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

This question already has answers here: Using a dictionary to count the items in a list (7 answers) Closed 5 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开发者_运维百科 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)

1. 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
2. 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()

---

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)

---

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

---

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]