[Tutor] How can I have type "function" in my script?

[xancorreu]

Al 07/05/12 21:07, En/na Dave Angel ha escrit:
> On 05/07/2012 02:24 PM, xancorreu wrote:
>> Hi,
>>
>> I have this script:
>>
>> from types import *
>>
> Bad idea.  Once you do that, you can silently overwrite globals in your
> own module with stuff that the current version of types happens to have
> in it.  Besides, it then becomes very hard to read your program and
> figure out which names you really did want to import.
>
> If you're just getting one or two names, such as in your case, better
> just do
>      import types
>
>> class Tag:
>>
>>      def __init__(self, nom, tipus, valor):
>>          self.nom = nom
>>          self.tipus = tipus
>>          self.valor = valor
>>
>>      def __str__(self):
>>          return "Nom: " + str(self.nom) + ", Tipus: " +
>> str(self.tipus)  + ", Valor: " + str(self.valor)
>>
>>
>> def main():
>>      a = Tag("descripció", str, "primera tasca")
>>      b = Tag("unmes", str, lambda x: x+1)
>>      print(a)
>>      print(b)
>>
>> if __name__ == '__main__':
>>          main()
>>
>>
>> All is ok, but when I substitute b = Tag("unmes", str, lambda x: x+1)
>> for b = Tag("unmes", function, lambda x: x+1) I receive an error that
>> function is not globally defined variable. How can I say that function
>> is the types.function? (the type of lambda x: x+1)
>>
> Where's the stack trace and the exact error message?
>
> types.function is undefined.  The types module does not expose a name
> called 'function,' at least not in python 3.2
>
> The type of a lambda is<class 'function'>, so it's not clear what you
> really want.
>
> Why don't you show the program as you actually run it (perhaps with both
> versions of the b= assignment), and the output and stack trace you got.
> Then explain just what you'd hoped to get, as output.
>

This is the code:

class Tag:

     def __init__(self, nom, tipus, valor):
         self.nom = nom
         self.tipus = tipus
         self.valor = valor

     def __str__(self):
         return "Nom: " + str(self.nom) + ", Tipus: " + str(self.tipus)  
+ ", Valor: " + str(self.valor)

class Task:
     _nombre = 0

     def __init__(self):
         self.tags = []
         Task._nombre = Task._nombre + 1
         self.num = Task._nombre


     def __str__(self):
         return "Número: " + str(self.num) + ", Tags: " + str(self.tags)

def main():
     a = Tag("descripció", str, "primera tasca")
     b = Tag("unmes", str, lambda x: x+1)
     c = Tag("twice", type(lambda: x: x), lambda x: 2*x)
     # en comptes de str ha de ser lambda
     print(a)
     print(b)
     print(b.valor(2))
     t = Task()
     print("Tasca 1:", t)
     t2 = Task()
     print("Tasca 2:", t2)

if __name__ == '__main__':
         main()


and it fails here:

$ python3 tasques.py
   File "tasques.py", line 26
     c = Tag("twice", type(lambda: x: x), lambda x: 2*x)
                                    ^
SyntaxError: invalid syntax


Really, I want to specify "manually" the type of lambda, but it does not 
work. How to do that?


Thanks,

>> I use python3
>>
>>
>> Thanks in advance,
>> Xan.