Item frequency count in Python

user_110572 0
Viewed 139708

Assume I have a list of words, and I want to find the number of times each word appears in that list.

An obvious way to do this is:

words = "apple banana apple strawberry banana lemon"
uniques = set(words.split())
freqs = [(item, words.split().count(item)) for item in uniques]
print(freqs)

But I find this code not very good, because the program runs through the word list twice, once to build the set, and a second time to count the number of appearances.

Of course, I could write a function to run through the list and do the counting, but that wouldn't be so Pythonic. So, is there a more efficient and Pythonic way?

counting frequency count python
14 Answers

The Counter class in the collections module is purpose built to solve this type of problem:

from collections import Counter
words = "apple banana apple strawberry banana lemon"
Counter(words.split())
# Counter({'apple': 2, 'banana': 2, 'strawberry': 1, 'lemon': 1})
sykora_538500

defaultdict to the rescue!

from collections import defaultdict

words = "apple banana apple strawberry banana lemon"

d = defaultdict(int)
for word in words.split():
    d[word] += 1

This runs in O(n).

user_430890 
freqs = {}
for word in words:
    freqs[word] = freqs.get(word, 0) + 1 # fetch and increment OR initialize

I think this results to the same as Triptych's solution, but without importing collections. Also a bit like Selinap's solution, but more readable imho. Almost identical to Thomas Weigel's solution, but without using Exceptions.

This could be slower than using defaultdict() from the collections library however. Since the value is fetched, incremented and then assigned again. Instead of just incremented. However using += might do just the same internally.

user_820110

Standard approach:

from collections import defaultdict

words = "apple banana apple strawberry banana lemon"
words = words.split()
result = defaultdict(int)
for word in words:
    result[word] += 1

print result

Groupby oneliner:

from itertools import groupby

words = "apple banana apple strawberry banana lemon"
words = words.split()

result = dict((key, len(list(group))) for key, group in groupby(sorted(words)))
print result
user_171600

If you don't want to use the standard dictionary method (looping through the list incrementing the proper dict. key), you can try this:

>>> from itertools import groupby
>>> myList = words.split() # ['apple', 'banana', 'apple', 'strawberry', 'banana', 'lemon']
>>> [(k, len(list(g))) for k, g in groupby(sorted(myList))]
[('apple', 2), ('banana', 2), ('lemon', 1), ('strawberry', 1)]

It runs in O(n log n) time.

user_409060

Without defaultdict:

words = "apple banana apple strawberry banana lemon"
my_count = {}
for word in words.split():
    try: my_count[word] += 1
    except KeyError: my_count[word] = 1
user_912480 
user_input = list(input().split(' '))

for word in user_input:

    print('{} {}'.format(word, user_input.count(word)))
user_163369640 
words = "apple banana apple strawberry banana lemon"
w=words.split()
e=list(set(w))       
word_freqs = {}
for i in e:
    word_freqs[i]=w.count(i)
print(word_freqs)

Hope this helps!

user_84569910 
list = input()  # Providing user input passes multiple tests
text = list.split()

for word in text:
    freq = text.count(word) 
    print(word, freq)
user_113801620

I had a similar assignment on Zybook, this is the solution that worked for me.

def build_dictionary(words):
    counts = dict()
    for word in words:
        if word in counts:
             counts[word] += 1
        else:
             counts = 1
    return counts
if __name__ == '__main__':
    words = input().split()
    your_dictionary = build_dictionary(words)
    sorted_keys = sorted(your_dictionary.keys())
    for key in sorted_keys:
        print(key + ':' + str(your_dictionary[key]))
user_188314800
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