Reputation: 53
I have an array of start and stop indices, like this:
[[0, 3], [4, 7], [15, 18]]
and i would like to construct a 2D numpy array where each row is a range from the corresponding pair of start and stop indices, as follows:
[[0, 1, 2],
[4, 5, 6],
[15, 16, 18]]
Currently, i am creating an empty array and filling it in a for loop:
ranges = numpy.empty((3, 3))
a = [[0, 3], [4, 7], [15, 18]]
for i, r in enumerate(a):
ranges[i] = numpy.arange(r[0], r[1])
Is there a more compact and (more importantly) faster way of doing this? possibly something that doesn't involve using a loop?
Upvotes: 5
Views: 2582
Reputation: 88226
If the ranges were to result in different lengths, for a vectorized approach you could use n_ranges
from the linked solution:
a = np.array([[0, 3], [4, 7], [15, 18]])
n_ranges(a[:,0], a[:,1], return_flat=False)
# [array([0, 1, 2]), array([4, 5, 6]), array([15, 16, 17])]
Which would also work with the following array:
a = np.array([[0, 3], [4, 9], [15, 18]])
n_ranges(*a.T, return_flat=False)
# [array([0, 1, 2]), array([4, 5, 6, 7, 8]), array([15, 16, 17])]
Upvotes: 2
Reputation: 375375
One way is to use broadcast to add the left hand edges to the base arange:
In [11]: np.arange(3) + np.array([0, 4, 15])[:, None]
Out[11]:
array([[ 0, 1, 2],
[ 4, 5, 6],
[15, 16, 17]])
Note: this requires all ranges to be the same length.
Upvotes: 4