Python根据子列表中的第一个元素将列表拆分为子列表

Question

I want to split a list that looks something like this:

list = [5, a, b, c, d, e, 2, a, b, 4, a ,b ,c ,d , ...]

into this:

list  = [ [5, a, b, c, d, e], [2, a, b] , [4, a ,b ,c ,d] ...]

The first element/number is variable, so no pattern to split it in even chunks. The chunks size or length should be based on that first element of the chunk. Also the alphabetic letters are just placeholders to make the example more readable, in reality the alphabetic letters are floats and numbers.

So the big list really looks something like this:

list = [5, 7, 3.2, 3.1, 4.6, 3, 2, 5.1, 7.1, 4, 5.12 ,3.4 ,4.8 ,12.1 , ...]

Answer 1

Simple way, read each chunk length n and include the next n values:

it = iter(l)
[[n, *islice(it, n)] for n in it]

Python 2 version:

it = iter(l)
[[n] + list(islice(it, n)) for n in it]

Or without islice :

it = iter(l)
[[n] + [next(it) for _ in range(n)] for n in it]

Demo:

>>> from itertools import islice
>>> l = [5, 'a', 'b', 'c', 'd', 'e', 2, 'f', 'g', 4, 'h' ,'i' ,'j' ,'k']
>>> it = iter(l)
>>> [[n, *islice(it, n)] for n in it]
[[5, 'a', 'b', 'c', 'd', 'e'], [2, 'f', 'g'], [4, 'h', 'i', 'j', 'k']]

Answer 2

You can try this one:

l = [5, 'a', 'b', 'c', 'd', 'e', 2, 'a', 'b', 4, 'a' ,'b' ,'c' ,'d']

pos = [(index, i) for index, i in enumerate(l) if type(i)==int]
l = [l[p[0]:p[0]+p[1]+1] for p in pos]
print(l)

Output:

[[5, 'a', 'b', 'c', 'd', 'e'], [2, 'a', 'b'], [4, 'a', 'b', 'c', 'd']]

Answer 3

Could also trying something simple like this:

l = [5, 'a', 'b', 'c', 'd', 'e', 2, 'f', 'g', 4, 'h' ,'i' ,'j' ,'k']

numbers = [i for i, e in enumerate(l) if isinstance(e, int)]

result = [l[s:e] for s, e in zip(numbers[:-1], numbers[1:])] + [l[numbers[-1]:]]

print(result)
# [[5, 'a', 'b', 'c', 'd', 'e'], [2, 'f', 'g'], [4, 'h', 'i', 'j', 'k']]