Python + Binary

pyther · 2009-10-14 19:30:48

I need to take the decimal 49 and figure out what makes the 49. In this case 32 + 16 + 1.

0 - Idle
1 - Transcoding
2 - Commercial Flagging
4 - Grabbing EPG data
8 - Recording - only valid if flag is 1
16 - Locked
32 - Jobs running or pending
64 - In a daily wakeup/shutdown period
128 - Less than 15 minutes to next wakeup period
255 - Setup is running

I have told to do this you'd use a bitmask, and this handy function will print out the bitmask (binary number)

def int2bitmask(some_int):
    bitmask = ""
    octal_dict={'0':'000','1':'001','2':'010','3':'011','4':'100','5':'101','6':'110','7':'111'}
    for c in oct(some_int)[1:]:
        bitmask = bitmask+octal_dict[c]
    return bitmask.lstrip('0').rjust(8,'0')

So now I need to take the bitmask and determine which decimal numbers there are. (for 49, its 32 + 16 + 1)

 
 1    1    0    0    0    1  (Binary for 49)
32   16    8    4    2    1 (Decimals)

As you can see 32, 16, and 1 are set.

I know this is a crappy explanation, but hopefully you get what I'm trying to do.
I'm not sure how to go about doing this. Any help would be much appreciated.

pyther · 2009-10-14 20:06:33

#!/usr/bin/env python

import sys

def int2bitmask(some_int):
    bitmask = ""
    octal_dict={'0':'000','1':'001','2':'010','3':'011','4':'100','5':'101','6':'110','7':'111'}
    for c in oct(some_int)[1:]:
        bitmask = bitmask+octal_dict[c]
    return bitmask.lstrip('0').rjust(8,'0')

def getValues(bin):
    
    #Assumes there are only 8 bits
    bin_dict={0:'128',1:'64',2:'32',3:'16',4:'8',5:'4',6:'2',7:'1',}
    
    nums=[]
    for (counter, x) in enumerate(bin):
        if x == '1':
            nums.append(bin_dict[counter])
    return nums


if __name__ == "__main__":
    print sys.argv[1]
    try:
        num=int(sys.argv[1])
    except NameError:
        print "Need an INT"
        sys.exit(2)

    print getValues(int2bitmask(num))

Well I wrote this and it seems to give me my desired results, but I am interested to see if there is a better to get the results.

Procyon · 2009-10-14 20:14:53

You need to read up on bitwise operators

for flag in map(lambda x: 2**x, range(9)):
  if 49 & flag: print flag

Dusty · 2009-10-14 20:21:43

The bitwise operators might help you clean up your code. They are the traditional way to compare bitmasks:

>>> 49 & 32
32
>>> 49 & 16
16
>>> 49 & 8
0
>>> 49 & 4
0
>>> 49 & 2
0
>>> 49 & 1
1

So you can do:

if 49 & 32:
# do something with the 32 flag

Further, if you want to generate the powers of 2, use the bitshift operator:
>>> for i in range(5):
... print i, 2 << i
...
0 2
1 4
2 8
3 16
4 32

I think with those tools, you can clean up your app quite a bit. :-)

Dusty

pyther · 2009-10-14 21:34:28

Well Damn that is easy! Much easier than what I was doing anyways!
Thanks!

Arch Linux

#1 2009-10-14 19:30:48

Python + Binary

#2 2009-10-14 20:06:33

Re: Python + Binary

#3 2009-10-14 20:14:53

Re: Python + Binary

#4 2009-10-14 20:21:43

Re: Python + Binary

#5 2009-10-14 21:34:28

Re: Python + Binary

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