Python 3

有两种方法可以做到这一点:

1)内置函数count()

sentence = 'Mary had a little lamb'
print(sentence.count('a'))`

2)不使用函数

sentence = 'Mary had a little lamb'    
count = 0

for i in sentence:
    if i == "a":
        count = count + 1

print(count)

“不使用计数查找字符串中需要的字符”方法。

import re

def count(s, ch):

   pass

def main():

   s = raw_input ("Enter strings what you like, for example, 'welcome': ")  

   ch = raw_input ("Enter you want count characters, but best result to find one character: " )

   print ( len (re.findall ( ch, s ) ) )

main()

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。

我不知道“最简单的”，但简单的理解可以做到:

>>> my_string = "Mary had a little lamb"
>>> sum(char == 'a' for char in my_string)
4

利用内置的和，生成器理解和bool是整数的子类的事实:如何乘字符等于'a'。

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)

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

print(count)