Pythonic way to use a list as an index (without numpy) [duplicate]

Question

What is a Pythonic way to use a list as an index in Python? With numpy you can do the following:

import numpy
a = numpy.zeros(10)
indices = [1,3,6]
a[indices] = 1
# This gives [0,1,0,1,0,0,1,0,0,0]

What is the simplest way to replicate this without using numpy?

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– match, Commented May 23, 2020 at 12:04

ForceBru · Accepted Answer · 2020-05-23 12:00:23Z

2

Iterate through the indices and update the list manually:

a = [0] * 10
for index in indices:
  a[index] = 1

answered May 23, 2020 at 12:00

ForceBru

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Comments

CanciuCostin · Accepted Answer · 2020-05-23 12:00:32Z

1

You could create an array of zeros on your own:

a=[0] * 10
>>> a
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

indices = [1,3,6]
for index in indices:
    a[index]=1

>>> a
[0, 1, 0, 1, 0, 0, 1, 0, 0, 0]

answered May 23, 2020 at 12:00

CanciuCostin

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JTateCC · Accepted Answer · 2020-05-23 11:59:37Z

0

maybe just use a loop and conditionally append to a list

a = []
indices = [1,3,6]
for i in range(10):
    if i in indices:
        a.append(1)
    else:
        a.append(0)

answered May 23, 2020 at 11:59

JTateCC

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dimay · Accepted Answer · 2020-05-23 12:01:03Z

0

You can do it with list comprehension

indices = [1,3,6]
a = [1 if i in indices else 0 for i in range(10) ]
# [0, 1, 0, 1, 0, 0, 1, 0, 0, 0]

answered May 23, 2020 at 12:01

dimay

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This is what I as going to suggest - a nice clean one-liner.