Initialize 2d array with variable size

Question

Using the following method:

myArray = [0,1] * NUM_ITEMS

Desired result (2d array):

[[0,1],[0,1],[0,1]...]

Actual result (extended 1d array):

[0,1,0,1,0,1...]

How can I achieve the desired result preferably without using numpy?

Before someone suggests [[0, 1]]*NUM_ITEMS, no, that doesn't work, even if it looks like it does. — user2357112
– user2357112, Commented Nov 20, 2018 at 22:09

chris · Accepted Answer · 2018-11-20 22:08:49Z

5

A list comprehension should do the trick:

>>> NUM_ITEMS = 5
>>> my_array = [[0, 1] for _ in range(NUM_ITEMS)]
>>> my_array
[[0, 1], [0, 1], [0, 1], [0, 1], [0, 1]]

answered Nov 20, 2018 at 22:08

chris

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timgeb · Accepted Answer · 2018-11-20 22:19:55Z

1

Since you tagged arrays, here's an alternative numpy solution using numpy.tile.

>>> import numpy as np
>>> NUM_ITEMS = 10
>>> np.tile([0, 1], (NUM_ITEMS, 1))
array([[0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1],
       [0, 1]])

answered Nov 20, 2018 at 22:19

timgeb

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