CODEWITHSUNDEEP
pythonarraysnumpy
Gabriel
Gabriel

Reputation: 42329

Sliced numpy array does not own its data

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

Upvotes: 0

Views: 582

Answers (2)

hpaulj
hpaulj

Reputation: 231385

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.

Upvotes: 1

f5r5e5d
f5r5e5d

Reputation: 3706

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

Upvotes: 1

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