[Python-Dev] Function Operators

2010-07-18 Thread Christopher Olah 
Dear python-dev,

In mathematical notation, f*g = z-f(g(z)) and f^n = f*f*f... (n
times). I often run into situations in python where such operators
could result in cleaner code. Eventually, I decided to implement it
myself and see how it worked in practice.

However, my intuitive implementation [1] doesn't seem to work. In
particular, despite what it says in function's documentation, function
does not seem to be in __builtin__. Furthermore, when I try to
implement this through type(f) (where f is a function) I get invalid
syntax errors.

I hope I haven't made some trivial error; I'm rather inexperienced as
a pythonist.

Christopher Olah


[1] Sketch:

def __builtin__.function.__mul__(self, f):
return lambda x: self(f(x))

def __builtin__.function.__pow__(self, n):
return lambda x: reduce(lambda a,b: [f for i in range(n)]+[x])
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Re: [Python-Dev] Function Operators

2010-07-18 Thread Reid Kleckner 
Usual disclaimer: python-dev is for the development *of* python, not
*with*.  See python-list, etc.

That said, def declares new functions or methods, so you can't put
arbitrary expressions in there like type(f).__mul__ .

You can usually assign to things like that though, but in this case
you run into trouble, as shown below:

 def func(): pass
...
 type(func)
class 'function'
 def compose(f, g):
... return lambda x: f(g(x))
...
 type(func).__mul__ = compose
Traceback (most recent call last):
  File stdin, line 1, in module
TypeError: can't set attributes of built-in/extension type 'function'

As the interpreter says, it doesn't like people mucking with operator
slots on built in types.

Finally, if you like coding in that very functional style, I'd
recommend Haskell or other ML derived languages.  Python doesn't
support that programming style very well by choice.

Reid

On Sun, Jul 18, 2010 at 8:34 AM, Christopher Olah
christopherolah...@gmail.com wrote:
 Dear python-dev,

 In mathematical notation, f*g = z-f(g(z)) and f^n = f*f*f... (n
 times). I often run into situations in python where such operators
 could result in cleaner code. Eventually, I decided to implement it
 myself and see how it worked in practice.

 However, my intuitive implementation [1] doesn't seem to work. In
 particular, despite what it says in function's documentation, function
 does not seem to be in __builtin__. Furthermore, when I try to
 implement this through type(f) (where f is a function) I get invalid
 syntax errors.

 I hope I haven't made some trivial error; I'm rather inexperienced as
 a pythonist.

 Christopher Olah


 [1] Sketch:

 def __builtin__.function.__mul__(self, f):
    return lambda x: self(f(x))

 def __builtin__.function.__pow__(self, n):
    return lambda x: reduce(lambda a,b: [f for i in range(n)]+[x])
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Re: [Python-Dev] Function Operators

2010-07-18 Thread Thomas Jollans 
On 07/18/2010 05:52 PM, Reid Kleckner wrote:
 Usual disclaimer: python-dev is for the development *of* python, not
 *with*.  See python-list, etc.

Moving to python-list. Please keep discussion there.

 
 That said, def declares new functions or methods, so you can't put
 arbitrary expressions in there like type(f).__mul__ .
 
 You can usually assign to things like that though, but in this case
 you run into trouble, as shown below:
 
 def func(): pass
 ...
 type(func)
 class 'function'
 def compose(f, g):
 ... return lambda x: f(g(x))
 ...
 type(func).__mul__ = compose
 Traceback (most recent call last):
   File stdin, line 1, in module
 TypeError: can't set attributes of built-in/extension type 'function'
 
 As the interpreter says, it doesn't like people mucking with operator
 slots on built in types.
 
 Finally, if you like coding in that very functional style, I'd
 recommend Haskell or other ML derived languages.  Python doesn't
 support that programming style very well by choice.
 
 Reid
 
 On Sun, Jul 18, 2010 at 8:34 AM, Christopher Olah
 christopherolah...@gmail.com wrote:
 Dear python-dev,

 In mathematical notation, f*g = z-f(g(z)) and f^n = f*f*f... (n
 times). I often run into situations in python where such operators
 could result in cleaner code. Eventually, I decided to implement it
 myself and see how it worked in practice.

 However, my intuitive implementation [1] doesn't seem to work. In
 particular, despite what it says in function's documentation, function
 does not seem to be in __builtin__. Furthermore, when I try to
 implement this through type(f) (where f is a function) I get invalid
 syntax errors.

 I hope I haven't made some trivial error; I'm rather inexperienced as
 a pythonist.

 Christopher Olah


 [1] Sketch:

 def __builtin__.function.__mul__(self, f):
return lambda x: self(f(x))

 def __builtin__.function.__pow__(self, n):
return lambda x: reduce(lambda a,b: [f for i in range(n)]+[x])


As Reid explained, you can't just muck around with built-in types like
that. However, you can use a different type.

If you're not familiar with Python decorators, look them up, and then
have a look at this simple implementation of what you were looking for:

 class mfunc:
... def __init__(self, func):
... self.func = func
... self.__doc__ = func.__doc__
... self.__name__ = func.__name__
... def __call__(self, *args, **kwargs):
... return self.func(*args, **kwargs)
... def __mul__(self, f2):
... @mfunc
... def composite(*a, **kwa):
... return self.func(f2(*a, **kwa))
... return composite
... def __pow__(self, n):
... if n  1:
... raise ValueError(n)
... elif n == 1:
... return self.func
... else:
... return self * (self ** (n-1))
...
 @mfunc
... def square(x): return x*x
...
 @mfunc
... def twice(x): return 2*x
...
 (square*twice)(1.5)
9.0
 addthree = mfunc(lambda x: x+3)
 addfifteen = (addthree ** 5)
 addfifteen(0)
15



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