Aaron Brady
3/2/2008 3:22:00 AM
On Mar 1, 8:50 pm, Michael Torrie <torr...@gmail.com> wrote:
> I need to use a lambda expression to bind some extra contextual data
> (should be constant after it's computed) to a call to a function. I had
> originally thought I could use something like this demo (but useless) code:
>
> funcs=[]
>
> def testfunc(a,b):
> print "%d, %d" % (a,b)
>
> for x in xrange(10):
> funcs.append(lambda p: testfunc(x+2,p))
>
> Now what I'd like is to call, for example, funcs[0](4) and it should
> print out "2,4". In other words I'd like the value of x+2 be encoded
> into the lambda somehow, for funcs[x]. However the disassembly shows
> this, which is reasonable, but not what I need:
>
> >>> dis.dis(funcs[0])
>
> 2 0 LOAD_GLOBAL 0 (testfunc)
> 3 LOAD_GLOBAL 1 (x)
> 6 LOAD_CONST 0 (2)
> 9 BINARY_ADD
> 10 LOAD_FAST 0 (p)
> 13 CALL_FUNCTION 2
> 16 RETURN_VALUE
>
> The LOAD_GLOBAL 1 (x) line is definitely a problem. For one it refers
> to a variable that won't be in scope, should this lambda be called from
> some stack frame not descended from the one where I defined it.
>
> So how can I create a lambda expression that calculates a constant based
> on an expression, rather than referring to the object itself? Can it be
> done?
>
> Michael
I hate that. Especially since ints are immutable.
>>> from functools import partial
>>> funcs= [ partial( print, x ) for x in range( 10 ) ]
>>> for func in funcs:
... func()
...
0
1
2
3
4
5
6
7
8
9
>>>
takes advantage of 2.5+ functools.partial and 3.0 print. You can use
sys.stdout instead of print in the example in 2.5, but without
partial, you're at:
class Val:
def __init__( self, val ):
self.val= val
def __call__( self ):
self.val+= 10
val= Val( 20 )
val()
. There's also a CellMaker implementation somewhere on Py-Ideas, but
it was outlawed in 18 provinces, and it just evaluates and compiles a
string anyway.
for x in xrange(10):
fun= exec( 'lambda p: testfunc(%i+2,p)'% x )
funcs.append( fun )