如何基于分隔符将列表拆分为子列表,类似于str.split()?

一个简单的生成器适用于您问题中的所有情况:

def split(sequence, sep):
    chunk = []
    for val in sequence:
        if val == sep:
            yield chunk
            chunk = []
        else:
            chunk.append(val)
    yield chunk

我的第一个Python程序:)

from pprint import pprint
my_array = ["a", "SEP", "SEP", "SEP"]
my_temp = []
my_final = []
for item in my_array:
  if item != "SEP":
    my_temp.append(item)
  else:
    my_final.append(my_temp);
    my_temp = []
pprint(my_final);

我不确定是否有一个简单的答案 itertools.groupby 解决方案,但这里有一个迭代方法应该可以工作:

def mySplit(iterable, sep):
    output = []
    sepcount = 0
    current_output = []
    for i, elem in enumerate(iterable):
        if elem != sep:
            sepcount = 0
            current_output.append(elem)
            if (i==(len(iterable)-1)):
                output.append(current_output)
        else:
            if current_output: 
                output.append(current_output)
                current_output = []

            sepcount+=1

            if (i==0) or (sepcount > 1):
                output.append([])
            if (i==(len(iterable)-1)):
                output.append([])

    return output

测试您的示例:

testLists = [
    ['a', 'SEP', 'b', 'c', 'SEP', 'SEP', 'd'],
    ["a", "SEP", "SEP", "SEP"],
    ["SEP"],
    ["a", "b", "c"]
]

for tl in testLists:
    print(mySplit(tl, sep="SEP"))
#[['a'], ['b', 'c'], [], ['d']]
#[['a'], [], [], []]
#[[], []]
#[['a', 'b', 'c']]

str.split(sep) :

for tl in testLists:
    print("".join(tl).split("SEP"))
#['a', 'bc', '', 'd']
#['a', '', '', '']
#['', '']
#['abc']

for tl in testLists:
    print([list(x) for x in "".join(tl).split("SEP")])
#[['a'], ['b', 'c'], [], ['d']]
#[['a'], [], [], []]
#[[], []]
#[['a', 'b', 'c']]

但是 mySplit() 函数更一般。

对于 list tuple 对象您可以使用以下各项:

def split(seq, sep):
    start, stop = 0, -1
    while start < len(seq):
        try:
            stop = seq.index(sep, start)
        except ValueError:
            yield seq[start:]
            break
        yield seq[start:stop]
        start = stop + 1
    else:
        if stop == len(seq) - 1:
            yield []

我不会用发电机,但它很快。

你可以用 itertools.takewhile :

def split(seq, sep):
    seq, peek = iter(seq), sep
    while True:
        try:
            peek = next(seq)
        except StopIteration:
            break
        yield list(it.takewhile(sep.__ne__, it.chain((peek,), seq)))
    if peek == sep:
        yield []

这个 it.chain seq 他筋疲力尽了。注意,如果需要,使用这种方法很容易生成生成器而不是列表。

l = ['a', 'SEP', 'b', 'c', 'SEP', 'SEP', 'd']

def sublist_with_words(word, search_list):
    res = []
    for i in range(search_list.count(word)):
        index = search_list.index(word)
        res.append(search_list[:index])
        search_list = search_list[index+1:]
    res.append(search_list)
    return res

当我尝试您提供的案例时:

print(sublist_with_words(word = 'SEP', search_list=l))
print(sublist_with_words(word = 'SEP', search_list=['a', 'b', 'c']))
print(sublist_with_words(word = 'SEP', search_list=['SEP']))

输出为:

[['a'], ['b', 'c'], [], ['d']]
[['a', 'b', 'c']]
[[], []]

itertools.takewhile @客人的方法简化了:

def split(seq, sep):
    from itertools import takewhile
    iterator = iter(seq)
    while subseq := list(takewhile(lambda x: x != sep, iterator)):
        yield subseq

请注意,它在第一个空子序列返回。

import re

def split_list(nums, n):
    nums_str = str(nums)
    splits = nums_str.split(f"{n},")

    patc = re.compile(r"\d+")
    group = []
    for part in splits:
        group.append([int(v) for v in patc.findall(part)])

    return group

if __name__ == "__main__":
    l = [1, 2, 3, 4, 3, 6, 7, 3, 8, 9, 10]
    n = 3
    split_l = split_list(l, n)
    assert split_l == [[1, 2], [4], [6, 7], [8, 9, 10]]