计算字符串中字符的出现次数

Question

计算字符串中字符出现次数的最简单方法是什么？

例如计算'a'出现的次数'Mary had a little lamb'

Answer 1

str.count(sub [,start [,end]])

返回sub范围中子字符串的非重叠出现次数[start, end].可选参数start,end并以切片表示法解释.

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

Answer 2

你可以使用count():

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

Answer 3

正如其他答案所说,使用字符串方法count()可能是最简单的,但如果你经常这样做,请查看collections.Counter:

from collections import Counter
my_str = "Mary had a little lamb"
counter = Counter(my_str)
print counter['a']

Answer 4

正则表达式可能？

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

Answer 5

myString.count('a');

更多信息在这里

Answer 6

"aabc".count("a")

Answer 7

str.count(a)是计算字符串中单个字符的最佳解决方案.但是如果你需要计算更多的字符,你必须读取整个字符串的次数与你想要计算的字符数一样多.

这项工作的更好方法是:

from collections import defaultdict

text = 'Mary had a little lamb'
chars = defaultdict(int)

for char in text:
    chars[char] += 1

因此,您将拥有一个dict,它返回字符串中每个字母的出现次数以及0它是否不存在.

>>>chars['a']
4
>>>chars['x']
0

对于不区分大小写的计数器,您可以通过子类化来覆盖mutator和accessor方法defaultdict(基类'是只读的):

class CICounter(defaultdict):
    def __getitem__(self, k):
        return super().__getitem__(k.lower())

    def __setitem__(self, k, v):
        super().__setitem__(k.lower(), v)


chars = CICounter(int)

for char in text:
    chars[char] += 1

>>>chars['a']
4
>>>chars['M']
2
>>>chars['x']
0

Answer 8

如果你想要不区分大小写(当然还有正则表达式的所有功能),正则表达式非常有用.

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,正则表达式版本的运行时间大约为十倍,这可能仅在my_string非常长或代码在深层循环内时才会出现问题.

Answer 9

这个简单直接的功能可能会有所帮助:

def check_freq(str):
    freq = {}
    for c in str:
       freq[c] = str.count(c)
    return freq

check_freq("abbabcbdbabdbdbabababcbcbab")
{'a': 7, 'b': 14, 'c': 3, 'd': 3}

Answer 10

使用次数：

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

Answer 11

我不知道“最简单”，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 sum、生成器理解以及 bool 是整数的子类这一事实：字符等于“a”的次数有多少。

Answer 12

a = 'have a nice day'
symbol = 'abcdefghijklmnopqrstuvwxyz'
for key in symbol:
    print key, a.count(key)