Giorgio wrote:
I have an update:
I can easily undertand why this example doesn't work:
def nochange(x):
x = 0
y = 1
nochange(y)
print y # Prints out 1
X is a local variable, and only gets modified in the function, that
doesn't return any value.
But it's very difficult for me to understand WHY this works:
def change(some_list):
some_list[1] = 4
x = [1,2,3]
change(x)
print x # Prints out [1,4,3]
some_list is a "local" list, isn't it? Maybe i can't have lists that
are only existing in a function?
Thankyou all
2010/2/22 Kent Johnson <ken...@tds.net <mailto:ken...@tds.net>>
On Mon, Feb 22, 2010 at 9:13 AM, Giorgio
<anothernetfel...@gmail.com <mailto:anothernetfel...@gmail.com>>
wrote:
> And, i have some difficulties understanding the other "strange"
example in
> that howto. Just scroll down to: "However, the point is that the
value
> of x is picked up from the environment at the time when the
function is
> defined. How is this useful? Let’s take an example — a function
which
> composes two other functions."
> He is working on a function that compose other 2 functions. This
is the
> final solution
> def compose(fun1, fun2):
> def inner(x, fun1=fun1, fun2=fun2):
> return fun1(fun2(x))
> return inner
> But also tries to explain why this example:
> # Wrong version
> def compose(fun1, fun2):
> def inner(x):
> return fun1(fun2(x))
> return inner
> def fun1(x):
> return x + " world!"
> def fun2(x):
> return "Hello,"
> sincos = compose(sin,cos) # Using the wrong version
> x = sincos(3)
> Won't work. Now, the problem is that the "inner" function gets
fun1 and fun2
> from other 2 functions.
> My question is: why? inner is a sub-function of compose, where
fun1 and fun2
> are defined.
It does work:
In [6]: def compose(fun1, fun2):
...: def inner(x):
...: return fun1(fun2(x))
...: return inner
...:
In [7]: def fun1(x):
...: return x + " world!"
...:
In [8]: def fun2(x):
...: return "Hello,"
...:
In [9]: from math import sin, cos
In [10]: sincos = compose(sin,cos) # Using the wrong version
In [11]:
In [12]: x = sincos(3)
In [13]:
In [14]: x
Out[14]: -0.8360218615377305
That is a very old example, from python 2.1 or before where nested
scopes were not supported. See the note "A Note About Python 2.1 and
Nested Scopes" - that is now the default behaviour.
Kent
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Email: anothernetfel...@gmail.com <mailto:anothernetfel...@gmail.com>
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