creating a new list with subset of list using index in python

Question

A list   a     a    b    c   3  4   d   6  7  8    I want a list using a subset of a using a 0 2  a 4   a 6      that is I want a list   a    b   4  6  7  8

User · Answer

The following definition might be more efficient than the first solution proposed

def new_list_from_intervals(original_list, *intervals):
    n = sum(j - i for i, j in intervals)
    new_list = [None] * n
    index = 0
    for i, j in intervals :
        for k in range(i, j) :
            new_list[index] = original_list[k]
            index += 1

    return new_list

then you can use it like below

new_list = new_list_from_intervals(original_list, (0,2), (4,5), (6, len(original_list)))

User · Answer

Try new list   a 0 2     a 4     a 6     Or more generally  something like this   from itertools import chain new list   list chain a 0 2    a 4    a 6       This works with other sequences as well  and is likely to be faster   Or you could do this   def chain elements or slices  elements or slices       new list          for i in elements or slices          if isinstance i  list               new list extend i          else              new list append i      return new list  new list   chain elements or slices a 0 2   a 4   a 6      But beware  this would lead to problems if some of the elements in your list were themselves lists  To solve this  either use one of the previous solutions  or replace a 4  with a 4 5   or more generally a n  with a n n 1

User · Answer

Suppose  a     a    b    c   3  4   d   6  7  8    and the list of indexes is stored in  b   0  1  2  4  6  7  8    then a simple one-line solution will be  c    a i  for i in b

[python] creating a new list with subset of list using index in python

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