如何使用标记元素拆分列表？

Question

我正在尝试使用一些元素作为标记在Python中划分列表。例如，请考虑列表：

["marker1", "elem1", "elem2", "marker2", "elem3"]

我希望将其分为2个子列表：

[["marker1", "elem1", "elem2"], ["marker2", "elem3"]]

如果第一个元素不是标记，则将标记之前的元素视为单独的sublist：

# From:
["elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5"]
# To:
[["elem1", "elem2"], ["marker1", "elem3"], ["marker2", "elem4", "elem5"]]

使用常规循环很容易执行：

lst = ["elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5"]

separated = []
sub_lst = []
for elem in lst:
    if elem[:6] == "marker" and sub_lst:
        separated.append(sub_lst)
        sub_lst = []
    sub_lst.append(elem)
if sub_lst:
    separated.append(sub_lst)

此代码 很容易长9行。我的问题是如何使用列表理解或任何其他功能样式在一行（左右）中执行此操作。也欢迎其他任何优雅的解决方案。

Answer 1

在开始或标记处创建（并包含）一个新的内部列表，否则将附加到当前的内部列表（< DMSULPOPUQTXHVZ7LM2FEUH@3DBF29/@dmuzsfijmmodcywyzyqcccccccccccccc3xrhx28M5IUH5Y0GMQMQAQMAEEZVDNGGEVDNGEVDNGEVDNGEV6BKNYYY2FYWUBD9JHZLGFV/7MIBLBCRBCRBCRBCRBCRBCRBCRBCR // avnxg“ rel =” nofollow noreferrer“ title =” python 3.8（预释放） - 在线尝试”>在线尝试！）：

a = None
separated = [a := [x] for x in lst if not a or x.startswith('marker') or a.append(x)]

另一个（Try it online!):

separated = [
    a
    for a in [None]
    for x in lst
    if not a or x.startswith('marker') or a.append(x)
    for a in [[x]]
]

Yet another (未经测试，现在无法测试）：

from more_itertools import split_before

separated = list(split_before(lst, lambda s: s.startswith('marker')))

Answer 2

您可以查找列表中标记元素的索引，然后根据这些位置进行订书机：

lst = ["elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5"]
idxs = [i for i, v in enumerate(lst) if type(v) == str and v.startswith('marker')]
separated = [lst[i:j] for i, j in zip([0]+idxs, idxs+[len(lst)]) if i < j]
# [['elem1', 'elem2'], ['marker1', 'elem3'], ['marker2', 'elem4', 'elem5']]

lst = ["marker1", "elem1", "elem2", "marker2", "elem3"]
idxs = [i for i, v in enumerate(lst) if type(v) == str and v.startswith('marker')]
separated = [lst[i:j] for i, j in zip([0]+idxs, idxs+[len(lst)]) if i < j]
# [['marker1', 'elem1', 'elem2'], ['marker2', 'elem3']]

从这个答案。

Answer 3

这就是 i 看起来像 [['elem1'，'elem2']，['marker1']，['elem3']，['marker2']，['elem4'， 'elem5']，['marker3']，['elem6'，'elem7']] ，但是在发电机中

，我们使用 zip 和 i i AS Generator 要将每个2个列表组合在一起 ['Elem1'，'Elem2'] + ['Marker1']

from itertools import groupby

lst = ["elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5", "marker3", "elem6", "elem7"]

i = (list(g) for _, g in groupby(lst, key=lambda x: x.startswith('marker')))

print([a + b for a, b in zip(i, i)])

edit1

from itertools import groupby, zip_longest

lst = ["marker0", "elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5", "marker3", "elem6", "elem7"]

i = (list(g) for _, g in groupby(lst, key=lambda x: x.startswith('marker')))

print([a + b for a, b in zip_longest(i, i, fillvalue=[])])

edit2

from itertools import groupby, zip_longest

lst = ["elem1", "elem2", "marker1", "elem3", "marker2", "elem4", "elem5", "marker3", "elem6", "elem7"]

i = iter([[]] + [list(g) for _, g in groupby(lst, key=lambda x: x.startswith('marker'))])

print([a + b for a, b in zip_longest(i, i, fillvalue=[])])