Way to know bytes of an int in Python [with bits class thrown in]

[dansolo]

Hi,

I need to know the byte which a number is made of. For example, let's suppose I have

i = 3000

I would like to know if there is a way to get the single byte of the number (0x0B and 0xB8).

Thanks in advance,
Daniele

[bartonc]

Hi,

I need to know the byte which a number is made of. For example, let's suppose I have

i = 3000

I would like to know if there is a way to get the single byte of the number (0x0B and 0xB8).

Thanks in advance,
Daniele

>>> aHex = hex(3000) # call built-in function to convert to hex
>>> print aHex
0xbb8
>>> bytes = aHex[2:].zfill(4) # slice off the '0x'
>>> print bytes
0bb8
>>> highByte = '0x%s' % bytes[0:2] #reformat with '0x' added to just the high byte
>>> print highByte
0x0b
>>> 

Let's see if you can get the low byte. We're here to help.

[dshimer]

Just a curiosity but is there any built-in that would convert to binary? For example:
3000='101110111000'

[dansolo]

First of all thanks,

the lowbyte comes from

lowByte = '0x%s' % bytes[2:4]

I've found also another solution:

# mod edit # added for completeness #
import struct
################################
def getByteLength(self, length):
        temp = struct.pack('!h', length)
        byteList = [firstLength, secondLength] = struct.unpack('BB', temp)
        return list(byteList)
 

to obtain a list in which the elements are the needed bytes

[bartonc]

Just a curiosity but is there any built-in that would convert to binary? For example:
3000='101110111000'

While it's true that
>>> int('101', 2)
5
>>>
There is no bit or byte type to call for conversion to base 2.

[bartonc]

While it's true that
>>> int('101', 2)
5
>>>
There is no bit or byte type to call for conversion to base 2.

How 'bout this little goodie:

def Byte2Bits(value):
    # Create a list of bits to convert big endian wise
    data = []
    outStr = ""
    for i in range(8):
        value, rem = divmod(value, 2)
        data.insert(0, rem)
 
    for pos, i in enumerate(data):
        if not (pos % 4):
            outStr += " "
        outStr += str(i)
    return outStr
 
 
def Base2Repr(value):
    # Create a list of bytes to convert big endian wise
    data = []
    outStr = ""
    while value:
        value, rem = divmod(value, 256)
        data.insert(0, rem)
    for i in data:
        outStr += Byte2Bits(i)
    return outStr[1:]
 
 
class bits(object):
    def __init__(self, value):
        self.value = value
 
    def __repr__(self):
        return Base2Repr(self.value)
 
 
 
if __name__ == "__main__":
    b = bits(0xf5a5)
    print b
 

1111 0101 1010 0101

[bvdet]

Just a curiosity but is there any built-in that would convert to binary? For example:
3000='101110111000'

I don't think so, but I made this up:

def ConvDecToBaseVar(num, base):
    if base > 10 or base < 2:
        raise ValueError, 'The base number must be between 2 and 10.'
    if num == 0: return 0
    ans = ''
    while num != 0:
        num, rem = divmod(num, base)
        ans =  str(rem)+ans
    return int(ans)
 
'''
>>> ConvDecToBaseVar(3000,2)
101110111000L
>>>
'''

[dshimer]

I was just curious about a built-in but that is absolutely Beautiful!

[bartonc]

I was just curious about a built-in but that is absolutely Beautiful!

It's beautiful, all right (ain't BV good). But what about nibble formatting?

[bvdet]

I was intrigued by a problem Motoma tackled and posted to Python Articles.
http://www.thescripts.com/forum/thread648799.html

[ghostdog74]

Just a curiosity but is there any built-in that would convert to binary? For example:
3000='101110111000'

def dectobin(number):
    if number < 1: return ""
    else:return dectobin(number/2) + str(number & 1) 
 

[bartonc]

def dectobin(number):
    if number < 1: return ""
    else:return dectobin(number/2) + str(number & 1) 
 

VERY nice, GD. So, let's see: If <edit: this isn't right [the integer division of number by 2]> number is odd, then the next [L]MSb is a one... Am I getting that right? And where/how did you dream up that algorithm?

[ghostdog74]

VERY nice, GD. So, let's see: If <edit: this isn't right [the integer division of number by 2]> number is odd, then the next [L]MSb is a one... Am I getting that right? And where/how did you dream up that algorithm?

the & operator is just to test the LSB of binary division.
say for example 14 divide by 2. in binary division of these 2 numbers,
1110 divide by 10 , as you slowly progress, you will reach the LSB and it will always be 0 or 1.
In & operations, a true and true is a true. So if the LSB is 0, & with 1 will be 0. (tested no remainder) . if LSB is 1, & with 1 is a 1, so there's remainder.

( in this case, 14/2 is remainder 0, so the bit tested is 0 )
and then we do 7/2 , 3/2 and so on.....the rest of the code is just to chain together all these 0's and 1's tested.
hope i didn't miss anything...

[Motoma]

I don't think so, but I made this up:

def ConvDecToBaseVar(num, base):
    if base > 10 or base < 2:
        raise ValueError, 'The base number must be between 2 and 10.'
    if num == 0: return 0
    ans = ''
    while num != 0:
        num, rem = divmod(num, base)
        ans =  str(rem)+ans
    return int(ans)
 
'''
>>> ConvDecToBaseVar(3000,2)
101110111000L
>>>
'''

Gotta love reusable code, eh?

[bvdet]

Gotta love reusable code, eh?

Yes. Here's some more recycling -

def ConvDecToBaseVar3(num, base):
    if base > 16 or base < 2:
        raise ValueError, 'The base number must be between 2 and 16.'
    dd = dict(zip(range(16), [hex(i).split('x')[1] for i in range(16)]))
    if num == 0: return ''
    num, rem = divmod(num, base)
    return ConvDecToBaseVar3(num, base)+dd[rem]

>>> ConvDecToBaseVar3(123456789, 16)
'75bcd15'
>>> ConvDecToBaseVar3(2**200, 16)
'100000000000000000000000000000000000000000000000000'
>>> ConvDecToBaseVar3(2**200, 7)
'141246066533632643213232344050606053061443446006544361632102630555343054'
>>> 

It's not suitable for very long numbers due to the recursion limit.

[bvdet]

This is considerably more efficient than the previous version:

def ConvDecToBaseVar3(num, base, dd=False):
    if base > 16 or base < 2:
        raise ValueError, 'The base number must be between 2 and 16.'
    if not dd:
        dd = dict(zip(range(16), [hex(i).split('x')[1] for i in range(16)]))
    if num == 0: return ''
    num, rem = divmod(num, base)
    return ConvDecToBaseVar3(num, base, dd)+dd[rem]

This avoids the recursion limit:

def ConvDecToBaseVar4(num, base):
    if num < 2: return str(num)
    if base > 16 or base < 2:
        raise ValueError, 'The base number must be between 2 and 16.'
    dd = dict(zip(range(16), [hex(i).split('x')[1] for i in range(16)]))
    ans = ''
    while num != 0:
        num, rem = divmod(num, base)
        ans =  str(rem)+ans
    return ans