can this be done without eval/exec?

[Schüle Daniel]

Hello group,
[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... lst.append(eval ("lambda:%i" % i))
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> lst[1]()[/color][/color][/color]
1[color=blue][color=green][color=darkred]
>>> lst[9]()[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]
[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... exec "tmp = lambda:%i" % i # assignment is not expression
.... lst.append(tmp)
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> lst[1]()[/color][/color][/color]
1[color=blue][color=green][color=darkred]
>>> lst[9]()[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

and now the obvious one (as I thought at first)
[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... lst.append(lamb da:i)
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>> i[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

I think I understand where the problem comes from
lambda:i seems not to be fully evalutated
it just binds object with name i and not the value of i
thus lst[0]() is not 0

are there other solutions to this problem
without use of eval or exec?

Regards, Daniel

[Chris Mellon]

Re: can this be done without eval/exec?

On 4/26/06, Schüle Daniel <uval@rz.uni-karlsruhe.de> wrote:[color=blue]
> Hello group,
>[color=green][color=darkred]
> >>> lst=[]
> >>> for i in range(10):[/color][/color]
> ... lst.append(eval ("lambda:%i" % i))
> ...[color=green][color=darkred]
> >>> lst[0]()[/color][/color]
> 0[color=green][color=darkred]
> >>> lst[1]()[/color][/color]
> 1[color=green][color=darkred]
> >>> lst[9]()[/color][/color]
> 9[color=green][color=darkred]
> >>>[/color][/color]
>[color=green][color=darkred]
> >>> lst=[]
> >>> for i in range(10):[/color][/color]
> ... exec "tmp = lambda:%i" % i # assignment is not expression
> ... lst.append(tmp)
> ...[color=green][color=darkred]
> >>> lst[0]()[/color][/color]
> 0[color=green][color=darkred]
> >>> lst[1]()[/color][/color]
> 1[color=green][color=darkred]
> >>> lst[9]()[/color][/color]
> 9[color=green][color=darkred]
> >>>[/color][/color]
>
> and now the obvious one (as I thought at first)
>[color=green][color=darkred]
> >>> lst=[]
> >>> for i in range(10):[/color][/color]
> ... lst.append(lamb da:i)
> ...[color=green][color=darkred]
> >>> lst[0]()[/color][/color]
> 9[color=green][color=darkred]
> >>> i[/color][/color]
> 9[color=green][color=darkred]
> >>>[/color][/color]
>
> I think I understand where the problem comes from
> lambda:i seems not to be fully evalutated
> it just binds object with name i and not the value of i
> thus lst[0]() is not 0
>
> are there other solutions to this problem
> without use of eval or exec?
>[/color]

Using a factory function & closures instead of lambda:[color=blue][color=green][color=darkred]
>>> def maker(x):[/color][/color][/color]
.... def inner_maker():
.... return x
.... return inner_maker
....[color=blue][color=green][color=darkred]
>>> lst = []
>>> for i in range(10):[/color][/color][/color]
.... lst.append(make r(i))
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> lst[5]()[/color][/color][/color]
5[color=blue][color=green][color=darkred]
>>> lst[9]()[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

[color=blue]
> Regards, Daniel
> --
> http://mail.python.org/mailman/listinfo/python-list
>[/color]

[Schüle Daniel]

Re: can this be done without eval/exec?

>> are there other solutions to this problem[color=blue][color=green]
>> without use of eval or exec?
>>[/color]
>
> Using a factory function & closures instead of lambda:[/color]
[color=blue][color=green][color=darkred]
>>> def a(x):[/color][/color][/color]
.... def b():
.... return x
.... return b
....[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... lst.append(a(i) )
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> lst[1]()[/color][/color][/color]
1[color=blue][color=green][color=darkred]
>>> lst[9]()[/color][/color][/color]
9[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

yes this works
I was playing a little more with this idea
and got into the next trouble :)
[color=blue][color=green][color=darkred]
>>> cnt=0
>>> def a():[/color][/color][/color]
.... def b():
.... return cnt
.... global cnt
.... cnt += 1
.... return b
....[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... lst.append(a())
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
10[color=blue][color=green][color=darkred]
>>> lst[1]()[/color][/color][/color]
10[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

I figured out what was wrong, here is corrected version
[color=blue][color=green][color=darkred]
>>> cnt = 0
>>> def a():[/color][/color][/color]
.... global cnt
.... tmp = cnt
.... def b():
.... return tmp
.... cnt += 1
.... return b
....[color=blue][color=green][color=darkred]
>>> lst=[]
>>> for i in range(10):[/color][/color][/color]
.... lst.append(a())
....[color=blue][color=green][color=darkred]
>>> lst[0]()[/color][/color][/color]
0[color=blue][color=green][color=darkred]
>>> lst[1]()[/color][/color][/color]
1[color=blue][color=green][color=darkred]
>>>[/color][/color][/color]

Regards, Daniel

[Kent Johnson]

Re: can this be done without eval/exec?

Schüle Daniel wrote:[color=blue]
> and now the obvious one (as I thought at first)
>[color=green][color=darkred]
> >>> lst=[]
> >>> for i in range(10):[/color][/color]
> ... lst.append(lamb da:i)
> ...[color=green][color=darkred]
> >>> lst[0]()[/color][/color]
> 9[color=green][color=darkred]
> >>> i[/color][/color]
> 9[color=green][color=darkred]
> >>>[/color][/color]
>
> I think I understand where the problem comes from
> lambda:i seems not to be fully evalutated
> it just binds object with name i and not the value of i
> thus lst[0]() is not 0[/color]

The problem is that variables in closures are not bound until the
variable goes out of scope. So each lambda is bound to the final value of i.[color=blue]
>
> are there other solutions to this problem
> without use of eval or exec?[/color]

The workaround is to use a default argument to bind the current value of i:
In [1]: lst = []

In [2]: for i in range(10):
...: lst.append(lamb da i=i: i)
...:
...:

In [3]: lst[0]()
Out[3]: 0

In [4]: lst[5]()
Out[4]: 5

A list comp makes this IMO cleaner:
In [5]: lst = [ lambda i=i: i for i in range(10) ]

In [6]: lst[0]()
Out[6]: 0

In [7]: lst[5]()
Out[7]: 5

Kent

[Schüle Daniel]

Re: can this be done without eval/exec?

Kent Johnson schrieb:[color=blue]
> Schüle Daniel wrote:[color=green]
>> and now the obvious one (as I thought at first)
>>[color=darkred]
>> >>> lst=[]
>> >>> for i in range(10):[/color]
>> ... lst.append(lamb da:i)
>> ...[color=darkred]
>> >>> lst[0]()[/color]
>> 9[color=darkred]
>> >>> i[/color]
>> 9[color=darkred]
>> >>>[/color]
>>
>> I think I understand where the problem comes from
>> lambda:i seems not to be fully evalutated
>> it just binds object with name i and not the value of i
>> thus lst[0]() is not 0[/color]
>
> The problem is that variables in closures are not bound until the
> variable goes out of scope. So each lambda is bound to the final value
> of i.[color=green]
>>
>> are there other solutions to this problem
>> without use of eval or exec?[/color]
>
> The workaround is to use a default argument to bind the current value of i:
> In [1]: lst = []
>
> In [2]: for i in range(10):
> ...: lst.append(lamb da i=i: i)
> ...:
> ...:
>
> In [3]: lst[0]()
> Out[3]: 0
>
> In [4]: lst[5]()
> Out[4]: 5
>
> A list comp makes this IMO cleaner:
> In [5]: lst = [ lambda i=i: i for i in range(10) ]
>
> In [6]: lst[0]()
> Out[6]: 0
>
> In [7]: lst[5]()
> Out[7]: 5
>
> Kent[/color]

many thanks for the explaination,
it look much simpler than my solutions too

Daniel