Let A be a numpy array like :
A = np.array([1, 2, 3, 4, 5])
I want to find the cleaner way to produce a new array with each value repeated two times:
B = np.array([1, 1, 2, 2, 3, 3, 4, 4, 5, 5])
Do you think this is the simpler way to do it ?
import numpy as np
B = np.tile(A,2).reshape(2,-1).flatten('F')
Use repeat()
In [1]: import numpy as np
In [2]: A = np.array([1, 2, 3, 4, 5])
In [3]: np.repeat(A,2)
Out[3]: array([1, 1, 2, 2, 3, 3, 4, 4, 5, 5])
You can use numpy.column_stack and numpy.ndarray.flatten:
In [12]: numpy.column_stack((A, A)).flatten()
Out[12]: array([1, 1, 2, 2, 3, 3, 4, 4, 5, 5])
Timing comparison:
In [27]: A = numpy.array([1, 2, 3, 4, 5]*1000)
In [28]: %timeit numpy.column_stack((A, A)).flatten()
10000 loops, best of 3: 44.7 µs per loop
In [29]: %timeit numpy.repeat(A, 2)
10000 loops, best of 3: 104 µs per loop
In [30]: %timeit numpy.tile(A,2).reshape(2,-1).flatten('F')
10000 loops, best of 3: 129 µs per loop
import timeit, etc.)
If you need to do this operation in a time critical region, the following code is the fastest (using Numpy 1.9 development version):
In [1]: A = numpy.array([1, 2, 3, 4, 5]*1000)
In [2]: %timeit numpy.array([A, A]).T.ravel('F')
100000 loops, best of 3: 6.44 µs per loop
Note that flatten would make an additional copy, so ravel should be used instead.
If you prefer readability, the column_stack and repeat functions are better:
In [3]: %timeit numpy.column_stack((A, A)).ravel()
100000 loops, best of 3: 15.4 µs per loop
In [4]: timeit numpy.repeat(A, 2)
10000 loops, best of 3: 53.9 µs per loop
