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Comments (3)

Trinityyi avatar Trinityyi commented on August 25, 2024 1

Thanks for the suggestions. Both seem valid. I'm assigning this to you, you can open a PR whenever you have the time.

from 30-seconds-of-python.

JahnaviChitta avatar JahnaviChitta commented on August 25, 2024

Hi, here is a different approach to count an element's occurrence in a list using list comprehension and lambda functions.

lst = [1, 2, 3, 4, 9, 2, 2, 1, "apple", "ball", "apple"]
elementToBeSearched = 2
print(len(list(filter(lambda x : x == elementToBeSearched, l))))

from 30-seconds-of-python.

vbrozik avatar vbrozik commented on August 25, 2024

@veronicaguo the collections module contains the class Counter intended exactly for this purpose. You can use
count = collections.Counter()
instead of
count = collections.defaultdict(int)

or you can even initialize the counter directly from the iterable and the function will consist of a single expression:

import collections

def count_by(iter1, map1=lambda x: x):
  return dict(collections.Counter(map(map1, iter1)))

@JahnaviChitta There is no need to first create the list and then to count it (it could be really bad if it had millions of items). This code is counting the items directly:
sum(1 for x in filter(lambda x : x == elementToBeSearched, lst))

...and Python generator expressions are much more powerful. The filtering can be part of the same expression:
sum(1 for x in lst if x == elementToBeSearched)

...but anyway for the solution to be efficient in most cases it has to count the all the values in a single pass like do the original functions above.

from 30-seconds-of-python.

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