使用数:

sentence = 'A man walked up to a door'
print(sentence.count('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)}

正则表达式?

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

a = "I walked today,"
c=['d','e','f']
count=0
for i in a:
    if str(i) in c:
        count+=1

print(count)

你可以使用.count():

>>> 'Mary had a little lamb'.count('a')
4

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。