How to implement the conversion of number systems with a fractional part in python?

The numbers in the binary system are multipliers in front of the corresponding powers of two, depending on the position. For example: 10.1 ₂ == 1⋅2 ¹ + 0⋅2 ⁰ + 1⋅2 ^-1 . The numbers in the fractional part are facing negative degrees.

In the octal system, the basis is 8, that is 2 ³ . Therefore, every three bits correspond to one octal digit. For example: 010 ₂ == 2 ₈ , 100 ₂ == 4 ₈ .

For convenience of converting, the fractional part can be added with zeros so that the length becomes a multiple of three: 10.1 ₂ == 10.100 ₂ == 2.4 ₈ . In Python 3:

from math import ceil def bin2oct_float(bits): n, dot, f = bits.partition('.') return bin2oct(n) + dot + bin2oct(f.ljust(3 * ceil(len(f) / 3), '0')) 

where bin2oct(bits) can "01" transform ( set("01").issuperset(bits) ):

 def bin2oct(bits): return bits and '{:o}'.format(int(bits, 2)) 

Example:

 >>> bin2oct_float('10.1') '2.4' 

You can not convert to int , but only on the lines of the operation to produce:

 def bin2oct(bits, bits2digit={'{:03b}'.format(i): str(i) for i in range(8)}): bits = bits.zfill(3 * ceil(len(bits) / 3)) return ''.join([bits2digit[bits[i:i+3]] for i in range(0, len(bits), 3)]) 

The code uses the idiom of traversing a sequence of exactly n elements at a time . The code does not support characters, spaces in the input: '+1.111' , '- 1.0' .

When transferring code to a language without built-in dictionaries, you can use analogs of nested lists in order to select an octal digit by three bits:

 def bits2digit(a, b, c): return [['01','23'], ['45','67']][a][b][c] 

Example:

 >>> bits2digit(1, 0, 0) '4'