Suppose I have
a = array([[1, 2],
[3, 4]])
and
b = array([1,1])
I'd like to use b in index a, that is to do a[b] and get 4 instead of [[3, 4], [3, 4]]
I can probably do
a[tuple(b)]
Is there a better way of doing it?
Thanks
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I don't think it is a problem. why you think a[tuple(b)] is bad?linjunhalida– linjunhalida2011-04-01 01:44:15 +00:00Commented Apr 1, 2011 at 1:44
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4 Answers
According the NumPy tutorial, the correct way to do it is:
a[tuple(b)]
Comments
Suppose you want to access a subvector of a with n index pairs stored in blike so:
b = array([[0, 0],
...
[1, 1]])
This can be done as follows:
a[b[:,0], b[:,1]]
For a single pair index vector this changes to a[b[0],b[1]], but I guess the tuple approach is easier to read and hence preferable.
1 Comment
Terry Brown
But which is faster I wonder? Wouldn't tuple() create a copy whereas the views above would not?
The above is correct. However, if you see an error like:
IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices
You may have your index array in floating type. Change it to something like this:
arr[tuple(a.astype(int))]
Comments
For performance, use
a[b[0], b[1]]
a[b[:,0], b[:,1]] # when b is a 2d array
unless d = tuple(b.T) is readily available (a[d] is the fastest)
time bench shows a[b[:,0], b[:,1]] is significantly faster than a[tuple(c)] both when b is 1d and 2d (omitted), where c == b.T
import numpy as np
import time
a = np.array([[1, 2], [3, 4]])s
b = np.array([1,1])
c = b.T # a[b[:,0], b[:,1]] == a[tuple(c)] when b is 2d
assert tuple(c) == tuple(b)
d = tuple(b)
t0 = time.time()
for _ in range(10000000):
a[tuple(c)]
t1 = time.time()
print(t1 - t0) # 7.781806468963623
t0 = time.time()
for _ in range(10000000):
a[b[0], b[1]]
t1 = time.time()
print(t1 - t0) # 2.8317997455596924
t0 = time.time()
for _ in range(10000000):
a[*b]
t1 = time.time()
print(t1 - t0) # 7.80025315284729
t0 = time.time()
for _ in range(10000000):
a[d]
t1 = time.time()
print(t1 - t0) # 1.3347711563110352
