How to split the list based on the value of the elements

Question

I have a list of values

lst = [75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 185, 186, 187, 188, 189, 190, 191, 192, 193, 195, 196, 197, 198, 199, 221, 222, 223, 224, 225, 226, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 309, 310, 311, 312, 313, 314, 315, 316, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 401, 402, 403, 404, 405, 406, 407, 408, 439, 440]

If you have a close look at the list, you can see that values are incrementing and then there is a big jump and then again values start to increment.

How can I split so in the end I can have

[[75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88], [118, 119, 120, 121, 122, 123, 124, 125, 126, 127], [155, 156, 157, 158, 159, 160, 161, 162, 163, 164]...[439, 440]]

To be clear: the rule is to split any time the next number is *more than* (previous number + 1)? — Karl Knechtel, Sep 16 '21 at 06:02
Does this answer your question? [Identify groups of continuous numbers in a list](https://stackoverflow.com/questions/2154249/identify-groups-of-continuous-numbers-in-a-list) — no comment, Sep 16 '21 at 06:06

mozway · Accepted Answer · 2021-09-16T06:34:54.063

3

Here is a solution with itertools. I used as condition to keep consecutive numbers (thanks @KarlKnechtel).

from itertools import groupby
diff = [item - i for i, item in enumerate(lst)]
[[x[1] for x in g] for i,g in groupby((zip(diff,lst)), lambda x: x[0])]

output:

[[75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88],
 [118, 119, 120, 121, 122, 123, 124, 125, 126, 127],
 [155, 156, 157, 158, 159, 160, 161, 162, 163, 164],
 [185, 186, 187, 188, 189, 190, 191, 192, 193],
 [195, 196, 197, 198, 199],
 [221, 222, 223, 224, 225, 226],
 [273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283],
 [309, 310, 311, 312, 313, 314, 315, 316],
 [371, 372, 373, 374, 375, 376, 377, 378, 379, 380],
 [401, 402, 403, 404, 405, 406, 407, 408],
 [439, 440]]

edited Sep 16 '21 at 06:34

answered Sep 16 '21 at 06:10

mozway

194,879
13
39
75

3

We don't need to compare adjacent elements to solve the problem. Based on the potential duplicate found by @don't talk just code, consider what happens if you subtract *the indices* from the values: `diff = [item - i for i, item in enumerate(lst)]`. You should find that it's possible to write a simpler `key` for the `groupby` after that. – Karl Knechtel Sep 16 '21 at 06:23
Using the same logic with `enumerate`, but as a one-liner thanks to [`more_itertools.groupby_transform`](https://more-itertools.readthedocs.io/en/stable/api.html#more_itertools.groupby_transform): `[g for k,g in more_itertools.groupby_transform(enumerate(lst), keyfunc=lambda p: p[1]-p[0], valuefunc=operator.itemgetter(1), reducefunc=list)]` – Stef Sep 16 '21 at 10:24
Actually, your solution is almost identical to the source code of `more_itertools.consecutive_groups`: https://more-itertools.readthedocs.io/en/stable/_modules/more_itertools/more.html#consecutive_groups – Stef Sep 16 '21 at 10:39

Corralien · Answer 2 · 2021-09-16T06:10:00.937

2

Try:

data = []
tmp = []
for i, j in zip(lst[:-1], lst[1:]):
    if j - i > 1:
        data.append(tmp)
        tmp = []
    else:
        tmp.append(i)
data.append(tmp + [lst[-1]])

Output:

>>> data
[[75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87],
 [118, 119, 120, 121, 122, 123, 124, 125, 126],
 [155, 156, 157, 158, 159, 160, 161, 162, 163],
 [185, 186, 187, 188, 189, 190, 191, 192],
 [195, 196, 197, 198],
 [221, 222, 223, 224, 225],
 [273, 274, 275, 276, 277, 278, 279, 280, 281, 282],
 [309, 310, 311, 312, 313, 314, 315],
 [371, 372, 373, 374, 375, 376, 377, 378, 379],
 [401, 402, 403, 404, 405, 406, 407],
 [439, 440]]

edited Sep 16 '21 at 06:10

answered Sep 16 '21 at 06:04

Corralien

109,409
8
28
52

Alternatively `list(more_itertools.split_when(lst, lambda x,y: y-x != 1))`: [documentation](https://more-itertools.readthedocs.io/en/stable/api.html#more_itertools.split_when); [source code](https://more-itertools.readthedocs.io/en/stable/_modules/more_itertools/more.html#split_when) – Stef Sep 16 '21 at 10:09

score 1 · Answer 3 · answered Sep 16 '21 at 06:06

1

ans = [[]]
for a, b in zip(lst[:-1], lst[1:]):
    ans[-1].append(a)
    if b-a > 1:
        ans .append([])
ans[-1].append(lst[-1])

We are checking brute force and then making new list and appending it to another parent list.

answered Sep 16 '21 at 06:06

user2736738

30,591
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How to split the list based on the value of the elements

3 Answers3