Python selecting elements in a list by indices

Question

I have a list in python that I want to get the a set of indexes out of and save as a subset of the original list:

templist = [[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]]

and I want this:

   sublist=[[1,    4,    7,                      16,      19,20]]

as an example.

I have no way of knowing ahead of time what the contents of the list elements will be . All I have is the indices that will always be the same.

Is there a single line way of doing this?

What do you mean? Python give each element in a list it's own index? — testname123, May 20 '14 at 15:10
so tell me if i understood this correctly: you have a list, you have some indices and you want to extract a sub list which contains elements of the list at those indices. — Chaitanya Nettem, May 20 '14 at 15:10
yes sir. They are scattered all over the index, so I'm suspicious that the `:` operator won't work since that does consecutive element selection — testname123, May 20 '14 at 15:12
are the indices always the same? I.e do you know that you will always want elements at indices 1,4,7,16,19,20 only? or can the indices vary too? — Chaitanya Nettem, May 20 '14 at 15:13
The process I'm using to generate the list also records a lot of uninteresting data that I don't care about. I only want to keep a few things of interest from what my process outputs, and it's output structure is the same every time. — testname123, May 20 '14 at 15:14

score 6 · Answer 1 · answered May 20 '14 at 15:15

6

Using operator.itemgetter:

>>> templist = [[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]]
>>> import operator
>>> f = operator.itemgetter(0,3,6,15,18,19)
>>> sublist = [list(f(templist[0]))]
>>> sublist
[[1, 4, 7, 16, 19, 20]]

answered May 20 '14 at 15:15

falsetru

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Chaitanya Nettem · Accepted Answer · 2021-05-14T12:07:50.247

4

Assuming you know what the indices to be selected are, it would work something like this:

indices = [1, 4, 7, 16, 19, 20]
templist = [[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21]]
sublist = []

for i in indices:
    sublist.append(templist[0][i])

This can also be expressed in the form of a list comprehension -

sublist = [templist[0][i] for i in indices]

edited May 14 '21 at 12:07

answered May 20 '14 at 15:19

Chaitanya Nettem

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score 2 · Answer 3 · answered May 20 '14 at 15:13

2

you can use list comprehension with enumerate:

indices = [1,2,3]
sublist = [element for i, element in enumerate(templist) if i in indices]

answered May 20 '14 at 15:13

ToonAlfrink

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score 2 · Answer 4 · answered Dec 18 '20 at 15:32

2

 mylist = ['A','B','C','D','E','F']
 idx = [0,1,3]
 [mylist[i] for i in idx]

['A', 'B', 'D']

answered Dec 18 '20 at 15:32

Farid Khafizov

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That doesn't answer OP's question at all, was this supposed to be another question? – JeffUK Dec 18 '20 at 16:59

score 1 · Answer 5 · answered May 20 '14 at 15:18

You can use list comprehension:

indices = set([1,2,3])
sublist = [el for i, el in enumerate(orig_list) if i in indices]

Or you can store indices in a list of True/False and use itertools.compress:

indices = [True, False, True]
sublist = itertools.compress(orig_list, indices)

Python selecting elements in a list by indices

5 Answers5

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