复制和过滤嵌套列表Python

Question

I have a list of lists that looks something like:

[[51502 2 0 ... 0 1 1]
 [8046 2 0 ... 1 1 2]
 ....
 [40701 1 1 ... 1 1 1]]

In my case, every first element in the nested list is "out-of-place" and I want to remove all of them.

# My goal: 
[[2 2 0 ... 0 1 1]
 [2 0 ... 1 1 2]
 ....
 [1 1 ... 1 1 1]]

I've tried np.delete(the_nested[i],0) using for-loop and it gave me the error of " could not broadcast input array "

I've also tried to delete it manually using del and pop , but as expected, numpy didn't allow it since static.

Could anyone provide an alternative solution?

Update: Type for the_nested is numpy.ndarray

Note: I'm sorry beforehand if this post turns out to be a duplicate (hopefully not!) :(

Answer 1

I think you can just index it out:

np.array(the_nested)[:,1:]

In your case:

the_nested = [[51502, 2, 0,  0, 1, 1],
              [8046 ,2 ,0 , 1 ,1 ,2],
              [40701, 1 ,1  ,1 ,1 ,1]]

>>> np.array(the_nested)[:,1:]
array([[2, 0, 0, 1, 1],
       [2, 0, 1, 1, 2],
       [1, 1, 1, 1, 1]])

Alternatively, with np.delete (no loop needed):

>>> np.delete(np.array(the_nested),0,axis=1)
array([[2, 0, 0, 1, 1],
       [2, 0, 1, 1, 2],
       [1, 1, 1, 1, 1]])

Answer 2

I see you have a list and if you don't want to use additional packages like numpy you can do something like this. But it involve looping. Also, it is not modifying the original list.

the_nested = [[51502, 2, 0,  0, 1, 1],
              [8046 ,2 ,0 , 1 ,1 ,2],
              [40701, 1 ,1  ,1 ,1 ,1]]

res = []

_ = [res.append(x[1:]) for x in the_nested]

# Output : 
[[2, 0, 0, 1, 1], [2, 0, 1, 1, 2], [1, 1, 1, 1, 1]]