I have a kinda ugly snippet of code that takes an array a, skips the i-th element np.hstack([a[:i], a[i+1:]) of this array, and appends it to the result array. I wonder whether there is any fancy way (without manual for-loops) to do the following:
[a, b, c] -> [[b, c], [a, c], [a, b]]
Here's one way with masking -
def skip_one(a):
n = len(a)
return np.repeat(a[None],n,axis=0)[~np.eye(n,dtype=bool)].reshape(n,-1)
Alternative #1 :
Save on memory with the replication, use np.broadcast_to(a,(n,n)) to replace np.repeat(a[None],n,axis=0).
Alternative #2 :
Replace last step with np.tile based code to bring in more compact-ness -
np.tile(a,(n,1))[~np.eye(n,dtype=bool)].reshape(n,-1)
Alternative #3 :
Use a custom made mask. Hence, replace the second step with -
np.broadcast_to(a,(n,n))[np.not_equal.outer(*[range(n)]*2)].reshape(n,-1)
Sample run -
In [22]: a
Out[22]: array([4, 7, 3, 8])
In [23]: skip_one(a)
Out[23]:
array([[7, 3, 8],
[4, 3, 8],
[4, 7, 8],
[4, 7, 3]])
reshape be in the tuple?not_equal more effective than using python != operator? I guess that not_equal exploits the knowledge of fixed array type making it faster, in comparison with searching for the implementation of __ne__
outer, which we need. So. it makes sense here.You could do what (I presume) you did in a list comprehension:
[np.hstack((data[:i],data[i+1:])) for i in range(len(data))]
