如何在字符串列表中找到主要字母

Question

I want to check for each position in the string what is the character that appears most often on that position. If there are more of the same frequency, keep the first one. All strings in the list are guaranteed to be of identical length!!!

I tried the following way:

print(max(((letter, strings.count(letter)) for letter in strings), key=lambda x:[1])[0])

But I get: mistul or qagic

And I can not figure out what's wrong with my code.

My list of strings looks like this:

Input: strings = ['mistul', 'aidteh', 'mhfjtr', 'zxcjer']

Output: mister

Explanation: On the first position, m appears twice. Second, i appears twice twice. Third, there is no predominant character, so we chose the first, that is, s . On the fourth position, we have t twice and j twice, but you see first t , so we stay with him, on the fifth position we have e twice and the last r twice.

Another examples:

Input: ['qagic', 'cafbk', 'twggl', 'kaqtc', 'iisih', 'mbpzu', 'pbghn', 'mzsev', 'saqbl', 'myead']

Output: magic

Input: ['sacbkt', 'tnqaex', 'vhcrhl', 'obotnq', 'vevleg', 'rljnlv', 'jdcjrk', 'zuwtee', 'xycbvm', 'szgczt', 'imhepi', 'febybq', 'pqkdfg', 'swwlds', 'ecmrut', 'buwruy', 'icjwet', 'gebgbq', 'djtfzr', 'uenleo']

Expected Output: secret

Some help?

Answer 1

Finally a use case for zip() :-)

If you like cryptic code, it could even be done in one statement:

def solve(strings):
    return ''.join([max([(letter, letters.count(letter)) for letter in letters], key=lambda x: x[1])[0] for letters in zip(*strings)])

But I prefer a more readable version:

def solve(strings):
    result = ''
    # "zip" the strings, so in the first iteration `letters` would be a list
    # containing the first letter of each word, the second iteration it would
    # be a list of all second letters of each word, and so on...
    for letters in zip(*strings):
        # Create a list of (letter, count) pairs:
        letter_counts = [(letter, letters.count(letter)) for letter in letters]
        # Get the first letter with the highest count, and append it to result:
        result += max(letter_counts, key=lambda x: x[1])[0]
    return result

# Test function with input data from question:
assert solve(['mistul', 'aidteh', 'mhfjtr', 'zxcjer']) == 'mister'
assert solve(['qagic', 'cafbk', 'twggl', 'kaqtc', 'iisih', 'mbpzu', 'pbghn',
              'mzsev', 'saqbl', 'myead']) == 'magic'
assert solve(['sacbkt', 'tnqaex', 'vhcrhl', 'obotnq', 'vevleg', 'rljnlv',
              'jdcjrk', 'zuwtee', 'xycbvm', 'szgczt', 'imhepi', 'febybq',
              'pqkdfg', 'swwlds', 'ecmrut', 'buwruy', 'icjwet', 'gebgbq',
              'djtfzr', 'uenleo']) == 'secret'

UPDATE

@dun suggested a smarter way of using the max() function, which makes the one-liner actually quite readable :-)

def solve(strings):
    return ''.join([max(letters, key=letters.count) for letters in zip(*strings)])

Answer 2

Using collections.Counter() is a nice strategy here. Here's one way to do it:

from collections import Counter

def most_freq_at_index(strings, idx):
  chars = [s[idx] for s in strings]
  char_counts = Counter(chars)
  return char_counts.most_common(n=1)[0][0]

strings = ['qagic', 'cafbk', 'twggl', 'kaqtc', 'iisih', 
           'mbpzu', 'pbghn', 'mzsev', 'saqbl', 'myead']

result = ''.join(most_freq_at_index(strings, idx) for idx in range(5))
print(result) 
## 'magic'

Answer 3

If you want something more manual without the magic of Python libraries you can do something like this:

def f(strings):
    dic = {}
    for string in strings:
        for i in range(len(string)):
            word_dic = dic.get(i, { string[i]: 0 })
            word_dic[string[i]] = word_dic.get(string[i], 0) + 1
            dic[i] = word_dic
    largest_string = max(strings, key = len)
    result = ""
    for i in range(len(largest_string)):
        result += max(dic[i], key = lambda x : dic[i][x])
    return result

strings = ['qagic', 'cafbk', 'twggl', 'kaqtc', 'iisih', 'mbpzu', 'pbghn', 'mzsev', 'saqbl', 'myead']
f(strings)
'magic'