How do I interpret the binary of a numpy array?

Question

Here describes the PyArrayObject struct.

In a Python session, I have the following:

>>> t_ar1 = np.linspace(0, 15, 16).reshape((4, 4))
>>> print(hex_fmt(string_at(id(t_ar1), 80).hex()))
0100000000000000
40b06eecbd7f0000
3053b973bf550000
02000000ffffffff
70eaa873bf550000
80eaa873bf550000
50c826adbd7f0000
00b66eecbd7f0000
01050000322c2033
0000000000000000

From my understanding, the third line is a pointer to the actual data of the array. Viewing the data there, I find

>>> print(hex_fmt(string_at(id(0x55bf73b95330), 96).hex()))
0200000000000000
4049a801be7f0000
0200000000000000
3053b933fd560100
489601adbd7f0000
10ab27adbd7f0000
0000000000000000
0000000000000000
d09501adbd7f0000
b0aa27adbd7f0000
0000000000000000
0000000000000000

Here, I'm expecting to see the floating point numbers 0.0 - 15.0 somewhere. However, I can't seem to find them. What's going on here?

string_at is from the ctypes module. hex_fmt just pretty-prints the hex (it's unimportant). — extremeaxe5
– extremeaxe5, Commented Apr 12, 2019 at 1:41

bubble · Accepted Answer · 2019-04-12 01:45:55Z

1

string_at is from ctypes module.

To get the data directly from a numpy array you need to interpret byte-string (obtained by means of string_at) as array of floating point numbers (8-byte doubles). So, we need to use struct module to interpret byte-string as array of numbers:

from ctypes import string_at
import numpy as np
import struct # needed to unpack data

t_ar1 = np.linspace(0, 15, 16).reshape((4, 4))
struct.unpack('16d', string_at(t_ar1.__array_interface__['data'][0], size=8 * 16))

(0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0)

edited Apr 12, 2019 at 1:45

answered Apr 12, 2019 at 1:30

bubble

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Comments

extremeaxe5 · Accepted Answer · 2019-04-13 01:47:21Z

0

It should be

string_at(0x55bf73b95330, 96)

Rather than what I had,

string_at(id(0x55bf73b95330), 96)

In this case, you'll get

>>> print(hex_fmt(string_at(0x55bf73b95330, 96).hex()))
0000000000000000
000000000000f03f
0000000000000040
0000000000000840
0000000000001040
0000000000001440
0000000000001840
0000000000001c40
0000000000002040
0000000000002240
0000000000002440
0000000000002640

As expected.

answered Apr 13, 2019 at 1:47

extremeaxe5

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