# numpy.tile() in Python

numpy.tile(arr, repetitions) : 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.

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 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 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 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)

Note :
These codes won’t run on online-ID. Please run them on your systems to explore the working
.
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