Sliced numpy array does not own its data

Question

I have to multiply two numpy arrays of different sizes (a and b), but I need to discard the first element of b before resizing.

What I see is that, if I use the full b array the resizing has no issues (but the result is not what I need, since it contains the first element of b). If I attempt to slice the first element off before applying resize I get a

ValueError: cannot resize this array: it does not own its data

Why does this happen, and how can I get around this in the most efficient way possible?

import numpy as np

# Some data
a = np.random.uniform(0., 1., 17)

# Works
b = np.array([56, 7, 343, 89, 234])
b.resize(a.shape)
d = b * a    

# Does not
b = np.array([56, 7, 343, 89, 234])[1:]
b.resize(a.shape)
d = b * a

I had the idea that copy() was not particularly efficient? This little block will be processed millions of times, so I need to make it as fast as possible. Also, why does the array not contain its data anymore after slicing? — Gabriel
– Gabriel, Commented Sep 9, 2017 at 17:21
If you care about performance, initialize a zero array and assign sliced array into it. — Divakar
– Divakar, Commented Sep 9, 2017 at 17:23
Also, why does the array not contain its data anymore after slicing? -Its a view into b, it doesn't have its own data. So, the in-place operation needed by resize won't work. — Divakar
– Divakar, Commented Sep 9, 2017 at 17:23
Is a normally (or always) larger than b? So the the resized b is padded with 0s?, d in that case will also be padded. — hpaulj
– hpaulj, Commented Sep 9, 2017 at 17:42

hpaulj · Accepted Answer · 2017-09-10 01:22:39Z

1

Another option is to just do the multiplication for the terms that matter:

n = len(b)-1
d = np.zeros_like(a)
d[:n] = b[1:] * a[:n]

for:

In [628]: a.shape
Out[628]: (1000000,)
In [629]: b=np.arange(100)

this is 2x faster than (which time about the same)

b.resize(len(a)+1,); d = b[1:] * a
b1 = b[1:].copy(); b1.resize(len(a)); d = b1 * a

Relative timings can vary with the size of b and a. While resize can be done in place, it probably will require a new data buffer, especially if it pads with a lot of zeros.

answered Sep 10, 2017 at 1:22

hpaulj

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Gabriel Over a year ago

Indeed, this answer is much faster for large arrays, and a bit slower for smaller ones. Since my actual arrays are on the smaller side I'll use the copy() method, but this answer is the fastest for large arrays. Thanks!

f5r5e5d · Accepted Answer · 2017-09-09 18:40:05Z

1

the other order? pad out b to one element longer b.resize((len(a)+1,))

then multiply d = b[1:] * a

b = np.array([56, 7, 343, 89, 234])
b.resize((len(a)+1,))
d = b[1:] * a

answered Sep 9, 2017 at 18:40

f5r5e5d

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Gabriel Over a year ago

Do you have any idea if this approach is faster to the copy() method mentioned by Divakar?

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