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()

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)}

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。

正则表达式非常有用，如果你想要区分大小写(当然还有regex的所有功能)。

my_string = "Mary had a little lamb"
# simplest solution, using count, is case-sensitive
my_string.count("m")   # yields 1
import re
# case-sensitive with regex
len(re.findall("m", my_string))
# three ways to get case insensitivity - all yield 2
len(re.findall("(?i)m", my_string))
len(re.findall("m|M", my_string))
len(re.findall(re.compile("m",re.IGNORECASE), my_string))

请注意，regex版本的运行时间大约是它的十倍，只有当my_string非常长或代码处于深度循环中时，这才可能成为问题。

myString.count('a');

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