如何在不使用库、除法、平方根、条件或内置函数的情况下找到 integer 或在 Python 中浮动的绝对值

Question

I wanted to know if it was possible to do such a thing in Python 3. Using abs() is of course not allowed. No importing is allowed as well.

Answer 1

{1e999: x, -1e999: -x}[x*1e999]

I'm serious. This works.

The 1e999 is interpreted as a float, but it overflows the available precision, so it's inf . But infinity can be positive or negative. By multiplying x with infinity, we can coerce it into one of two values based on its sign, without using a condition operator. Then we can select x or its negation from a lookup table.

It's equivalent to the following, but doesn't require the import:

from math import inf

{inf: x, -inf: -x}[x*inf]

Answer 2

If you consider the built-in classes ( int , str , etc) to be built-in "functions". You could do something like the following

num.__class__(('%s' % num).lstrip('-'))

Convert the number to a string, strip the negative sign and then use the original numbers class to convert back

Answer 3

Convert to a string with a sign and slice off the first character. The __class__ is not technically a method, so no methods.

x.__class__(f'{x:+}'[1:])

Answer 4

Assume the first character of the string representation is '-' and return the negation. But don't negate when you're wrong.

def abs(x):
    try:
        return {'-': -x}[f'{x}'[0]]
    except:
        return x

Answer 5

Format a string to always include the sign and then look up its first character in a table.

{'-': -x, '+': x}[f'{x:+}'[0]]

Answer 6

The abs() builtin just calls the .__abs__() method. No imports, no builtins, no conditions, no division, no square roots. And no abs() .

x.__abs__()