Reputation: 34091
In a large code base, I am using np.broadcast_to
to broadcast arrays (just using simple examples here):
In [1]: x = np.array([1,2,3])
In [2]: y = np.broadcast_to(x, (2,1,3))
In [3]: y.shape
Out[3]: (2, 1, 3)
Elsewhere in the code, I use third-party functions that can operate in a vectorized way on Numpy arrays but that are not ufuncs. These functions don't understand broadcasting, which means that calling such a function on arrays like y
is inefficient. Solutions such as Numpy's vectorize
aren't good either because while they understand broadcasting, they introduce a for
loop over the array elements which is then very inefficient.
Ideally, what I'd like to be able to do is to have a function, which we can call e.g. unbroadcast
, that returns an array with a minimal shape that can be broadcasted back to the full size if needed. So e.g.:
In [4]: z = unbroadcast(y)
In [5]: z.shape
Out[5]: (1, 1, 3)
I can then run the third-party functions on z
, then broadcast the result back to y.shape
.
Is there a way to implement unbroadcast
that relies on Numpy's public API? If not, are there any hacks that would produce the desired result?
Upvotes: 4
Views: 877
Reputation: 35080
This is probably equivalent to your own solution, only a bit more built-in. It uses as_strided
in numpy.lib.stride_tricks
:
import numpy as np
from numpy.lib.stride_tricks import as_strided
x = np.arange(16).reshape(2,1,8,1) # shape (2,1,8,1)
y = np.broadcast_to(x,(2,3,8,5)) # shape (2,3,8,5) broadcast
def unbroadcast(arr):
#determine unbroadcast shape
newshape = np.where(np.array(arr.strides) == 0,1,arr.shape) # [2,1,8,1], thanks to @Divakar
return as_strided(arr,shape=newshape) # strides are automatically set here
z = unbroadcast(x)
np.all(z==x) # is True
Note that in my original answer I didn't define a function, and the resulting z
array had (64,0,8,0)
as strides
, whereas the input has (64,64,8,8)
. In the current version the returned z
array has identical strides to x
, I guess passing and returning the array forces a creation of a copy. Anyway, we could always set the strides manually in as_strided
to get identical arrays under all circumstances, but this doesn't seem necessary in the above setup.
Upvotes: 3
Reputation: 34091
I have a possible solution, so will post it here (however if anyone has a better one, please feel free to reply too!). One solution is to check the strides
argument of arrays, which will be 0 along broadcasted dimensions:
def unbroadcast(array):
slices = []
for i in range(array.ndim):
if array.strides[i] == 0:
slices.append(slice(0, 1))
else:
slices.append(slice(None))
return array[slices]
This gives:
In [14]: unbroadcast(y).shape
Out[14]: (1, 1, 3)
Upvotes: 4