ValueError: invalid literal for int() with base 10: '0.00'

Question

I have a string in Python like that:

l = "0.00 0.00"

And I want to convert it in a list of two numbers.

The following instruction does not work:

int(l.strip(" \n").split(" ")[0])

Apparently the function int() can convert string like 0 or 00 to an int, but it does not work with 0.0.

Is there a way to convert 0.0?

A.

What would you expect 0.9 to be? What about 9.9? Truncate [0, 9] or round [1, 10]? — dawg
– dawg, Commented May 26, 2014 at 17:30

Andrey · Accepted Answer · 2014-05-26 16:53:20Z

11

The easiest way to first convert to Decimal:

from decimal import Decimal
int(Decimal('0.00'))

if you are sure that fractional part is always zero then faster would be to use float

int(float('0.00'))

answered May 26, 2014 at 16:53

Andrey

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P0W · Accepted Answer · 2014-05-26 16:58:10Z

0

For floating point conversion from a string you can use following, here assumption is that you need int for integral values in string

 def conv(st):
    try:
        return int(st)
    except ValueError: #If you get a ValueError
        return float(st)

 conv(l.strip(" \n").split(" ")[0])

answered May 26, 2014 at 16:58

P0W

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