How to format a string using variables instead of numbers within curly braces?

Question

I'm using Python 3.7 and I have stored a value inside a variable. This variable holds the value of padding which I want to use within curly braces for string formatting. The code explains what I am trying to do.

def print_formatted(number):
    for i in range(1, number + 1):
        binum = bin(i).replace('0b', '')
        ocnum = oct(i).replace('0o', '')
        hexnum = hex(i).replace('0x', '')

        length = len(bin(number).replace('0b', ''))
        print('{0:>length} {1:>length} {2:>length} {3:>length}'.format(i, ocnum, hexnum, binum)) # Error here

This is the code that I have been trying to run. What I am trying to do is to right align the numbers by padding it by the value of the length of the last binary number.

ValueError: Invalid format specifier

This is the error I get. What am I doing wrong?

Answer 1

You can use f-strings and also format specifiers to avoid use of the hex, oct and bin builtins and then string slicing and use int.bit_length() instead of taking the length of the binary string, eg:

def print_formatted(number):
    # get number of bits required to store number
    w = number.bit_length()
    for n in range(1, number + 1):
        # print each number as decimal, then octal, then hex, then binary with padding
        print(f'{n:>{w}} {n:>{w}o} {n:>{w}x} {n:>{w}b}')

Running print_formatted(20) will give you:

    1     1     1     1
    2     2     2    10
    3     3     3    11
    4     4     4   100
    5     5     5   101
    6     6     6   110
    7     7     7   111
    8    10     8  1000
    9    11     9  1001
   10    12     a  1010
   11    13     b  1011
   12    14     c  1100
   13    15     d  1101
   14    16     e  1110
   15    17     f  1111
   16    20    10 10000
   17    21    11 10001
   18    22    12 10010
   19    23    13 10011
   20    24    14 10100