Python，字符串中字符的特定计数 [英] Python, specific count of chars in string

问题描述

我正在尝试计算python中字符串中出现的次数。我想输入一个二进制输入，例如 001101。然后计算1s，0s，11s，00s等的数目。我只希望它输出1 1和1 11s，并且不让它们单独计数，除非它们是单独存在的。

我也尝试实现此功能找到，但是我遇到了同样的问题。

任何帮助将不胜感激，谢谢。

解决方案

您可以使用 itertools.groupby 和 collections.Counter ：

 从itertools导入groupby 
从集合导入计数器
 
s ='001101011'
c = Counter（''。joinby（g）for _，g in groupby（s）
 
 c.get（'11'）
＃2 
 c.get（'1'）
＃1 
 c.get（'111'，0）＃使用默认值正确捕获计数0 
＃0 
  

这会将字符串分组为仅包含相等字符的子字符串，并对这些子字符串执行计数。

I am trying to count the number of occurrences in a string in python. I would like to take a binary input, say '001101'. Then count the number of 1s, 0s, 11s, 00s etc.

I have tried to implement this by using count, but this will output that there are 3 1s, when i only want it to output 1 1, and 1 11s and for it to not count them individually, unless they are on their own.

I have also tried to implement this with find, but i am having the same problem.

Any help would be appreciated, thanks.

解决方案

You can do the following, using itertools.groupby and collections.Counter:

from itertools import groupby
from collections import Counter

s = '001101011'
c = Counter(''.join(g) for _, g in groupby(s))

c.get('11')
# 2
c.get('1')
# 1
c.get('111', 0)  # use default value to capture count 0 properly
# 0

This groups the string into substrings consisting only of equal chars and performs the counting on those substrings.

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