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

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

Answer 2

#!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()

Answer 3

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)

Answer 4

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]

Answer 5

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

Answer 6

To count the number of appearances:

from collections import defaultdict

appearances = defaultdict(int)

for curr in a:
    appearances[curr] += 1

To remove duplicates:

a = set(a)