如何将列表拆分为以分隔符开头的子列表？ [关闭]

Question

• 我想将列表拆分为以分隔符开头的子列表
  • 必须保留分隔符
  • 分隔符必须是每个子列表的第一个字符

例子：

delimiter = "x" 
input = ["x","a","x","x",1,2,3,"a","a","x","e"]
output = [["x","a"], ["x"], ["x",1,2,3,"a","a"], ["x","e"]]

• 除了以下解决方案，请参阅Python splitting a list based on a delimiter word
  • 问题类似，但预期输出略有不同。例如，那里的最佳解决方案在索引 0 处返回并清空列表。

Answer 1

第 1 步：查找列表中出现 variable 的索引：

idx = [ix for ix, val in enumerate(input) if val==variable]

第 2 步：使用列表切片生成子列表：

res = [input[i:j] for i,j in zip(idx, idx[1:]+[len(input)])]

输出

print(res)
# [['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

Answer 2

您可以使用以下函数来获得所需的输出。

试试这个：

def splitList(inputList, delim):
    finalList = []
    chunk = []
    for val in inputList:
        if val == delim:
            finalList.append(chunk)
            chunk = [delim]
        else:
            chunk.append(val)
    finalList.append(chunk)        
    return finalList[1:]


my_input_list = ["x","a","x","x",1,2,3,"a","a","x","e"]
my_output = splitList(my_input_list, "x")    #just call the function with input_list and delimiter to split onto.


print(my_output)

>>> [["x","a"],["x"],["x",1,2,3,"a","a"],["x","e"]]

Answer 3

使用堆栈

variable = "x" 
input1 = ["x","a","x","x",1,2,3,"a","a","x","e"]

sol = []
tmp = []
for char in input1:
    if char==variable and tmp:
        sol.append(tmp)
        tmp = [char]
    else:
        tmp.append(char)
if tmp:
    sol.append(tmp)
    
print(sol)

输出

 [['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

Answer 4

使用itertools.groupby

例如：

from itertools import groupby

variable = "x" 
data = ["x","a","x","x",1,2,3,"a","a","x","e"]
output = []
for k, v in groupby(data, lambda x: x==variable):   #--->[(True, ['x']), (False, ['a']), (True, ['x', 'x']), (False, [1, 2, 3, 'a', 'a']), (True, ['x']), (False, ['e'])]
    v = list(v)
    if k:
        for i in v:
            output.append([i])
    else:
        output[-1].extend(v)
print(output)

输出：

[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

Answer 5

• input是python的方法，不要用它作为变量名
• 如果列表中的第一个字符不是分隔符，此解决方案也可以使用
  • 给定：['a', 'b', 'c', 'x', 'a', 'x', 'x', 1, 2, 3, 'a', 'a', 'x', 'e', 'x']
  • 返回：[['a', 'b', 'c'], ['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e'], ['x']]

from typing import List  # for type annotations


def sublist_by_delimiter(flat_list: list, delimiter: str) -> List[list]:
    result = list()  # main list
    chunk = list()  # inner list to 
    len_flat_list = len(flat_list)
    for i, v in enumerate(flat_list, 1):  # iterate through t, begin enumerating at 1
        if (v == delimiter) & (i != 1):  # except for the first delimiter 
            result.append(chunk)  # append chunk to result
            chunk = [v]  # create new chunk beginning with v
            if i == len_flat_list:  # if the last value in the list is delimiter
                result.append(chunk)
        elif (i == len_flat_list):  # for the last list in lines
            chunk.append(v)  # append that line to inner
            result.append(chunk)  # append chunk to result
        else:
            chunk.append(v)  # append each v to chunk where v isn't delimiter
            
    return result
            

t = ['x', 'a', 'x', 'x', 1, 2, 3, 'a', 'a', 'x', 'e', 'x']  # an extra x has been added at the end for testing
delim = 'x'
sublist_by_delimiter(t, delim)
[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e'], ['x']]

使用collections.defaultdict

• 从 python 3.7 开始，dicts 保证根据插入进行排序，因此dict.values() 将在返回时进行排序。
• 此解决方案对于任何想要拥有片段字典的人来说都是一个不错的选择
  • 将return list(dd.values()) 更改为return dd
• 如果列表中的第一个字符不是分隔符，此解决方案也可以使用
  • 给定：['a', 'b', 'c', 'x', 'a', 'x', 'x', 1, 2, 3, 'a', 'a', 'x', 'e', 'x']
  • 返回：[['a', 'b', 'c'], ['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e'], ['x']]

from collection import defaultdict

def sublist_by_delimiter(flat_list: list, delimiter: str) -> List[list]:
    dd = defaultdict(list)
    counter = 0
    for v in flat_list:
        if v == delimiter:
            counter += 1
            dd[counter].append(v)
        else:
            dd[counter].append(v)
    return list(dd.values())


sublist_by_delimiter(t, 'x')
[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e'], ['x']]

使用dict

• 3.61 s ± 9.85 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 用于 25M 元素列表
  • 与defaultdict:3.74 s ± 53.7 ms
• 如前所述，如果第一个字符不是分隔符，此解决方案将生成 KeyError。

def sublist_by_delimiter(flat_list: list, delimiter: str) -> List[list]:
    dd = dict(list)
    counter = 0
    for v in flat_list:
        if v == delimiter:
            counter += 1
            if dd.get(counter) == None:
                dd[counter] =  [v]
        else:
            dd[counter].append(v)
    return list(dd.values())

Answer 6

您可以使用re 尝试这种简单的方法：

import re
variable = "x" 
inp = ["x","a","x","x",1,2,3,"a","a","x","e"]
st=''.join(list(map(str,inp)))
regex=f'({variable}[^{variable}]*)'
ls=[[k if not k.isdigit() else int(k) for k in l] for l in re.findall(regex,st)]
print(ls)

输出：

ls
[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

Answer 7

你可以使用双指针，试试这个：

var = "x"
l = ["x", "a", "x", "x", 1, 2, 3, "a", "a", "x", "e"]
r = []
left, right = 0, 0
while left < len(l)-1:
    left = right
    if l[left] == var:
        right += 1
        if right < len(l)-1:
            while l[right] != "x":
                right += 1
        else:
            right = len(l)
        r.append(l[left:right])
    left += 1
print(r)

Answer 8

试试：

indices = [i for i, v in enumerate(input) if v =='x']
res = [input[i:j] for i, j in zip([0]+indices, indices+[None])][1:]
res

---

[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

使用itertools's groupby：

class get_indices(object):
    def __init__(self, value):
        self.value = value
        self.i = 0

    def __call__(self, value):  # For masking
        self.i += (value == self.value)
        return self.i

res = [list(g) for _, g in groupby(input, key=get_indices('x'))]
res

---

[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e']]

Answer 9

这篇文章是 %%timeit 对截至本文发布时可用的解决方案的比较

• 感谢所有回答的人，感谢您的贡献。有一些很好的答案。我今天早些时候（2020-07-01）正在处理一个类似的问题，但是有一个列表列表，其中分隔符在间歇列表中的索引 0 处。我认为解决这个问题并非易事。
• 既然有这么多的解决方案，我认为比较会有所帮助
• 样本数据列表l有25_000_000个元素，第一个值为'x'
• 此测试中的所有函数都正确返回[['x', 'a'], ['x'], ['x', 1, 2, 3, 'a', 'a'], ['x', 'e'], ['x']]，对于['x', 'a', 'x', 'x', 1, 2, 3, 'a', 'a', 'x', 'e', 'x']

测试数据

import random

random.seed(25)
l = [random.choice(['x', 'a', 'e', 1, 2, 3]) for _ in range(25000000)]
l[0] = 'x'
print(f'Length of list l: {len(l)}')
print(f'First 10 values of list l: {l[:10]}')

Length of list l: 25000000
First 10 values of list l: ['x', 'x', 'a', 'e', 3, 1, 'x', 'e', 'x', 'e']

%%timeit 测试

%%timeit
pygirl(l)
2.75 s ± 14.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
pygirl2(l)
9 s ± 79.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
trenton(l)
4.78 s ± 36.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
trenton2(l)
3.74 s ± 53.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
trenton3(l)
3.6 s ± 16.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
mrnobody33(l)
9.68 s ± 37.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
rakesh(l)
5.78 s ± 91.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
ansfourtytwo(l)
2.69 s ± 26 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
sahasrara62(l)
2.63 s ± 27.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
prashant(l)
2.64 s ± 9.89 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%%timeit
kevin(l)
# results in
---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-11-270e2f7daf8d> in kevin(test_list)
    101             right += 1
    102             if right < len(test_list)-1:
--> 103                 while test_list[right] != "x":
    104                     right += 1
    105             else:

IndexError: list index out of range

功能

def pygirl(test_list):
    indices = [i for i, v in enumerate(test_list) if v =='x']
    return [test_list[i:j] for i, j in zip([0]+indices, indices+[None])][1:]


class get_indices(object):
    def __init__(self, value):
        self.value = value
        self.i = 0

    def __call__(self, value):  # For masking
        self.i += (value == self.value)
        return self.i
    
    
def pygirl2(test_list):
    return [list(g) for _, g in groupby(test_list, key=get_indices('x'))]


def trenton(test_list):
    result = list()
    chunk = list()
    delimiter = 'x'
    len_test_list = len(test_list)
    for i, v in enumerate(test_list, 1):
        if (v == delimiter) & (i != 1):
            result.append(chunk)
            chunk = [v]
            if i == len_test_list:
                result.append(chunk)
        elif (i == len_test_list):
            chunk.append(v)
            result.append(chunk)
        else:
            chunk.append(v)       
    return result


def trenton2(test_list):
    dd = defaultdict(list)  # defaultdict
    delim = 'x'
    counter = 0
    for v in test_list:
        if v == delim:
            counter += 1
            dd[counter].append(v)
        else:
            dd[counter].append(v)
    return list(dd.values())


def trenton3(test_list):
    dd = dict()  # regular dict
    delim = 'x'
    counter = 0
    for v in test_list:
        if v == delim:
            counter += 1
            if dd.get(counter) == None:
                dd[counter] = [v]
        else:
            dd[counter].append(v)
        
    return list(dd.values())


def mrnobody33(test_list):
    variable = "x" 
    st=''.join(list(map(str,test_list)))
    regex=f'({variable}[^{variable}]*)'
    return [[k if not k.isdigit() else int(k) for k in v] for v in re.findall(regex,st)]


def rakesh(test_list):
    variable = "x" 
    output = []
    for k, v in groupby(test_list, lambda x: x==variable):
        v = list(v)
        if k:
            for i in v:
                output.append([i])
        else:
            output[-1].extend(v)
    return output


def ansfourtytwo(test_list):
    variable = 'x'
    idx = [ix for ix, val in enumerate(test_list) if val==variable]
    return [test_list[i:j] for i,j in zip(idx, idx[1:]+[len(test_list)])]


def sahasrara62(test_list):
    variable = "x" 
    sol = []
    tmp = []
    for char in test_list:
        if char==variable and tmp:
            sol.append(tmp)
            tmp = [char]
        else:
            tmp.append(char)
    if tmp:
        sol.append(tmp)
    return sol


def prashant(test_list):
    delim = 'x'
    finalList = []
    chunk = []
    for val in test_list:
        if val == delim:
            finalList.append(chunk)
            chunk = [delim]
        else:
            chunk.append(val)
    finalList.append(chunk)        
    return finalList[1:]


def kevin(test_list):
    var = "x"
    r = []
    left, right = 0, 0
    while left < len(test_list)-1:
        left = right
        if test_list[left] == var:
            right += 1
            if right < len(test_list)-1:
                while test_list[right] != "x":
                    right += 1
            else:
                right = len(test_list)
            r.append(test_list[left:right])
        left += 1
    return r