* operator--as in *args?

[7stud]

Hi,

I can't find any documentation for the * operator when applied in
front of a name. Is it a pointer?

What about the @ operator?

Are there python names for these operators that would make searching
for documentation on them more fruitful?

Thanks

[Dustan]

On Mar 18, 3:53 pm, "7stud" <bbxx789_0...@y ahoo.comwrote:

Hi,

I can't find any documentation for the * operator when applied in
front of a name. Is it a pointer?

What about the @ operator?

Are there python names for these operators that would make searching
for documentation on them more fruitful?

Thanks

For the *star, see apply here:
http://docs.python.org/lib/non-essen...-in-funcs.html

For example:
aFunction(*(1,2 ,3))
is equivalent to:
aFunction(1,2,3 )

For the @ symbol, I assume you mean function decorators:
http://docs.python.org/ref/function.html

For example:
@someDecorator
def someFunc(): pass
is equivalent to:
def someFunc(): pass
someFunc = someDecorator(s omeFunc)

[7stud]

On Mar 18, 3:40 pm, "Dustan" <DustanGro...@g mail.comwrote:

For example:
aFunction(*(1,2 ,3))
is equivalent to:
aFunction(1,2,3 )

That's the context I've seen it in, but written like this:

someFunc(*args)

I played around with it a little bit, and it appears the * operator
unpacks a list, tuple, or dictionary so that each element of the
container gets assigned to a different parameter variable. Although
with a dictionary, only the keys appear to be assigned to the
parameter variables, e.g.:

def g(a,b,c):
print a, b, c

dict = {"x":10, "y":20, "z":30}
g(*dict)

Is that right?

[Alex Martelli]

7stud <bb**********@y ahoo.comwrote:

On Mar 18, 3:40 pm, "Dustan" <DustanGro...@g mail.comwrote:

For example:
aFunction(*(1,2 ,3))
is equivalent to:
aFunction(1,2,3 )

That's the context I've seen it in, but written like this:

someFunc(*args)

I played around with it a little bit, and it appears the * operator
unpacks a list, tuple, or dictionary so that each element of the
container gets assigned to a different parameter variable. Although
with a dictionary, only the keys appear to be assigned to the
parameter variables, e.g.:

def g(a,b,c):
print a, b, c

dict = {"x":10, "y":20, "z":30}
g(*dict)

Is that right?

As far as it goes, yes. More generally, with any iterable x, the *x
construct in function call will pass as positional arguments exactly
those items which (e.g.) would be printed by the loop:
for item in x: print x

[[this applies to iterators, generators, genexps, and any other iterable
you may care to name -- not just lists, tuples, dicts, but also sets,
files open for reading [the items are the lines], etc, etc]].
Alex

[Gabriel Genellina]

En Sun, 18 Mar 2007 18:40:32 -0300, Dustan <Du**********@g mail.com>
escribió:

>I can't find any documentation for the * operator when applied in
front of a name. Is it a pointer?

For the *star, see apply here:
http://docs.python.org/lib/non-essen...-in-funcs.html

Also, you can read this section on the Tutorial:
http://docs.python.org/tut/node6.htm...00000000000000
or the more technical Language Reference:
http://docs.python.org/ref/calls.html and
http://docs.python.org/ref/function.html

>Are there python names for these operators that would make searching
for documentation on them more fruitful?

Uhhhm, none that I can think of :(

--
Gabriel Genellina

[Gabriel Genellina]

En Sun, 18 Mar 2007 22:21:41 -0300, Alex Martelli <al***@mac.come scribió:

7stud <bb**********@y ahoo.comwrote:

>I played around with it a little bit, and it appears the * operator
unpacks a list, tuple, or dictionary so that each element of the
container gets assigned to a different parameter variable. Although
with a dictionary, only the keys appear to be assigned to the
parameter variables, e.g.:

def g(a,b,c):
print a, b, c

dict = {"x":10, "y":20, "z":30}
g(*dict)

Is that right?

As far as it goes, yes. More generally, with any iterable x, the *x
construct in function call will pass as positional arguments exactly
those items which (e.g.) would be printed by the loop:
for item in x: print x

[[this applies to iterators, generators, genexps, and any other iterable
you may care to name -- not just lists, tuples, dicts, but also sets,
files open for reading [the items are the lines], etc, etc]].

But the language reference says "sequence", not "iterable"
(http://docs.python.org/ref/calls.html) and a dictionary is not a
sequence. With Python 2.1 it was an error; it is not with 2.3 (I can't
test with 2.2 right now)

Python 2.1.3 (#35, Apr 8 2002, 17:47:50) [MSC 32 bit (Intel)] on win32
Type "copyright" , "credits" or "license" for more information.

>>def f(*args, **kw):

.... print "args",args
.... print "kw",kw
....

>>d = {"a":1, "b":2, "c":3}
f(**d)

args ()
kw {'b': 2, 'c': 3, 'a': 1}

>>f(*d)

Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: f() argument after * must be a sequence

If allowing f(*d) is actually the intended behavior, maybe the wording in
the reference should be updated. If not, f(*d) should still raise an error.

--
Gabriel Genellina

[Alex Martelli]

Gabriel Genellina <ga*******@yaho o.com.arwrote:
...

As far as it goes, yes. More generally, with any iterable x, the *x
construct in function call will pass as positional arguments exactly
those items which (e.g.) would be printed by the loop:
for item in x: print x

[[this applies to iterators, generators, genexps, and any other iterable
you may care to name -- not just lists, tuples, dicts, but also sets,
files open for reading [the items are the lines], etc, etc]].

But the language reference says "sequence", not "iterable"

Yes, Python's docs often say "sequence" where they in fact mean
"iterable". I count (on all the .tex files under Doc/ in a current SVN
tree) 854 occurrences of "sequence" versus 251 occurrences of
"iterable", and I suspect the correct ratio would be rather close to the
reverse:-).

If allowing f(*d) is actually the intended behavior, maybe the wording in
the reference should be updated. If not, f(*d) should still raise an error.

Patches to the docs will doubtlessly be welcome (though fixing 1 out of
a suspected 600 or so misuses won't make a huge difference, it's still
definitely better than nothing:-).
Alex

[7stud]

On Mar 18, 7:52 pm, "Gabriel Genellina" <[e-mail].a r>
wrote:

def f(*args, **kw):

... print "args",args
... print "kw",kw
...>>d = {"a":1, "b":2, "c":3}

>f(**d)

Whoa! **? And applied to a function parameter? Back to the drawing
board.

On Mar 18, 7:21 pm, [e-mail] (Alex Martelli) wrote:

More generally, with any iterable x, the *x
construct in function call will pass as positional arguments exactly
those items which (e.g.) would be printed by the loop:
for item in x: print x

Thanks.

On Mar 18, 7:52 pm, "Gabriel Genellina" <[e-mail].a r>
wrote:

def f(*args, **kw):

... print "args",args
... print "kw",kw
...>>d = {"a":1, "b":2, "c":3}

>f(**d)

Whoa! **? And applied to a function parameter name? Back to the
drawing board.

The following is what I discovered for anyone else that is interested:

More on Defining Functions(GvR tutorial, section 4.7)
--Keyword Argument Lists
--Arbitrary Argument Lists
--Unpacking Argument Lists

1) Applying ** to a function parameter:

def g(a, b, **xtraArgs):
print xtraArgs
g(b=20, c="big", a="hello", d=10)

#output: {'c': 'big', 'd': 10}

**xtraArgs will be a dictionary of all arguments sent to the function
with the syntax "keyword=va lue" if the keyword does not match any of
the function parameter names.

2)Applying * to a function parameter:

def g(a, *xtraArgs):
print xtraArgs
g(10, 20, 30, 40)

#output: (20, 30, 40)

*xtraArgs will be a tuple containing all arguments sent to the
function with the syntax "value" that do not match any function
parameter.

3) A function parameter with the syntax *xtraArgs1 must come before a
function parameter with the syntax **xtraArgs2:

def g(a, b, *xtraArgs1, **xtraArgs2):
print xtraArgs1
print xtraArgs2

g(10, 20, 30, 35, d=40, e="hello")

#(30, 35)
#{'e': 'hello', 'd': 40}

4) Using * to unpack a list, tuple, etc.(any iterable collection--see
earlier posts):

def g(a, b, c):
print a,b,c

tpl = (1, 2, 3)
g(*tpl)

#output: 1 2 3

5) Using ** to unpack dictionaries:

def g(a, b, c):
print a,b,c

dict = {"b":"world" , "c":"goodby e", "a":"hello" }
g(**dict)

#output: hello world goodbye