I am defining an array of two's, with one's on either end. In MATLAB this can be acheived by
x = [1 2*ones(1,3) 1]
In Python, however, numpy gives something quite different:
import numpy
numpy.array([[1],2*numpy.ones(3),[1]])
What is the most efficient way to perform this MATLAB command in Python?
In [33]: import numpy as np
In [34]: np.r_[1, 2*np.ones(3), 1]
Out[34]: array([ 1., 2., 2., 2., 1.])
Alternatively, you could use hstack:
In [42]: np.hstack(([1], 2*np.ones(3), [1]))
Out[42]: array([ 1., 2., 2., 2., 1.])
In [45]: %timeit np.r_[1, 2*np.ones(300), 1]
10000 loops, best of 3: 27.5 us per loop
In [46]: %timeit np.hstack(([1], 2*np.ones(300), [1]))
10000 loops, best of 3: 26.4 us per loop
In [48]: %timeit np.append([1],np.append(2*np.ones(300)[:],[1]))
10000 loops, best of 3: 28.2 us per loop
Thanks to DSM for pointing out that pre-allocating the right-sized array from the very beginning, can be much much faster than appending, using r_ or hstack on smaller arrays:
In [49]: %timeit a = 2*np.ones(300+2); a[0] = 1; a[-1] = 1
100000 loops, best of 3: 6.79 us per loop
In [50]: %timeit a = np.empty(300+2); a.fill(2); a[0] = 1; a[-1] = 1
1000000 loops, best of 3: 1.73 us per loop
r_ can be 4+ times slower than simply doing 2*ones(n+2) and then patching the edges. This won't be an issue unless it's in an inner loop, of course.
r_ with hstack and got roughly comparable results.a = 2*np.ones(3+2); a[0] = 1; a[-1] = 1;, which for me (1.6.2) is consistently about 4 times faster.r_ used here.Use numpy.ones instead of just ones:
numpy.array([[1],2*numpy.ones(3),[1]])
