(acc) else acc+[0 for _ in range(e-len(acc))]+[1], L, [])
[0, 4, 4, 2, 1, 2]
An explanation:
we start with an empty acc list;
if the next element e of L is lower than the size of acc, we just update this element: v+(i==e) means v+1 if the index i of acc is the current element e, otherwise the previous value v;
if the next element e of L is greater or equals to the size of acc, we have to expand acc to host the new 1.
The elements do not have to be sorted (itertools.groupby). You'll get weird results if you have negative numbers.

Another approach of doing this, albeit by using a heavier but powerful library - NLTK.
import nltk
fdist = nltk.FreqDist(a)
fdist.values()
fdist.most_common()

Found another way of doing this, using sets.
#ar is the list of elements
#convert ar to set to get unique elements
sock_set = set(ar)
#create dictionary of frequency of socks
sock_dict = {}
for sock in sock_set:
sock_dict[sock] = ar.count(sock)

For an unordered list you should use:
[a.count(el) for el in set(a)]
The output is
[4, 4, 2, 1, 2]

Yet another solution with another algorithm without using collections:
def countFreq(A):
n=len(A)
count=[0]*n # Create a new list initialized with '0'
for i in range(n):
count[A[i]]+= 1 # increase occurrence for value A[i]
return [x for x in count if x] # return non-zero count

num=[3,2,3,5,5,3,7,6,4,6,7,2]
print ('\nelements are:\t',num)
count_dict={}
for elements in num:
count_dict[elements]=num.count(elements)
print ('\nfrequency:\t',count_dict)

You can use the in-built function provided in python
l.count(l[i])
d=[]
for i in range(len(l)):
if l[i] not in d:
d.append(l[i])
print(l.count(l[i])
The above code automatically removes duplicates in a list and also prints the frequency of each element in original list and the list without duplicates.
Two birds for one shot ! X D

This approach can be tried if you don't want to use any library and keep it simple and short!
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
marked = []
b = [(a.count(i), marked.append(i))[0] for i in a if i not in marked]
print(b)
o/p
[4, 4, 2, 1, 2]

Retrieve the first element from Counter in O(n) time

I have such a nums list
In [72]: nums
Out[72]: [4, 1, 2, 1, 2]
try to get the unique number from the list
n [72]: nums
Out[72]: [4, 1, 2, 1, 2]
In [73]: c = Counter(nums)
In [74]: c
Out[74]: Counter({4: 1, 1: 2, 2: 2})
I can see the result from the counter, it is 4:1, but cannot retrieve it in O(1) time
In [79]: list(c)[0]
Out[79]: 4 #O(n) time
Is it possible to get 4 in O(1)time

According to the comments to the question, you want to get the elements that have a count of 1. But it is still not clear what you want to get exactly, as the term "the first element" is unclear in the context of a Counter, which is a dict and no defined order internally.
Here are a few options (I used str instead of int to make it clearer which are the values and which are their counts):
>>> import collections
>>> input_str = 'awlkjelqkdjlakd'
>>> c = collections.Counter(input_str)
>>> c
Counter({'l': 3, 'k': 3, 'a': 2, 'j': 2, 'd': 2, 'w': 1, 'e': 1, 'q': 1})
Get all elements that have count of 1 (takes O(k), where k is the number of different elements):
>>> [char for char, count in c.items() if count == 1]
['w', 'e', 'q']
Get one (random, not specified) element that has count of 1 (takes O(k), because the list has to be built):
>>> [char for char, count in c.items() if count == 1][0]
'w'
This can be improved by using a generator, so the full list will not be built; the generator will stop when the first element with count 1 is found, but there is no way to know if that will be first or last or in the middle ...
>>> g = (char for char, count in c.items() if count == 1)
>>> g
<generator object <genexpr> at 0x7fd520e82f68>
>>> next(g)
'w'
>>> next(char for char, count in c.items() if count == 1)
'w'
Now, if you want to find the count of the first element of your input data (in my example input_str), that is done in O(1) because it is a list item access and then a dict lookup:
>>> elem = input_str[0]
>>> elem
'a'
>>> c[elem]
2
But I cannot give a more concrete answer without more information on what exactly you need.

How to assign certain scores from a list to values in multiple lists and get the sum for each value in python?

Could you explain how to assign certain scores from a list to values in multiple lists and get the total score for each value?
score = [1,2,3,4,5] assigne a score based on the position in the list
l_1 = [a,b,c,d,e]
assign a=1, b=2, c=3, d=4, e=5
l_2 = [c,a,d,e,b]
assign c=1, a=2, d=3, e=4, b=5
I am trying to get the result like
{'e':9, 'b': 7, 'd':7, 'c': 4, 'a': 3}
Thank you!

You can zip the values of score to each list, which gives you a tuple of (key, value) for each letter-score combination. Make each zipped object a dict. Then use a dict comprehension to add the values for each key together.
d_1 = dict(zip(l_1, score))
d_2 = dict(zip(l_2, score))
{k: v + d_2[k] for k, v in d_1.items()}
# {'a': 3, 'b': 7, 'c': 4, 'd': 7, 'e': 9}

You better use zip function:
dic = {'a':0, 'b': 0, 'c':0, 'd': 0, 'e': 0}
def score(dic, *args):
for lst in args:
for k, v in zip(lst, range(len(lst))):
dic[k] += v+1
return dic
l_1 = ['a','b','c','d','e']
l_2 = ['c','a','d','e','b']
score(dic, l_1, l_2)

Instead of storing your lists in separate variables, you should put them in a list of lists so that you can iterate through it and calculate the sums of the scores according to each key's indices in the sub-lists:
score = [1, 2, 3, 4, 5]
lists = [
['a','b','c','d','e'],
['c','a','d','e','b']
]
d = {}
for l in lists:
for i, k in enumerate(l):
d[k] = d.get(k, 0) + score[i]
d would become:
{'a': 3, 'b': 7, 'c': 4, 'd': 7, 'e': 9}

from collections import defaultdict
score = [1,2,3,4,5] # note: 0 no need to use this list if there is no scenario like [5,6,9,10,4]
l_1 = ['a','b','c','d','e']
l_2 = ['c','a','d','e','b']
score_dict = defaultdict(int)
'''
for note: 0
if your score is always consecutive
like score = [2,3,4,5,6] or [5,6,7,8,9]...
you don't need to have seperate list of score you can set
start = score_of_char_at_first_position_ie_at_zero-th_index
like start = 2, or start = 5
else use this function
def add2ScoreDict( lst):
for pos_score, char in zip(score,lst):
score_dict[char] += pos_score
'''
def add2ScoreDict( lst):
for pos, char in enumerate( lst,start =1):
score_dict[char] += pos
# note: 1
add2ScoreDict( l_1)
add2ScoreDict( l_2)
#print(score_dict) # defaultdict(<class 'int'>, {'a': 3, 'b': 7, 'c': 4, 'd': 7, 'e': 9})
score_dict = dict(sorted(score_dict.items(), reverse = True, key=lambda x: x[1]))
print(score_dict) # {'e': 9, 'b': 7, 'd': 7, 'c': 4, 'a': 3}
edit 1:
if you have multiple lists put them in list_of_list = [l_1, l_2] so that you don't have to call func add2ScoreDict yourself again and again.
# for note: 1
for lst in list_of_list:
add2ScoreDict( lst)

You could zip both lists with score as one list l3 then you could use dictionary comprehension with filterto construct your dicitonary. The key being index 1 of the the newly formed tuples in l3, and the value being the sum of all index 0's in l3 after creating a sublist that is filtered for only matching index 0's
score = [1,2,3,4,5]
l_1 = ['a', 'b', 'c', 'd', 'e']
l_2 = ['c', 'a', 'd', 'e', 'b']
l3 = [*zip(score, l_1), *zip(score,l_2)]
d = {i[1]: sum([j[0] for j in list(filter(lambda x: x[1] ==i[1], l3))]) for i in l3}
{'a': 3, 'b': 7, 'c': 4, 'd': 7, 'e': 9}
Expanded Explanation:
d = {}
for i in l3:
f = list(filter(lambda x: x[1] == i[1], l3))
vals = []
for j in f:
vals.append(j[0])
total_vals = sum(vals)
d[i[1]] = total_vals

The simplest way is probably to use a Counter from the Python standard library.
from collections import Counter
tally = Counter()
scores = [1, 2, 3, 4, 5]
def add_scores(letters):
for letter, score in zip(letters, scores):
tally[letter] += score
L1 = ['a', 'b', 'c', 'd', 'e']
add_scores(L1)
L2 = ['c', 'a', 'd', 'e', 'b']
add_scores(L2)
print(tally)
>>> python tally.py
Counter({'e': 9, 'b': 7, 'd': 7, 'c': 4, 'a': 3})
zip is used to pair letters and scores, a for loop to iterate over them and a Counter to collect the results. A Counter is actually a dictionary, so you can write things like
tally['a']
to get the score for letter a or
for letter, score in tally.items():
print('Letter %s scored %s' % (letter, score))
to print the results, just as you would with a normal dictionary.
Finally, small ells and letter O's can be troublesome as variable names because they are hard to distinguish from ones and zeros. The Python style guide (often referred to as PEP8) recommends avoiding them.

In Python, I have a table with columns and lines; how do I count how many repetitions there are? [duplicate]

Given a single item, how do I count occurrences of it in a list, in Python?
A related but different problem is counting occurrences of each different element in a collection, getting a dictionary or list as a histogram result instead of a single integer. For that problem, see Using a dictionary to count the items in a list.

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})
Silly answer, sum
There are good builtin answers, but this example is slightly instructive. Here we sum all the occurences where the character, c, is equal to 'b':
>>> sum(c == 'b' for c in l)
4
Not great for this use-case, but if you need to have a count of iterables where the case is True it works perfectly fine to sum the boolean results, since True is equivalent to 1.
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
As established before,
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()

list.count(x) returns the number of times x appears in a list
see:
http://docs.python.org/tutorial/datastructures.html#more-on-lists

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])

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

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]]

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.

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.

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}

# 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)]

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

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}

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}

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

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'

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)

Unsort a list back to original sequence

aList = [2, 1, 4, 3, 5]
aList.sort()
=[1, 2, 3, 4, 5]
del aList[2]
=[1, 2, 4, 5]
**unsort the list back to original sequence with '3' deleted**
=[2, 1, 4, 5]
In reality I have a list of tuples that contain (Price, Quantity, Total).
I want to sort the list, allow the user to delete items in the list and
then put it back in the original order minus the deleted items.
One thing to note is that the values in the tuples can repeat in the list,
such as:
aList = [(4.55, 10, 45.5), (4.55, 10, 45.5), (1.99, 3, 5.97), (1.99, 1, 1.99)]

You cannot unsort the list but you could keep the original unsorted index to restore positions.
E.g.
from operator import itemgetter
aList = [(4.55, 10, 45.5), (4.55, 10, 45.5), (1.99, 3, 5.97), (1.99, 1, 1.99)]
# In keyList:
# * every element has a unique id (it also saves the original position in aList)
# * list is sorted by some criteria specific to your records
keyList = sorted(enumerate(aList), key = itemgetter(1))
# User want to delete item 1
for i, (key, record) in enumerate(keyList):
if key == 1:
del keyList[i]
break
# "Unsort" the list
theList = sorted(keyList, key = itemgetter(0))
# We don't need the unique id anymore
result = [record for key, record in theList]
As you can see this works with duplicate values.

Unsorting can be done
This approach is like others - the idea is to keep the original indices to restore the positions. I wanted to add a clearer example on how this is done.
In the example below, we keep track of the original positions of the items in a by associating them with their list index.
>>> a = [4, 3, 2, 1]
>>> b = [(a[i], i) for i in range(len(a))]
>>> b
[(4, 0), (3, 1), (2, 2), (1, 3)]
b serves as a mapping between the list values and their indices in the unsorted list.
Now we can sort b. Below, each item of b is sorted by the first tuple member, which is the corresponding value in the original list.
>>> c = sorted(b)
>>> c
[(1, 3), (2, 2), (3, 1), (4, 0)]
There it is... sorted.
Going back to the original order requires another sort, except using the second tuple item as the key.
>>> d = sorted(c, key=lambda t: t[1])
>>> d
[(4, 0), (3, 1), (2, 2), (1, 3)]
>>>
>>> d == b
True
And now it's back in its original order.
One use for this could be to transform a list of non sequential values into their ordinal values while maintaining the list order. For instance, a sequence like [1034 343 5 72 8997] could be transformed to [3, 2, 0, 1, 4].
>>> # Example for converting a list of non-contiguous
>>> # values in a list into their relative ordinal values.
>>>
>>> def ordinalize(a):
... idxs = list(range(len(a)))
... b = [(a[i], i) for i in idxs]
... b.sort()
... c = [(*b[i], i) for i in idxs]
... c.sort(key=lambda item: item[1])
... return [c[i][2] for i in idxs]
...
>>> ordinalize([58, 42, 37, 25, 10])
[4, 3, 2, 1, 0]
Same operation
>>> def ordinalize(a):
... idxs = range(len(a))
... a = sorted((a[i], i) for i in idxs)
... a = sorted(((*a[i], i) for i in idxs),
... key=lambda item: item[1])
... return [a[i][2] for i in idxs]

You can't really do an "unsort", the best you can do is:
aList = [2, 1, 4, 3, 5]
aList.remove(sorted(aList)[2])
>>> print aList
[2, 1, 4, 5]

Try this to unsort a sorted list
import random
li = list(range(101))
random.shuffle(li)

Here's how I recommend to sort a list, do something, then unsort back to the original ordering:
# argsort is the inverse of argsort, so we use that
# for undoing the sorting.
sorter = np.argsort(keys)
unsorter = np.argsort(sorter)
sorted_keys = np.array(keys)[sorter]
result = do_a_thing_that_preserves_order(sorted_keys)
unsorted_result = np.array(result)[unsorter]

I had the same use case and I found an easy solution for that, which is basically random the list:
import random
sorted_list = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k']
unsorted_list = random.sample(sorted_list, len(sorted_list))