Coverting hex to bin and putting in another array Python 3

Question

I'm working on a Python 3 project. My code is more longer but I prepared for you a sample basic idea of my code which works.

arr = []
arr2 = []
number = (["01", "02", "03" ])
arr = number
print(arr) # Output : ["01", "02", "03"]

Question: How to convert this numbers into another array with converting hex to bin? Note: My expected output is for arr2 : ["00000001", "00000010", "00000011"] And when I print(arr2[0]) I want to see 00000001

Possible duplicate of [Python int to binary?](https://stackoverflow.com/questions/699866/python-int-to-binary) — Space Impact, Oct 26 '18 at 11:16

score 4 · Accepted Answer · answered Oct 26 '18 at 11:12

4

You can use the string formatter in a list comprehension:

['{:08b}'.format(int(n, 16)) for n in number]

This returns:

['00000001', '00000010', '00000011']

answered Oct 26 '18 at 11:12

blhsing

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Thank you so much, This is I want to see, you make me happy :) – Asell Oct 26 '18 at 11:17

score 1 · Answer 2 · answered Oct 26 '18 at 11:19

1

Use the bin function. Convert an int to binary. the [2:] "cuts" the b0 (the function returns etc 03 -> b011) if you want to fill the number with 8 zeros use the zfill(8) function

number = (["01", "02", "03" ])
    arr = []
    for i in number:
        i = int(i)
        i = str(bin(i)[2:]).zfill(8)
        print(i)
        arr.append(str(i))


    print(arr)

answered Oct 26 '18 at 11:19

Anagnostou John

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Thank you so much for answer – Asell Oct 26 '18 at 11:26

Bogdan · Answer 3 · 2018-10-26T11:14:37.773

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as a one of possible ways:

hex_var = "02"
length = 8

a = bin(int(hex_var, 16))[2:]
print(str(a).zfill(length))

edited Oct 26 '18 at 11:14

answered Oct 26 '18 at 11:07

Bogdan

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Thank you so much for answer – Asell Oct 26 '18 at 11:25

Coverting hex to bin and putting in another array Python 3

3 Answers3