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

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…