Python使用子列表拆分列表

Question

I have a problem with my list. I have a list with many sublist. This looks like this:

L=[[1,5],[1,1,2,8,5,6],[6,46,35,86,24,3,34,46,23,35],[12,14,53,24,41,53],[1,3,6,4,3,8],[2,5,54,4,61,72,65,54],[43,6,2,6,51,3,43,13,64,52,5,8,3,57,52]]

What I want is something like this:

L1=[1,5]
L2=[1,1,2,8,5,6]
L3=[6,46,35,86,24,3,34,46,23,35]
L4=[12,14,53,24,41,53]
L5=[1,3,6,4,3,8]
...

I tried it with itertools.groupby() , but this just gave me :

L=[[[1,5],[1,1,2,8,5,6],[1,3,6,4,3,8]],[6,46,35,86,24,3,34,46,23,35],[12,14,53,24,41,53],[2,5,54,4,61,72,65,54],[43,6,2,6,51,3,43,13,64,52,5,8,3,57,52]]

How can I do what I want?

Answer 1

You don't need to create variables L1 , L2 , L3 , etc. List indexing already does what you need:

L=[[1,5],
   [1,1,2,8,5,6],
   [6,46,35,86,24,3,34,46,23,35],
   [12,14,53,24,41,53],
   [1,3,6,4,3,8],
   [2,5,54,4,61,72,65,54],
   [43,6,2,6,51,3,43,13,64,52,5,8,3,57,52]]

print L[0]  # prints [1, 5]
print L[4]  # prints [1, 3, 6, 4, 3, 8]

# If you want the first element of the first list in L, you use
L[0][0]

This has the advantage of working regardless of the size of L . You don't need to make a zillion variables if L is huge, or rewrite your program for every possible size of L.

Answer 2

best I can do for you - create a dict with keys like L1, L2:

>>> {'L{}'.format(i): x for i, x in enumerate(L, 1)}
{'L1': [1, 5],
 'L2': [1, 1, 2, 8, 5, 6],
 'L3': [6, 46, 35, 86, 24, 3, 34, 46, 23, 35],
 'L4': [12, 14, 53, 24, 41, 53],
 'L5': [1, 3, 6, 4, 3, 8],
 'L6': [2, 5, 54, 4, 61, 72, 65, 54],
 'L7': [43, 6, 2, 6, 51, 3, 43, 13, 64, 52, 5, 8, 3, 57, 52]}

Answer 3

Firstly, itertools.groupby() won't help you here.

You can use unpacking here, a fantastic feature in python (in my opinion):

L1, L2, L3, L4, L5, L6, L7 = L

But if you're intending to have something like this, then you're looking at the wrong approach. Consider a dictionary:

d = {}
for i, j in enumerate(L, 1):
    d['L{}'.format(i)] = j

Printing d gives:

{'L6': [2, 5, 54, 4, 61, 72, 65, 54], 'L7': [43, 6, 2, 6, 51, 3, 43, 13, 64, 52, 5, 8, 3, 57, 52], 'L4': [12, 14, 53, 24, 41, 53], 'L5': [1, 3, 6, 4, 3, 8], 'L2': [1, 1, 2, 8, 5, 6], 'L3': [6, 46, 35, 86, 24, 3, 34, 46, 23, 35], 'L1': [1, 5]}

Answer 4

You can use indexes!!

 L1=L[0]
 L2=L[1]

and so on....

 print L1
 >> [1,5]