On 21/05/2015 23:31, MRAB wrote:
On 2015-05-21 23:20, John Pote wrote:
Hi everyone.
I recently had the problem of converting from an integer to its
representation as a list of binary bits, each bit being an integer 1 or
0, and vice versa. E.G.
0x53
becomes
[ 0, 1, 0, 1, 0, 0, 1, 1 ]
This I wanted to do for integers of many tens, if not hundreds, of bits.
Python very nicely expands integers to any size required, great feature.
Just wondered if there was a neat way of doing this without resorting to
a bit bashing loop.
Looking forward to some interesting answers,
John
I don't know how efficient you want it to be, but:
>>> number = 0x53
>>> bin(number)
'0b1010011'
>>> bin(number)[2 : ]
'1010011'
>>> list(map(int, bin(number)[2 : ]))
[1, 0, 1, 0, 0, 1, 1]
Thanks for the replies. Interesting that the offered solutions involve
converting to a binary text string and then the individual chars back to
ints. I had thought this would be a route to solve this problem but it
seemed a bit 'heavy' hence I thought it worthwhile asking the question.
My solution to converting a list of 1s and 0s back to an int is
listLen = len( binList )
n = sum( [ binList[i]*( 2**(listLen-1 - i) ) for i in
range(listLen)] )
In response to Ben Finney's question, I haven't done homework for 40
years! Genuine problem, I had decided that the clearest way to write the
algorithm I was working on was to use lists of 1s and 0s rather than
normal ints.
Thanks for the help,
John
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