Efficient standard basis vector with numpy

Question

Given an index and a size, is there a more efficient way to produce the standard basis vector:

import numpy as np
np.array([1.0 if i == index else 0.0 for i in range(size)])

Answer 1

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.]])

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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

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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

Answer 2

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...