Efficient standard basis vector with numpy
In [2]: import numpy as np
In [9]: size = 5
In [10]: index = 2
In [11]: np.eye(1,size,index)
Out[11]: array([[ 0., 0., 1., 0., 0.]])
Hm, unfortunately, using np.eye for this is rather slow:
In [12]: %timeit np.eye(1,size,index)
100000 loops, best of 3: 7.68 us per loop
In [13]: %timeit a = np.zeros(size); a[index] = 1.0
1000000 loops, best of 3: 1.53 us per loop
Wrapping np.zeros(size); a[index] = 1.0
in a function makes only a modest difference, and is still much faster than np.eye
:
In [24]: def f(size, index):
....: arr = np.zeros(size)
....: arr[index] = 1.0
....: return arr
....:
In [27]: %timeit f(size, index)
1000000 loops, best of 3: 1.79 us per loop
x = np.zeros(size)
x[index] = 1.0
at least i think thats it...
>>> t = timeit.Timer('np.array([1.0 if i == index else 0.0 for i in range(size)]
)','import numpy as np;size=10000;index=5123')
>>> t.timeit(10)
0.039461429317952934 #original method
>>> t = timeit.Timer('x=np.zeros(size);x[index]=1.0','import numpy as np;size=10000;index=5123')
>>> t.timeit(10)
9.4077963240124518e-05 #zeros method
>>> t = timeit.Timer('x=np.eye(1.0,size,index)','import numpy as np;size=10000;index=5123')
>>> t.timeit(10)
0.0001398340635319073 #eye method
looks like np.zeros is fastest...