import numpy as np
data = np.array([0, 0, 0, 0, 0, 0, 0, 1], dtype=np.uint8)
data.view(np.uint64)
What I would expect is the binary would be:
0b0000000000000000000000000000000000000000000000000000000000000001
but instead the 8 bit groups are reversed.
np.array([72057594037927936], dtype=np.uint64)
which is:
0b0000000100000000000000000000000000000000000000000000000000000000
Why is that? Is there computation being done that reverses this or is this just the layout?
Your assumptions are correct with the how binary data is order in your array. Instead of viewing as uint64, you can view as 8 bytes (V8 dtype):
import numpy as np
np.array([0, 0, 0, 0, 0, 0, 0, 1], dtype=np.uint8).view('V8')[0]
# void(b'\x00\x00\x00\x00\x00\x00\x00\x01')
However your CPU is using little endian ordering of bytes to represent integers. This means that when viewing the bytes as uint64, you get a really big number.
You can check this as follows using the struct package:
import struct
import sys
print(sys.bybteorder)
# 'little'
# view these bytes as uint64 with little endian gives a big number
struct.unpack('<Q', b'\x00\x00\x00\x00\x00\x00\x00\x01')[0]
# 72057594037927936
# view these bytes as uint64 with big endian gives 1
struct.unpack('>Q', b'\x00\x00\x00\x00\x00\x00\x00\x01')[0]
# 1
# view these bytes as uint64 with native endian gives a big number with your CPU
struct.unpack('=Q', b'\x00\x00\x00\x00\x00\x00\x00\x01')[0]
# 72057594037927936