How to count the frequency of the elements in an unordered list?

Question

I need to find the frequency of elements in an unordered list

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]

output->

b =         


        

Answer 1

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()

Answer 2

In Python 2.7 (or newer), you can use collections.Counter:

import collections
a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
counter=collections.Counter(a)
print(counter)
# Counter({1: 4, 2: 4, 3: 2, 5: 2, 4: 1})
print(counter.values())
# [4, 4, 2, 1, 2]
print(counter.keys())
# [1, 2, 3, 4, 5]
print(counter.most_common(3))
# [(1, 4), (2, 4), (3, 2)]

If you are using Python 2.6 or older, you can download it here.

Answer 3

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

Answer 4

Note: You should sort the list before using groupby.

You can use groupby from itertools package if the list is an ordered list.

a = [1,1,1,1,2,2,2,2,3,3,4,5,5]
from itertools import groupby
[len(list(group)) for key, group in groupby(a)]

Output:

[4, 4, 2, 1, 2]

Answer 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))

Answer 6

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]))