如何从现有列表创建字典？ 键值必须是该键的重复次数

Question

I'm trying to create a "most wanted letter" program which takes a string as input and output should be the most repeated letter in that string. Currently, I can't figure out how to create a dictionary (eg dict.items():[("a", 2), ("b", 5)...] ).

def checkio(text: str) -> str:

    input_text = text
    lowercase_input = str.lower(input_text)
    x = list('')
    for i in lowercase_input:
        x.append(i)

Answer 1

You may use collections.Counter to do it directly.
Consider the following example.

from collections import Counter

s = "example"
c = Counter(s)
# Counter({'e': 2, 'x': 1, 'a': 1, 'm': 1, 'p': 1, 'l': 1})

You can also get the most common letter with

c.most_common(1) #[('e', 2)]

Note also that in case the input is a sentence you may want to avoid whitespace and in this case you can do something like s = str.join("",s.split()) .

Answer 2

Without any imports, you can use a regular dictionary.

def checkio(text: str) -> str:
    lowercase_input = text.lower()
    x = {}
    for i in lowercase_input:
        x[i] = x.get(i, 0) + 1
    return max(x.items(), key=lambda x: x[1])

key, value = checkio('asdafsasaaaa')  # ('a', 7)

Explanation

dict.get has an optional second parameter, which we set to 0 . This means if the key does not exist in the dictionary x.get(i, 0) will return 0 . max has an optional key parameter which accepts an anonymous ( lambda ) function. Since dict.items() returns an iterable of (key, value) pairs, we can calculate the maximum by looking at the value component (1st index).

Performance

This is inefficient versus Counter + most_common as shown by @abc . Specifically, most_common uses heapq to reduce time complexity. See also Python collections.Counter: most_common complexity .