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

正则表达式?

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

其他回答

正则表达式?

import re
my_string = "Mary had a little lamb"
len(re.findall("a", my_string))
a = 'have a nice day'
symbol = 'abcdefghijklmnopqrstuvwxyz'
for key in symbol:
    print(key, a.count(key))

最简单的方法是一行代码:

'Mary had a little lamb'.count("a")

但是如果你想用这个也可以:

sentence ='Mary had a little lamb'
   count=0;
    for letter in sentence :
        if letter=="a":
            count+=1
    print (count)

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。

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)}