Python 对按顺序出现的相同字符进行计数

我试图以文件格式返回最常用的x字符。例如，此函数应返回[4,1,2]

例如，如果字符串是XXXXX YXYXXX'，它应该返回[5，2，3]

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我的函数没有返回正确的列表。任何帮助都将不胜感激。谢谢，您没有初始化序列之间的total值，因此它会继续计数

def count_squences(string):

    i= 0 
    total = 0
    total_char_list = []

    while i < len(string):
        print(string[i])

        if string[i] == "x":
            total += 1
        if string[i] == "y":

            total_char_list.append(total)
            total = 0

        i = i + 1

    return total_char_list


print(count_squences("xxxxyyxyxx"))

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当您遇到y字符时，您不会重置计数器，并且如果在您找到y字符时至少有一个x字符计数，则您只应将y字符附加到total_char_列表中。y字符也可以重复：

my_str = "xxxxyyxyxx"
[len(z) for z in re.split("y+", my_str)]

接下来，您确实需要使用for循环来循环序列。您可以通过以下方式获得单个字符：

while ...:
    # ...

if total:
    # x characters at the end
    total_char_list.append(total)

您可以通过以下方式加快速度：

groupby将一个iterable输入（如字符串）划分为每个组的独立迭代器，其中组定义为具有相同keyvalue结果的任何连续值。默认键函数只返回值，因此groupbystring为您提供相同的连续字符组。char是重复的字符，组中u的sum1采用迭代器的长度

然后可以使其更通用，并计算所有组：

from itertools import groupby

def count_squences(string):
    return [sum(1 for _ in group) for char, group in groupby(string) if char == 'x']

正则表达式也可以这样做：

def count_all_sequences(string):
    counts = {}
    for char, group in groupby(string):
        counts.setdefault(char, []).append(sum(1 for _ in group))
    return counts

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编辑函数格式：

import re

def count_all_sequences(string):
    counts = {}
    # (.)(\1*) finds repeated characters; (.) matching one, \1 matching the same
    # This gives us (first, rest) tuples, so len(rest) + 1 is the total length
    for char, group in re.findall(r'(.)(\1*)', string):
        counts.setdefault(char, []).append(len(group) + 1)
    return counts

---

返回[4,1,2]

什么不起作用？def count_xs:return[lent for t in re.findall'x+'，s]也将完成您想要的功能。当有连续的y字符时，这将附加多个零。感谢@MartijnPieters，我修复了它+另一个修复

from itertools import groupby

def count_squences(string):
    return [sum(1 for _ in group) for char, group in groupby(string) if char == 'x']

def count_all_sequences(string):
    counts = {}
    for char, group in groupby(string):
        counts.setdefault(char, []).append(sum(1 for _ in group))
    return counts

import re

def count_all_sequences(string):
    counts = {}
    # (.)(\1*) finds repeated characters; (.) matching one, \1 matching the same
    # This gives us (first, rest) tuples, so len(rest) + 1 is the total length
    for char, group in re.findall(r'(.)(\1*)', string):
        counts.setdefault(char, []).append(len(group) + 1)
    return counts

def count_sequences(string):
    return [len(x) for x in re.findall(r"x+", string)]

count_sequences("xxxxyyxyxx")