在python 3中将十六进制转换为数组中的bin

Question

I'm working on Visual Studio about Python Project.I have an array and user input for converting hex to dec and dec to bin. I use both of them in my project (hex,dec,bin). Here basic sample of my code:

dynamicArrayBin = [ ]
dynamicArrayHex = [ ] 
hexdec = input("Enter the hex number to binary "); 
dynamicArrayHex = [hexdec[idx:idx+2]  for idx in range(len(hexdec)) if idx%2 == 0]
binary = '{:08b}'.format(int(dynamicArrayHex[0] , 16))

So, when the user enter 01 for input, the code gives 00000001 . I want to separate this result of elements 0 0 0 0 0 0 0 1 and put in dynamicArrayBin=[] . After a while when I call dynamicArrayBin=[0] ,it should show 0 .

Is there any way to do it?

Answer 1

If you want a list of binary digits for a hexadecimal input, there is no need to split out the input into bytes first (which is what your code currently does, convert each 2 characters of hex input into an integer covering the range 0-255).

Just convert the whole hex input to an integer, and format it to binary from there:

integer_value = int(hexdec, 16)
byte_length = (len(hexdec) + 1) // 2  # to help format the output binary
binary_representation = format(integer_value, '0{}b'.format(byte_length * 8))

The binary_representation value is a string of '0' and '1' characters, and since strings are sequences, there is no need to turn that into a list unless you must be able to mutate individual characters.

So:

print(binary_representation[0])

works and prints 0 or 1 .

If you must have a list, you can do so with list(binary_representation)) .

Demo:

>>> hexdec = 'deadbeef'  # 4 bytes, or 32 bits
>>> integer_value = int(hexdec, 16)
>>> byte_length = (len(hexdec) + 1) // 2  # to help format the output binary
>>> binary_representation = format(integer_value, '0{}b'.format(byte_length * 8))
>>> integer_value
3735928559
>>> byte_length
4
>>> binary_representation
'11011110101011011011111011101111'
>>> binary_representation[4]
'1'
>>> binary_representation[2]
'0'
>>> list(binary_representation)
['1', '1', '0', '1', '1', '1', '1', '0', '1', '0', '1', '0', '1', '1', '0', '1', '1', '0', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1']

If all you wanted was the first bit of a hexadecimal value, then there is a faster method:

if len(hexdec) % 2:  # odd number of hex characters, needs a leading 0
    hexdec = '0'     # doesn't matter what the rest of the hex value is
print('1' if hexdec[0].lower() in '89abcdef' else '0')

because the first 4 bits of the binary representation are determined entirely by the first hexadecimal character, and the very first bit is set for hex values 8 through to F .

Answer 2

You can do something like below

hexLst = ['ABC123EFFF', 'ABC123EFEF', 'ABC123EEFF']
binLst = [bin(int(n, 16))[2:] for n in hexLst]
print(binLst)

Which will give you a output

['1010101111000001001000111110111111111111', '1010101111000001001000111110111111101111', '1010101111000001001000111110111011111111']

then you can make a list from that

dynamicArrayBin=[list(b) for b in binLst]
print(dynamicArrayBin)

Output

[['1', '0', '1', '0', '1', '0', '1', '1', '1', '1', '0', '0', '0', '0', '0', '1', '0', '0', '1', '0', '0', '0', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1', '1'], ['1', '0', '1', '0', '1', '0', '1', '1', '1', '1', '0', '0', '0', '0', '0', '1', '0', '0', '1', '0', '0', '0', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '0', '1', '1', '1', '1'], ['1', '0', '1', '0', '1', '0', '1', '1', '1', '1', '0', '0', '0', '0', '0', '1', '0', '0', '1', '0', '0', '0', '1', '1', '1', '1', '1', '0', '1', '1', '1', '0', '1', '1', '1', '1', '1', '1', '1', '1']]