CODEWITHSUNDEEP
pythoncountfrequencycounting
Daniyar
Daniyar

Reputation: 1720

Item frequency count in Python

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?

Upvotes: 70

Views: 156214

Answers (15)

jesse
jesse

Reputation: 1

Here is my solution. No imports, just a simple nested loop.

words = input().split(" ")

for word in words:

    word_count = 0
    for word2 in words:

       if word2.lower() == word.lower():
           word_count += 1
    print(f'{word} {word_count}')

Upvotes: 0

B1029
B1029

Reputation: 1

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]))

Upvotes: 0

Gadi
Gadi

Reputation: 1152

Use reduce() to convert the list to a single dict.

from functools import reduce

words = "apple banana apple strawberry banana lemon"
reduce( lambda d, c: d.update([(c, d.get(c,0)+1)]) or d, words.split(), {})

returns

{'strawberry': 1, 'lemon': 1, 'apple': 2, 'banana': 2}

Upvotes: 0

PanamaPHat
PanamaPHat

Reputation: 51

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

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

Upvotes: 0

dB_19
dB_19

Reputation: 21

user_input = list(input().split(' '))

for word in user_input:

    print('{} {}'.format(word, user_input.count(word)))

Upvotes: 2

Varun Shaandhesh
Varun Shaandhesh

Reputation: 79

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!

Upvotes: 1

nosklo
nosklo

Reputation: 223172

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

Upvotes: 11

sykloid
sykloid

Reputation: 101416

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})

Upvotes: 152

javaidiot
javaidiot

Reputation: 1

I happened to work on some Spark exercise, here is my solution.

tokens = ['quick', 'brown', 'fox', 'jumps', 'lazy', 'dog']

print {n: float(tokens.count(n))/float(len(tokens)) for n in tokens}

**#output of the above **

{'brown': 0.16666666666666666, 'lazy': 0.16666666666666666, 'jumps': 0.16666666666666666, 'fox': 0.16666666666666666, 'dog': 0.16666666666666666, 'quick': 0.16666666666666666}

Upvotes: 0

Prabhu S
Prabhu S

Reputation: 34

The answer below takes some extra cycles, but it is another method

def func(tup):
    return tup[-1]


def print_words(filename):
    f = open("small.txt",'r')
    whole_content = (f.read()).lower()
    print whole_content
    list_content = whole_content.split()
    dict = {}
    for one_word in list_content:
        dict[one_word] = 0
    for one_word in list_content:
        dict[one_word] += 1
    print dict.items()
    print sorted(dict.items(),key=func)

Upvotes: -1

Antonio
Antonio

Reputation: 11

Can't you just use count?

words = 'the quick brown fox jumps over the lazy gray dog'
words.count('z')
#output: 1

Upvotes: 0

hopla
hopla

Reputation: 3382

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.

Upvotes: 12

Thomas Weigel
Thomas Weigel

Reputation: 159

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

Upvotes: 3

Kenan Banks
Kenan Banks

Reputation: 212208

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).

Upvotes: 95

Nick Presta
Nick Presta

Reputation: 28715

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.

Upvotes: 7

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