不超过这个IMHO -你可以添加上或下的方法

def count_letter_in_str(string,letter):
    return string.count(letter)

python - 3. x:

"aabc".count("a")

Str.count (sub[， start[， end]]) 返回子字符串sub在范围[start, end]中不重叠出现的次数。可选参数start和end被解释为片表示法。

不使用Counter()， count和regex获得所有字符计数的另一种方法

counts_dict = {}
for c in list(sentence):
  if c not in counts_dict:
    counts_dict[c] = 0
  counts_dict[c] += 1

for key, value in counts_dict.items():
    print(key, value)

正则表达式?

import re
my_string = "Mary had a little lamb"
len(re.findall("a", my_string))

要获得所有字母的计数，请使用集合。计数器:

>>> from collections import Counter
>>> counter = Counter("Mary had a little lamb")
>>> counter['a']
4

这个简单直接的函数可能会有帮助:

def check_freq(x):
    freq = {}
    for c in set(x):
       freq[c] = x.count(c)
    return freq

check_freq("abbabcbdbabdbdbabababcbcbab")
{'a': 7, 'b': 14, 'c': 3, 'd': 3}

如果需要理解:

def check_freq(x):
    return {c: x.count(c) for c in set(x)}