Python Print Octal Without ‘0o’

Problem Formulation

If you print a hexadecimal number, Python uses the prefix '0o' to indicate that it’s a number in the octal system and not in the decimal system like normal integers.

# 0o52

However, if you already know that the output numbers are octal, you don’t necessarily need the '0o' prefix.

How to print oct numbers without the '0o' prefix?

Method 1: Slicing

To skip the prefix, use slicing and start with index 2 on the octal string. For example, to skip the prefix '0o' on the result of x = oct(42) ='0o52', use the slicing operation x[2:] that results in just the octal number '52' without the prefix '0o'.

x = oct(42)

# 0o52

# 52

Feel free to dive into the oct() built-in function in this video tutorial:

But what if you actually want to replace the prefix '0o' with the prefix '00' so that the resulting string has the same length?

Method 2: Slicing + zfill()

The Python string.zfill() method fills the string from the left with '0' characters. In combination with slicing from the third character, you can easily construct an octal string without leading '0o' characters and with leading '0' characters up to the length passed into the string.zfill(length) method.

# 0052

Alternatively, if you want to create a string with 8 characters, use string.zfill(8):

# 00000042

You can learn more about zfill() in this video about Python string methods:

Method 3: Negative Octal Numbers

If you need to handle negative octal numbers, the above methods do not work because the oct number now needs to replace the second and third character '-0o'. For example, the hexadecimal number hex(-42) is '-0o52'. You cannot simply skip the first two characters to obtain the correct result, can you? At the same time, if you always skipped or replaced the second and third characters, it wouldn’t work for positive numbers either. So what to do?

To print a positive or negative octal without the '0o' or '-0o' prefixes, you can simply use the string.replace('o', '0') method and replace each occurrence of alphabetical 'o' with numerical '0'. The resulting string is mathematically correct because leading '0's don’t change the value of the number.

# Negative Octal
print(oct(-42).replace('o', '0'))
# -0052

# Positive Octal
print(oct(42).replace('o', '0'))
# 0052

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