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add an octal_to_binary converter #8940

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49 changes: 49 additions & 0 deletions conversions/octal_to_binary.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,49 @@
"""
Convert a octal value to its binary equivalent
>>> octal_to_binary("17")
'1111'
>>> octal_to_binary("52523")
'101010101010011'
>>> octal_to_binary("532.51")
'101011010.101001'
Comment on lines +3 to +8
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These doctests should go inside a docstring within the function, not at the top of the file.

"""


from operator import index
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@tianyizheng02 tianyizheng02 Aug 12, 2023

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Suggested change
from operator import index

This import is never used, and it's causing pre-commit to fail.



def octal_to_binary(octal_string: str) -> str:
if not all(char in "01234567." for char in octal_string):
raise ValueError("Non-octal value was passed to the function")
if not octal_string:
raise ValueError("Empty string was passed to the function")
binary_string = ""
dict_of_octal_to_binary = {
"0": "000",
"1": "001",
"2": "010",
"3": "011",
"4": "100",
"5": "101",
"6": "110",
"7": "111",
}
indexl = 0
for i in octal_string:
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Can we do this with a comprehension instead of manually appending to a string?

if i == ".":
binary_string += "."
continue
if binary_string == "":
for char in dict_of_octal_to_binary[i]:
if char != "0" or indexl >= 1:
binary_string += char
indexl += 1
Comment on lines +36 to +40
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Instead of having a special case for handling 0's at the start of the binary string, why not just append to the binary string as normal and handle the starting 0's at the very end?

else:
binary_string += dict_of_octal_to_binary[i]
return binary_string


if __name__ == "__main__":
from doctest import testmod

testmod()