CODEWITHSUNDEEP
python
90abyss
90abyss

Reputation: 7337

Python: Find element in the list which has frequency = 2

Let's say I have a list:

Input: ids = [123, 456, 123]

I need to find the element whose frequency is 2.

Output: 123

Best way to do this in python?

Upvotes: 0

Views: 188

Answers (3)

Iron Fist
Iron Fist

Reputation: 10951

This is can be done also with a built-in method, filter to filter out only those elements who shall meet your requirement, but better do it on a set rather than list to not repeat it for the same element:

>>> ids = [123, 456, 123]
>>> s = set(ids)
>>> filter(lambda x:ids.count(x)==2, s)
>>> list(filter(lambda x:ids.count(x)==2, s)) #If Python3
[123]

Upvotes: 0

TigerhawkT3
TigerhawkT3

Reputation: 49320

Make a set and count each item (or make a Counter or defaultdict), then find which one has a count of 2.

ids = [123, 456, 123]
freq = 2
set_ids = set(ids)
counts = {item:ids.count(item) for item in set_ids}
for item in counts:
    if counts[item] == freq:
        print(item)

 

import collections
ids = [123, 456, 123]
freq = 2
counted_ids = collections.Counter(ids)
for item in counted_ids:
    if counted_ids[item] == freq:
        print(item)

 

import collections
ids = [123, 456, 123]
freq = 2
counts = collections.defaultdict(int)
for item in ids:
    counts[item] += 1
for item in counts:
    if counts[item] == freq:
        print(item)

Upvotes: 0

wim
wim

Reputation: 362557

from collections import Counter
counter = Counter(ids)
[k for k,v in counter.items() if v == 2]

Upvotes: 7

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