parameters to lambda's executed at run time.

[wyleu]

I'm trying to supply parameters to a function that is called at a
later time as in the code below:

llist = []

for item in range(5):
llist.append(lambda: func(item))

def func(item):
print item

for thing in llist:
thing()

which produces the result

IDLE 1.2.1

>>================================ RESTART ================================

<function <lambdaat 0xb716356c>
<function <lambdaat 0xb71635a4>
<function <lambdaat 0xb71635dc>
<function <lambdaat 0xb7163614>
<function <lambdaat 0xb716364c>

>>================================ RESTART ================================

4
4
4
4
4

>>>

How can one allocate a different parameter to each instance of the
function rather than all of them getting the final value of the loop?

[Diez B. Roggisch]

wyleu wrote:

I'm trying to supply parameters to a function that is called at a
later time as in the code below:

llist = []

for item in range(5):
llist.append(lambda: func(item))

def func(item):
print item

for thing in llist:
thing()

which produces the result

IDLE 1.2.1

>>>================================ RESTART
================================

<function <lambdaat 0xb716356c>
<function <lambdaat 0xb71635a4>
<function <lambdaat 0xb71635dc>
<function <lambdaat 0xb7163614>
<function <lambdaat 0xb716364c>

>>>================================ RESTART
================================

4
4
4
4
4

>>>>

How can one allocate a different parameter to each instance of the
function rather than all of them getting the final value of the loop?

That's a FAQ. Python creates a closure for you that will retain the last
value bound. To prevent that, you need to create a named paramter like
this:

lambda item=item: func(item)

That will bind the current item value at the lambda creation time.

Diez

[Boris Borcic]

One way :

>>from functools import partial
def func(item) : print item

>>llist = [partial(func,item) for item in range(5)]
for thing in llist : thing()

0
1
2
3
4
wyleu wrote:

I'm trying to supply parameters to a function that is called at a
later time as in the code below:

llist = []

for item in range(5):
llist.append(lambda: func(item))

def func(item):
print item

for thing in llist:
thing()

which produces the result

IDLE 1.2.1

>>>================================ RESTART ================================

<function <lambdaat 0xb716356c>
<function <lambdaat 0xb71635a4>
<function <lambdaat 0xb71635dc>
<function <lambdaat 0xb7163614>
<function <lambdaat 0xb716364c>

>>>================================ RESTART ================================

4
4
4
4
4

How can one allocate a different parameter to each instance of the
function rather than all of them getting the final value of the loop?

[Marco Mariani]

Boris Borcic wrote:

One way :

>>from functools import partial
>>def func(item) : print item

>>llist = [partial(func,item) for item in range(5)]
>>for thing in llist : thing()

0
1
2
3
4

Another way:
class Func(object):
def __init__(self, item):
self.item = item
def __call__(self):
print self.item

llist = [Func(item) for item in range(5)]
for thing in llist: thing()

[castironpi]

On May 6, 5:17*am, "Diez B. Roggisch" <[e-mail]:

wyleu wrote:

I'm trying to supply parameters to a function that is called at a
later time as in the code below:

llist = []

for item in range(5):
* * llist.append(lambda: func(item))

def func(item):
* * print item

for thing in llist:
* * thing()

which produces the result

IDLE 1.2.1

>>================================ RESTART
================================

<function <lambdaat 0xb716356c>
<function <lambdaat 0xb71635a4>
<function <lambdaat 0xb71635dc>
<function <lambdaat 0xb7163614>
<function <lambdaat 0xb716364c>

>>================================ RESTART
================================

4
4
4
4
4

How can one allocate a different parameter to each instance of the
function rather than all of them getting the final value of the loop?

That's a FAQ. Python creates a closure for you that will retain the last
value bound. To prevent that, you need to create a named paramter like
this:

lambda item=item: func(item)

That will bind the current item value at the lambda creation time.

Diez- Hide quoted text -

- Show quoted text -

I am getting lambda creation-time bindings on 2.5.

>>g= [ lambda h= i: h for i in range( 10 ) ]
[ e( ) for e in g ]

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

>>g= [ lambda: i for i in range( 10 ) ]
[ e( ) for e in g ]

[9, 9, 9, 9, 9, 9, 9, 9, 9, 9]