index string by character

Question

I need to determine the most frequent character of a list of strings by each index in Python.

Example

list1 = [‘one’, ‘two’, ‘twin’, ‘who’]

the most frequent character between all the strings at index 0 is ‘t’
the most frequent character between all the strings at index 1 is ‘w’
the most frequent character between all the strings at index 2 is ‘o’
the most frequent character between all the strings at index 3 is ‘n'(because there is only one character at the index 3

from that I want to create a final string ‘twon’ which shows the most frequent character of all the strings by each index

how can I perform this on python without importing any library?
thanks

Asked By: nuovo2000ita

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Source

Answer 1

from collections import Counter

list1 = ['one', 'two', 'twin', 'who']

idx =0  #t 
print(Counter(list(zip(*[i for i in list1 if len(i)>3]))[idx]).most_common()[0][0])

without importing any libraries

list1 = ['one', 'two', 'twin', 'who']

idx =0  #t 
l = list(list(zip(*[i for i in list1 if len(i)>3]))[idx])
print(max(set(l), key = l.count))

Answered By: Dmitriy Neledva

Answer 2

from itertools import zip_longest

list1 = ['one', 'two', 'twin', 'who']

chars = {}
for i, item in enumerate(zip_longest(*list1)):
    set1 = set(item)
    if None in set1:
        set1.remove(None)
    chars[i] = max(set1, key=item.count)

Without importing any library:

list1 = ['one', 'two', 'twin', 'who']

width = len(max(list1, key=len))

chars = {}

for i, item in enumerate(zip(*[s.ljust(width) for s in list1])):
    set1 = set(item)
    if ' ' in set1:
        set1.remove(' ')
    chars[i] = max(set1, key=item.count)

Output:

chars

{
  0: 't',
  1: 'w',
  2: 'o', 
  3: 'n'
}

"".join(chars.values())

'twon'

Answered By: Hamid Rasti

index string by character

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