Python int() of a string that is a float number

Question

In all probability a stupid question, but I was wondering why python can't make a integer out of a string that is actually a float number.

>>> int(1.0)
1
>>> int(float('1.0'))
1

But

>>> int('1.0')
Traceback (most recent call last):
  File "<pyshell#17>", line 1, in <module>
    int('1.0')
ValueError: invalid literal for int() with base 10: '1.0'

Can anyone clarify why it cant be done in one step?

Answer 1

Can anyone clarify why it cant be done in one step?

To quote the Zen of Python: Explicit is better than implicit.

I was wondering why python can't make a integer out of a string that is actually a float number.

In line with Python's philosophy, if a string contains a float number which you then want to turn into an int, it's your responsibility to spell things out. The language won't try and second-guess your intentions.

Answer 2

Straight from the docs about int :

If x is not a number or if base is given, then x must be a string or Unicode object representing an integer literal in radix base.

And here is how an integer literal in radix base is defined:

longinteger    ::=  integer ("l" | "L")
integer        ::=  decimalinteger | octinteger | hexinteger | bininteger
decimalinteger ::=  nonzerodigit digit* | "0"
octinteger     ::=  "0" ("o" | "O") octdigit+ | "0" octdigit+
hexinteger     ::=  "0" ("x" | "X") hexdigit+
bininteger     ::=  "0" ("b" | "B") bindigit+
nonzerodigit   ::=  "1"..."9"
octdigit       ::=  "0"..."7"
bindigit       ::=  "0" | "1"
hexdigit       ::=  digit | "a"..."f" | "A"..."F"

As you can see . is not present in the lexical definition.

more on int and integer literals

Also worth mentioning that what you wish is naturally achievable in languages of static typing , while Python employs dynamic typing