numpy.tile() in Python

Last Updated : 28 Mar, 2022

The numpy.tile() function constructs a new array by repeating array - 'arr', the number of times we want to repeat as per repetitions. The resulted array will have dimensions max(arr.ndim, repetitions) where, repetitions is the length of repetitions. If arr.ndim > repetitions, reps is promoted to arr.ndim by pre-pending 1’s to it. If arr.ndim < repetitions, reps is promoted to arr.ndim by pre-pending new axis. Syntax :

numpy.tile(arr, repetitions)

Parameters :

array       : [array_like]Input array. 
repetitions : No. of repetitions of arr along each axis.

Return :

An array with repetitions of array - arr as per d, number of times we want to repeat arr

Code 1 :

Python

# Python Program illustrating
# numpy.tile()

import numpy as geek

#Working on 1D
arr = geek.arange(5)
print("arr : \n", arr)

repetitions = 2
print("Repeating arr 2 times : \n", geek.tile(arr, repetitions))

repetitions = 3
print("\nRepeating arr 3 times : \n", geek.tile(arr, repetitions))
# [0 1 2 ..., 2 3 4] means [0 1 2 3 4 0 1 2 3 4 0 1 2 3 4]
# since it was long output, so it uses [ ... ]

Output :

arr : 
 [0 1 2 3 4]
Repeating arr 2 times : 
 [0 1 2 3 4 0 1 2 3 4]

Repeating arr 3 times : 
 [0 1 2 ..., 2 3 4]

Code 2 :

Python

# Python Program illustrating
# numpy.tile()

import numpy as geek

arr = geek.arange(3)
print("arr : \n", arr)

a = 2  
b = 2  
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

a = 3  
b = 2   
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

a = 2
b = 3  
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

Output :

arr : 
 [0 1 2]

Repeating arr : 
 [[0 1 2 0 1 2]
 [0 1 2 0 1 2]]
arr Shape : 
 (2, 6)

Repeating arr : 
 [[0 1 2 0 1 2]
 [0 1 2 0 1 2]
 [0 1 2 0 1 2]]
arr Shape : 
 (3, 6)

Repeating arr : 
 [[0 1 2 ..., 0 1 2]
 [0 1 2 ..., 0 1 2]]
arr Shape : 
 (2, 9)

Code 3 : (repetitions == arr.ndim) == 0

Python

# Python Program illustrating
# numpy.tile()

import numpy as geek

arr = geek.arange(4).reshape(2, 2)
print("arr : \n", arr)

a = 2  
b = 1  
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

a = 3  
b = 2   
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

a = 2
b = 3  
repetitions = (a, b)
print("\nRepeating arr : \n", geek.tile(arr, repetitions))
print("arr Shape : \n", geek.tile(arr, repetitions).shape)

Output :

arr : 
 [[0 1]
 [2 3]]

Repeating arr : 
 [[0 1]
 [2 3]
 [0 1]
 [2 3]]
arr Shape : 
 (4, 2)

Repeating arr : 
 [[0 1 0 1]
 [2 3 2 3]
 [0 1 0 1]
 [2 3 2 3]
 [0 1 0 1]
 [2 3 2 3]]
arr Shape : 
 (6, 4)

Repeating arr : 
 [[0 1 0 1 0 1]
 [2 3 2 3 2 3]
 [0 1 0 1 0 1]
 [2 3 2 3 2 3]]
arr Shape : 
 (4, 6)

References : https://docs.scipy.org/doc/numpy/reference/generated/numpy.tile.html Note : These codes won’t run on online IDE's. Please run them on your systems to explore the working .

numpy.tile() in Python

Mohit Gupta

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numpy.tile() in Python

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