Mix lambda and list comprehension?

[Peter Barth]

Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
---

[lambda x:x+y for y in range(10)][9](2) 11 [lambda x:x+y for y in range(10)][4](2)

11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter

[Raymond Hettinger]

Try this:

[lambda x, y=y:x+y for y in range(10)][4](2)

6
It is important to bind y in a closure at the time
the lambda is defined. Otherwise, y remains unbound
until you invoke the function call. At that time, the most
recent value of y is the last value in the range loop (namely, 9).
Raymond Hettinger

"Peter Barth" <pe*********@t-online.de> wrote in message
news:6f**************************@posting.google.c om...

Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
--- [lambda x:x+y for y in range(10)][9](2) 11 [lambda x:x+y for y in range(10)][4](2)

11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter

[Michele Simionato]

pe*********@t-online.de (Peter Barth) wrote in message news:<6f**************************@posting.google. com>...

Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
---

[lambda x:x+y for y in range(10)][9](2) 11 [lambda x:x+y for y in range(10)][4](2) 11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter

It is a scope issue. The last value for y is used for all
the created lambdas. All lambdas users are bitten by that,
soon or later. The solution is to make y local to the
lambda function, with the optional argument trick:

[lambda x,y=y:x+y for y in range(10)][4](2)

6

Michele

[Peter Barth]

Thanks a lot, works fine.
However, the solution does not really feel "pythonesque".
Is it considered a usability bug or fine as is?
- Peter

"Raymond Hettinger" <vz******@verizon.net> wrote in message news:<hM***************@nwrdny01.gnilink.net>...

Try this:

[lambda x, y=y:x+y for y in range(10)][4](2) 6
It is important to bind y in a closure at the time
the lambda is defined. Otherwise, y remains unbound
until you invoke the function call. At that time, the most
recent value of y is the last value in the range loop (namely, 9).
Raymond Hettinger

"Peter Barth" <pe*********@t-online.de> wrote in message
news:6f**************************@posting.google.c om...

Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
---

>> [lambda x:x+y for y in range(10)][9](2)

11>> [lambda x:x+y for y in range(10)][4](2)

11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter

[Michele Simionato]

bo**@oz.net (Bengt Richter) wrote in message news:<bf**********@216.39.172.122>...

On 15 Jul 2003 05:41:02 -0700, mi**@pitt.edu (Michele Simionato) wrote:

pe*********@t-online.de (Peter Barth) wrote in message news:<6f**************************@posting.google. com>...

Hi,
trying to mix lambda expresions and list comprehension
doesn't seem to work.
---
>>> [lambda x:x+y for y in range(10)][9](2) 11 >>> [lambda x:x+y for y in range(10)][4](2)
11
---
I expected the second expression to return 6.
What did I do wrong? Any hints?
Thanks
- Peter

It is a scope issue. The last value for y is used for all
the created lambdas. All lambdas users are bitten by that,
soon or later. The solution is to make y local to the
lambda function, with the optional argument trick:

> [lambda x,y=y:x+y for y in range(10)][4](2)

6

or you could capture y as constants in the lambdas ;-)

>>> [eval('lambda x:x+%s'%y) for y in range(10)][4](2)

6
Regards,
Bengt Richter

Thanks to God, there is a smile in your post!!

;)

Michele