What is "lambda x=x : ... " ?

[zslevi]

I'm reading this page: http://www.ps.uni-sb.de/~duchier/pyt...inuations.html
and I've found a strange usage of lambda:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?
I thought it bounds the outer variables, so I experimented a little
bit:

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

[zslevi]

I've figured it out, it is default argument.
print y()
gives 13 as result.

It's a bit evil though.
I hope this post will be useful some newbie like i'm now someday :)

On Jan 10, 7:25 pm, "[e-mail]" <[e-mail]:

I'm reading this page:http://www.ps.uni-sb.de/~duchier/pyt...inuations.html
and I've found a strange usage of lambda:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?
I thought it bounds the outer variables, so I experimented a little
bit:

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

[kirby.urner]

You're talking about syntax from the bad old days
when the scope rules were different.

If not too archeological for your tastes, download
and boot a 1.5 and see what happens.

Less empirically, here're some key references:
http://www.python.org/doc/2.2.3/whatsnew/node9.html
http://www.python.org/dev/peps/pep-0227/

The change came in 2.2 with from __future__ support
in 2.1.

Kirby
4D
On Jan 10, 11:25 am, "[e-mail]" <[e-mail]:

I'm reading this page:http://www.ps.uni-sb.de/~duchier/pyt...inuations.html
and I've found a strange usage of lambda:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?
I thought it bounds the outer variables, so I experimented a little
bit:

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

[Tim Chase]

What does "y=y" and "c=c" mean in the lambda function?

the same thing it does in a function definition:

def myfunct(a, b=42, y=3.141):
pass

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

You're mostly correct, as it does pull it out of the local
context. Try

x = 3
y = lambda x=x: x+10
print y(2)
print y()

to get "12" then "13" back.

-tkc

[Fredrik Lundh]

zs****@gmail.com wrote:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?

they bind the argument "y" to the *object* currently referred to by the
outer "y" variable. for example,

y = 10
f = lambda y=y: return y
y = 11

calling f() will return 10 no matter what the outer "y" is set to.

in contrast, if you do

y = 10
f = lambda: y
y = 11

calling f() will return whatever "y" is set to at the time of the call.

or in other words, default arguments bind to values, free variables bind
to names.

I thought it bounds the outer variables, so I experimented a little
bit:

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

it does, but you're overriding the bound value by passing in a value. try:

x = 3
y = lambda x=x : x+10
y()
x = 10
y()

instead.

</F>

[Mike Meyer]

On Thu, 10 Jan 2008 10:25:27 -0800 (PST) "zs****@gmail.com" <le*******@gmail.comwrote:

I'm reading this page: http://www.ps.uni-sb.de/~duchier/pyt...inuations.html
and I've found a strange usage of lambda:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?

Older versions of python didn't make variables in an outer scope
visible in the inner scope. This was the standard idiom to work
around that.

<mike
--
Mike Meyer <mw*@mired.org> http://www.mired.org/consulting.html
Independent Network/Unix/Perforce consultant, email for more information.

[Fredrik Lundh]

Mike Meyer wrote:

>What does "y=y" and "c=c" mean in the lambda function?

Older versions of python didn't make variables in an outer scope
visible in the inner scope. This was the standard idiom to work
around that.

lexically scoped free variables and object binding are two different
things, and have different semantics. the former does not always
replace the latter.

</F>

[Mike Meyer]

On Thu, 10 Jan 2008 19:59:23 +0100 Fredrik Lundh <fr*****@pythonware.comwrote:

Mike Meyer wrote:

What does "y=y" and "c=c" mean in the lambda function?

Older versions of python didn't make variables in an outer scope
visible in the inner scope. This was the standard idiom to work
around that.

lexically scoped free variables and object binding are two different
things, and have different semantics. the former does not always
replace the latter.

And?

<mike

--
Mike Meyer <mw*@mired.org> http://www.mired.org/consulting.html
Independent Network/Unix/Perforce consultant, email for more information.

[Dustan]

On Jan 10, 12:36 pm, "[e-mail]" <[e-mail]:

I've figured it out, it is default argument.
print y()
gives 13 as result.

It's a bit evil though.

Why? It's the same syntax as with functions:

x=3
def y(x=x):
return x+10

print y(2) # prints 12
print y() # prints 13

[Scott David Daniels]

zs****@gmail.com wrote:

I'm reading this page: http://www.ps.uni-sb.de/~duchier/pyt...inuations.html
and I've found a strange usage of lambda:

####################
Now, CPS would transform the baz function above into:

def baz(x,y,c):
mul(2,x,lambda v,y=y,c=c: add(v,y,c))

###################

What does "y=y" and "c=c" mean in the lambda function?
I thought it bounds the outer variables, so I experimented a little
bit:

#################
x = 3
y = lambda x=x : x+10

print y(2)
##################

It prints 12, so it doesn't bind the variable in the outer scope.

Primary use:
funcs = [lambda x=x: x+2 for x in range(10)]
print funcs[3]()

_or_

funcs = []
for x in range(10):
funcs.append(lambda x=x: x+2)
print funcs[3]()
Try these w/o the default binding.

--Scott David Daniels
Sc***********@Acm.Org