Handling binary literal in python

Question

I am trying to translate code from php to python. I have a list of binary literals -

['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']

This is equal to 0000000000000000000000000000000000000000000000000000010001001100 when concatenated.

int("0000000000000000000000000000000000000000000000000000010001001100",2)

gives 1100

How do i make this from the list. Unable to concatenate binary literals.

Unstated here is that each piece represents eight bits, and may be missing leading zeros. — Ned Batchelder
– Ned Batchelder, Commented Dec 9, 2012 at 22:35

Ned Batchelder · Accepted Answer · 2012-12-09 22:33:21Z

1

>>> l = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
>>> int("".join("%02x" % int(x,0) for x in l), 16)
1100

Python understand 0b0101 as a binary literal, so I use int('0b0101', 0) to convert each piece to an int. Then I format it in a convenient format (two digits of hex), concatenate them, and interpret them as a hex integer.

answered Dec 9, 2012 at 22:33

Ned Batchelder

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2 Comments

Eric Over a year ago

Where is int(..., 0) documented? Never seen it before.

Ned Batchelder Over a year ago

Right there in the docs: docs.python.org/2/library/functions.html#int "Base 0 means to interpret the string exactly as an integer literal, so that the actual base is 2, 8, 10, or 16."

Scout · Accepted Answer · 2012-12-09 22:35:02Z

1

You need the zfill method to pad your elements with the right quantity of zeros

li = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
zero_padded = [x[2:].zfill(8) for x in li]
print ''.join(zero_padded)

Outputs

0000000000000000000000000000000000000000000000000000010001001100

answered Dec 9, 2012 at 22:35

Scout

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Comments

Eric · Accepted Answer · 2012-12-09 22:29:03Z

0

Just strip off the 0b:

binary = ''.join(x[2:] for x in yourlist)
print binary
print int(binary, 2)

answered Dec 9, 2012 at 22:29

Eric

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2 Comments

Aditya Singh Over a year ago

Outputs 588. I need to concatenate and get the ascii value

Eric Over a year ago

@AdityaSingh: Your expected output is wrong. You've added a zero somewhere.

jimhark · Accepted Answer · 2012-12-09 22:47:37Z

0

Is this ugly enough for you:

int(''.join([i for sl in [s[2:].zfill(8) for s in l] for i in sl]),2)

(It seems to work.)

Maybe this is more readable:

l = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
n = 0
for s in l:
    n = n*256 + int(s,2)
print n

edited Dec 9, 2012 at 22:47

answered Dec 9, 2012 at 22:39

jimhark

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