“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()

myString.count('a');

更多信息请点击这里

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。

使用数:

sentence = 'A man walked up to a door'
print(sentence.count('a'))
# 4

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'。

spam = 'have a nice day'
var = 'd'


def count(spam, var):
    found = 0
    for key in spam:
        if key == var:
            found += 1
    return found
count(spam, var)
print 'count %s is: %s ' %(var, count(spam, var))