How can I replicate numpy array, so that it would be repeated (as a whole array) n times?
So with an example array:
import numpy as np
x = np.arange(0, 5)
I want to create an array like below, without a need of manually typing np.arange(0, 5) n times:
x_3times = np.concatenate([np.arange(0, 5), np.arange(0, 5), np.arange(0, 5)])
or with a set length of output (e.g. 12)?
x_12 = np.concatenate([np.arange(0, 5), np.arange(0, 5), np.arange(0, 5)])[0:12]
You can use np.tile.
>> x_3times = np.tile(x, 3)
>> x_3times
array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4])
For repeating till some particular limit, use np.resize
>> x_12 = np.resize(x, 12)
>> x_12
array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1])
np.tileSimply try list comprehension:
x_3times = np.concatenate([np.arange(0, 5) for x in range(3)])
where the number 3 can be substituted by any number n.
edit
if you want to limit the length by any number, you can simply do:
cutoff = 12
x_3times = np.concatenate([np.arange(0, 5) for x in range(3)])[:cutoff]
which will result to:
array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1])
However, this isn't a very efficient line of code, especially when dealing with large numbers. Another answer would probably be to make a generator:
def generator(arr, n, cutoff=None):
length = len(arr)
if cutoff:
for i in range(cutoff):
yield arr[i%length]
else:
for _ in range(n):
for i in arr:
yield i
array = np.array([x for x in generator(np.arange(0, 5), 3, 12)])
x_3times