How to split a list using marker elements?

Question

I'm trying to split a list in Python using some elements as markers. For example, consider the list:

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

I wish to split it into 2 sublists:

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

If the first element is not a marker, the elements before the marker shall be considered as a separate sublist:

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

It is easy to do using a regular loop:

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)

This code is 9 lines long. My question is how to do that in one line (or so) using list comprehension or any other functional style. Any other elegant solutions are welcome as well.

I don't like 9 lines of code. And I feel that functional style here would be beneficial. — Dmitry Kuzminov, Jun 12 '22 at 23:50

score 1 · Answer 1 · answered Jun 13 '22 at 00:43

You could look up the indexes of the marker elements in the list, and then take sublists based on those positions:

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']]

Adapted from this answer.

Kelly Bundy · Accepted Answer · 2022-06-13T01:07:35.407

1

Create (and include) a new inner list at the start or at a marker, otherwise append to the current inner list (Try it online!):

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

Another (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 (untested, can't test now):

from more_itertools import split_before

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

edited Jun 13 '22 at 01:07

answered Jun 13 '22 at 00:47

Kelly Bundy

23,480
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Good solution, but due to the walrus operator it works on Python 3.8+ only. – Dmitry Kuzminov Jun 13 '22 at 00:53
1

@DmitryKuzminov Which everybody should be using by now :-) – Kelly Bundy Jun 13 '22 at 00:55
@DmitryKuzminov Added a version without walrus. – Kelly Bundy Jun 13 '22 at 01:01
`split_before` is what I was looking for. This is the most elegant solution that does exactly what I was requesting. – Dmitry Kuzminov Jun 13 '22 at 05:47

Hanna · Answer 3 · 2022-06-13T00:45:35.970

0

this is how i look like [['elem1', 'elem2'],['marker1'],['elem3'],['marker2'],['elem4', 'elem5'],['marker3'],['elem6', 'elem7']] but in a generator ofcourse

so we use zip and i as generator to compine every 2 lists together ['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=[])])

edited Jun 13 '22 at 00:45

answered Jun 12 '22 at 23:59

Hanna

1,071
1
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14

https://stackoverflow.com/a/52591060/17246406 – Hanna Jun 13 '22 at 00:04
2

While markers are assigned to previous sublists instead of being headers for the next ones, the `groupby` is a good approach to solve my problem. Thanks. – Dmitry Kuzminov Jun 13 '22 at 00:12
1

Doesn't seem to work in case of no starting `marker`? – Daniel Hao Jun 13 '22 at 00:13
@DmitryKuzminov adding a `'marker0'` at the beginning of the list will solve it – Hanna Jun 13 '22 at 00:26
@DanielHao Thanks I didn't think of that, but i guess `zip_longest(i, i, fillvalue=[])` will do it – Hanna Jun 13 '22 at 00:30
1

@AlbertoHanna, adding a `'marker0'` at the beginning of the list makes this solution less practical than a regular loop. – Dmitry Kuzminov Jun 13 '22 at 00:35

How to split a list using marker elements?

3 Answers3