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Accepted
hellog888
Asked:2022-09-08 20:49:24 +0000 UTC2022-09-08 20:49:24 +0000 UTC

单词中出现频率最高的字母。Python

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from collections import Counter
import re
def checkio(text: str) -> str:
    text1 = re.findall('[^\W]', text.lower())
    a = (''.join(text1))
    text2 = re.findall('[a-z]', a)
    a1 = (''.join(text2))
    return(Counter(a1).most_common(1)[0][0])

print(checkio("Hello World!"))          # == "l", "Hello test"
print(checkio("How do you do?"))        # == "o", "O is most wanted"
print(checkio("One"))                   # == "e", "All letter only once."
print(checkio("Oops!"))                 # == "o", "Don't forget about lower case."
print(checkio("AAaooo!!!!"))            # == "a", "Only letters."
print(checkio("abe"))                   # == "a", "The First."
print(checkio("a" * 9000 + "b" * 1000)) # == "a", "Long."

如果“One”应该是“e”，而“abe”应该是“a”，如何解决这个问题？

这里问了同样的问题：https ://ru.stackoverflow.com/questions/1036053/Most-frequent-letter-in-text

python

1 个回答

  1. Best Answer

    据我了解，您不仅需要按频率对结果进行排序，还需要按字母表中哪个字母排在第一位：

    return sorted(Counter(a1).items(), key=lambda x: (-x[1], x[0]))[0][0]

    那么结果就是你想要的：

    l
o
e
o
a
a
a

    PS 而是不能排序，直接取最小值，这样更优一点。但也使用相同的键：

    return min(Counter(a1).items(), key=lambda x: (-x[1], x[0]))[0]
    • 2

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