How can I split a 8 byte array into two four byte arrays in Python?

Question

what would be the fastest way to put the first 4 bytes of an 8 byte array in one array and the last 4 bytes in another one.

My approach was to create a for loop and then extract everything. Like this

for i in range(0,7):
    if i < 4:
        ...
    else
        ...

There has to be something more efficient. What am I missing?

what object do you call byte array? and what do you optimise, number of lines, excetuion time, something else? — alko
– alko, Commented Dec 9, 2013 at 22:13
Your for loop doesn't do anything with the last byte. You should type in range(0,7) at a Python 2 command line (or list(range(0,7)) at a Python 3 command line) and see what you get. — John Y
– John Y, Commented Dec 9, 2013 at 22:21

Tim Peters · Accepted Answer · 2013-12-09 22:15:54Z

1

Try

hi, lo = some_array[:4], some_array[4:]

answered Dec 9, 2013 at 22:15

Tim Peters

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Comments

Lazybeem · Accepted Answer · 2013-12-09 22:17:19Z

1

a = range(0,8)
b = a[:4]
c = a[4:]

Easiest way I know.

answered Dec 9, 2013 at 22:17

Lazybeem

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Steve Barnes · Accepted Answer · 2013-12-09 22:15:34Z

0

Use slices:

A = [1,2,3,4,5,6,7,8]
A[0:4]
A[4:8]

answered Dec 9, 2013 at 22:15

Steve Barnes

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Martino Dino · Accepted Answer · 2013-12-09 22:17:38Z

0

List slicing is your friend here ;)

array = [0,1,2,3,4,5,6,7]

firstpart,secondpart = (array[:4],array[4:])

print firstpart
[0, 1, 2, 3]
print secondpart
[4, 5, 6, 7]

answered Dec 9, 2013 at 22:17

Martino Dino

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