Convert a string to integer with decimal in Python

Question

I have a string in the format: \'nn.nnnnn\' in Python, and I\'d like to convert it to an integer.

Direct conversion fails:

>>> s = \'23.4567         


        

Answer 1

How about this?

>>> s = '23.45678'
>>> int(float(s))
23

Or...

>>> int(Decimal(s))
23

Or...

>>> int(s.split('.')[0])
23

I doubt it's going to get much simpler than that, I'm afraid. Just accept it and move on.

Answer 2

What sort of rounding behavior do you want? Do you 2.67 to turn into 3, or 2. If you want to use rounding, try this:

s = '234.67'
i = int(round(float(s)))

Otherwise, just do:

s = '234.67'
i = int(float(s))

Answer 3

"Convert" only makes sense when you change from one data type to another without loss of fidelity. The number represented by the string is a float and will lose precision upon being forced into an int.

You want to round instead, probably (I hope that the numbers don't represent currency because then rounding gets a whole lot more complicated).

round(float('23.45678'))

Answer 4

>>> s = '23.45678'
>>> int(float(s))
23
>>> int(round(float(s)))
23
>>> s = '23.54678'
>>> int(float(s))
23
>>> int(round(float(s)))
24

You don't specify if you want rounding or not...

Answer 5

round(float("123.789"))

will give you an integer value, but a float type. With Python's duck typing, however, the actual type is usually not very relevant. This will also round the value, which you might not want. Replace 'round' with 'int' and you'll have it just truncated and an actual int. Like this:

int(float("123.789"))

But, again, actual 'type' is usually not that important.

Answer 6

The expression int(float(s)) mentioned by others is the best if you want to truncate the value. If you want rounding, using int(round(float(s)) if the round algorithm matches what you want (see the round documentation), otherwise you should use Decimal and one if its rounding algorithms.