Efficient standard basis vector with numpy

2020-06-01 05:29发布

问题:

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)])

回答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.]])

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


回答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...



回答3:

I'm not sure if this is faster, but it's definitely more clear to me.

a = np.zeros(size)
a[index] = 1.0


回答4:

It may not be the fastest, but the method scipy.signal.unit_impulse generalizes the above concept to numpy arrays of any shape.



回答5:

Often, you need not one but all basis vectors. If this is the case, consider np.eye:

basis = np.eye(3)
for vector in basis:
  ...

Not exactly the same, but closely related: This even works to get a set of basis matrices with a bit of tricks:

>>> d, e = 2, 3    # want 2x3 matrices
>>> basis = np.eye(d*e,d*e).reshape((d*e,d,e))
>>> print(basis)
[[[ 1.  0.  0.]
  [ 0.  0.  0.]]

 [[ 0.  1.  0.]
  [ 0.  0.  0.]]

 [[ 0.  0.  1.]
  [ 0.  0.  0.]]

 [[ 0.  0.  0.]
  [ 1.  0.  0.]]

 [[ 0.  0.  0.]
  [ 0.  1.  0.]]

 [[ 0.  0.  0.]
  [ 0.  0.  1.]]]

and so on.



回答6:

Another way to implement this is :

>>> def f(size, index):
...     return (np.arange(size) == index).astype(float)
... 

Which gives a slightly slower execution time :

>>> timeit.timeit('f(size, index)', 'from __main__ import f, size, index', number=1000000)
2.2554846050043125


标签: python numpy