如何计算字符串中字符出现的次数?

如。“a”在“Mary had a little lamb”中出现了4次。


当前回答

使用数:

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

其他回答

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。

正则表达式?

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

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'。

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。

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

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)