获取列表的每个片段

Question

I've been through itertools inside and out and I cannot figure out how to do the following. I want to take a list.

x = [1,2,3,4,5,6,7,8] and I want to get a new list:

y = [[1],[1,2],[1,2,3],.......[2],[2,3],[2,3,4].....[8]]

I need a list of all slices, but not combinations or permutations.

x = list(zip(x[::2], x[1::2])) is close, but doesn't do exactly what I'm hoping

Answer 1

Use combinations not of x , but of the range of possible slice indices (including one past the end, thus len(x)+1 , since slices are exclusive on the end) to make the slice end points, then use them to slice x :

from itertools import combinations

y = [x[s:e] for s, e in combinations(range(len(x)+1), 2)]

That gets exactly what you're going for as straightforwardly as possible. If you want (possibly) faster map based code, you can rephrase it as ( list wrapper unnecessary on Python 2):

from itertools import combinations, starmap

y = list(map(x.__getitem__, starmap(slice, combinations(range(len(x)+1), 2))))

which gets the same result, but without any Python bytecode execution per-item, which might run faster (implementation dependent).

Answer 2

You can utilize list comprehension if you insist on a one-liner:

> x=[1,2,3,4]
> [x[a:b+1] for a in range(len(x)) for b in range(len(x)) if a<=b]

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [2], [2, 3], [2, 3, 4], [3], [3, 4], [4]]

Or you can even get rid of that if :

> [x[a:b+1] for a in range(len(x)) for b in range(a, len(x))]

Answer 3

You can try this:

x = [1,2,3,4,5,6,7,8]
y = [x[b:i+1] for b in range(len(x)) for i in range(len(x))]
final_list = list(filter(lambda x:x, y))

Output:

[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5], [1, 2, 3, 4, 5, 6], [1, 2, 3, 4, 5, 6, 7], [1, 2, 3, 4, 5, 6, 7, 8], [2], [2, 3], [2, 3, 4], [2, 3, 4, 5], [2, 3, 4, 5, 6], [2, 3, 4, 5, 6, 7], [2, 3, 4, 5, 6, 7, 8], [3], [3, 4], [3, 4, 5], [3, 4, 5, 6], [3, 4, 5, 6, 7], [3, 4, 5, 6, 7, 8], [4], [4, 5], [4, 5, 6], [4, 5, 6, 7], [4, 5, 6, 7, 8], [5], [5, 6], [5, 6, 7], [5, 6, 7, 8], [6], [6, 7], [6, 7, 8], [7], [7, 8], [8]]

Answer 4

I think this is a good aproach, an iterative way, I could understand it well:

lst = [1,2,3,4,5,6,7,8]

res = []
ln= len(lst)


for n in range(ln):
  for ind in range(n+1, ln+1):
    res.append(lst[n:ind])