why is this different?

[Schüle Daniel]

Hello snakes :)

In [38]: f = [lambda:i for i in range(10)]
In [39]: ff = map(lambda i: lambda : i, range(10))
In [40]: f[0]()
Out[40]: 9
In [41]: f[1]()
Out[41]: 9
In [42]: ff[0]()
Out[42]: 0
In [43]: ff[1]()
Out[43]: 1

I don't understand why in the first case f[for all i in 0..9]==9
what is different from (more usefull)

In [44]: f = ["%i" % i for i in range(10)]
In [45]: f
Out[45]: ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
doing it like this works again

In [54]: def d(x):
....: return lambda:x
....:

In [55]: f = [d(i) for i in range(10)]
In [56]: f[0]()
Out[56]: 0
In [57]: f[1]()
Out[57]: 1

in a C programmer sence I would say there seems to be no "sequence
point" which would separate "now" from "next"

Regards, Daniel

[Gabriel Genellina]

On 7 dic, 22:53, Schüle Daniel <[e-mail]:

In [38]: f = [lambda:i for i in range(10)]
In [39]: ff = map(lambda i: lambda : i, range(10))
In [40]: f[0]()
Out[40]: 9
In [41]: f[1]()
Out[41]: 9
In [42]: ff[0]()
Out[42]: 0
In [43]: ff[1]()
Out[43]: 1

I don't understand why in the first case f[for all i in 0..9]==9

In the first case, i is a free variable. That means that Python will
get it from other place (the global namespace, likely [surely?])

>>f=[lambda:i for i in range(10)]
f[0]()

9

>>i=123
f[0]()

123

>>print f[0].func_closure

None

In the second case, the inner i is a free variable, but local to its
enclosing scope (outer lambda). It's a closure:

>>ff = map(lambda i: lambda : i, range(10))
ff[4]()

4

>>ff[4].func_closure

(<cell at 0x00ACEA90: int object at 0x00995344>,)

>>i=321
ff[4]()

4

what is different from (more usefull)

In [44]: f = ["%i" % i for i in range(10)]
In [45]: f
Out[45]: ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

This is a simple expression evaluated at each iteration

doing it like this works again

In [54]: def d(x):
....: return lambda:x
....:

x inside lambda is a free variable, but since it's local to d, this
becomes a closure:

>>d(1)

<function <lambdaat 0x00A35EF0>

>>d(1).func_closure

(<cell at 0x00A2A4F0: int object at 0x00995338>,)

In [55]: f = [d(i) for i in range(10)]
In [56]: f[0]()
Out[56]: 0
In [57]: f[1]()
Out[57]: 1

This is similar to the ff example above

in a C programmer sence I would say there seems to be no "sequence
point" which would separate "now" from "next"

No, the problem is that C has no way to express a closure; this is a
functional concept absolutely extraneous to the C language.

--
Gabriel Genellina

[Schüle Daniel]

Gabriel Genellina schrieb:
Gabriel Genellina schrieb:

On 7 dic, 22:53, Schüle Daniel <[e-mail]:

>In [38]: f = [lambda:i for i in range(10)]
In [39]: ff = map(lambda i: lambda : i, range(10))
In [40]: f[0]()
Out[40]: 9
In [41]: f[1]()
Out[41]: 9
In [42]: ff[0]()
Out[42]: 0
In [43]: ff[1]()
Out[43]: 1

I don't understand why in the first case f[for all i in 0..9]==9

In the first case, i is a free variable. That means that Python will
get it from other place (the global namespace, likely [surely?])

>>>f=[lambda:i for i in range(10)]
f[0]()

9

>>>i=123
f[0]()

123

>>>print f[0].func_closure

None

In the second case, the inner i is a free variable, but local to its
enclosing scope (outer lambda). It's a closure:

>>>ff = map(lambda i: lambda : i, range(10))
ff[4]()

4

>>>ff[4].func_closure

(<cell at 0x00ACEA90: int object at 0x00995344>,)

>>>i=321
ff[4]()

4

>what is different from (more usefull)

In [44]: f = ["%i" % i for i in range(10)]
In [45]: f
Out[45]: ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

This is a simple expression evaluated at each iteration

>doing it like this works again

In [54]: def d(x):
....: return lambda:x
....:

x inside lambda is a free variable, but since it's local to d, this
becomes a closure:

>>>d(1)

<function <lambdaat 0x00A35EF0>

>>>d(1).func_closure

(<cell at 0x00A2A4F0: int object at 0x00995338>,)

>In [55]: f = [d(i) for i in range(10)]
In [56]: f[0]()
Out[56]: 0
In [57]: f[1]()
Out[57]: 1

This is similar to the ff example above

>in a C programmer sence I would say there seems to be no "sequence
point" which would separate "now" from "next"

No, the problem is that C has no way to express a closure; this is a
functional concept absolutely extraneous to the C language.

On 7 dic, 22:53, Schüle Daniel <[e-mail]:

>In [38]: f = [lambda:i for i in range(10)]
In [39]: ff = map(lambda i: lambda : i, range(10))
In [40]: f[0]()
Out[40]: 9
In [41]: f[1]()
Out[41]: 9
In [42]: ff[0]()
Out[42]: 0
In [43]: ff[1]()
Out[43]: 1

I don't understand why in the first case f[for all i in 0..9]==9

In the first case, i is a free variable. That means that Python will
get it from other place (the global namespace, likely [surely?])

>>>f=[lambda:i for i in range(10)]
f[0]()

9

>>>i=123
f[0]()

123

>>>print f[0].func_closure

None

intersting
I think I got it

[]

I have two more examples, but now I understand the difference

In [70]: x = [eval("lambda:i") for i in range(10)]
In [71]: y = [eval("lambda:%i" % i) for i in range(10)]

I think [71] is most obvious what the programmer intends

Thx

[Marc 'BlackJack' Rintsch]

In <el**********@news2.rz.uni-karlsruhe.de>, Schüle Daniel wrote:

I have two more examples, but now I understand the difference

In [70]: x = [eval("lambda:i") for i in range(10)]
In [71]: y = [eval("lambda:%i" % i) for i in range(10)]

I think [71] is most obvious what the programmer intends

But unnecessary use of `eval()` is evil. You can spell it this way:

z = [lambda x=i: x for i in range(10)]

The `x` is a local name and default values are evaluated and bound when
the function is created.

Ciao,
Marc 'BlackJack' Rintsch

[Gabriel Genellina]

On 7 dic, 22:53, Schüle Daniel <[e-mail]:

In [38]: f = [lambda:i for i in range(10)]
In [39]: ff = map(lambda i: lambda : i, range(10))
In [40]: f[0]()
Out[40]: 9
In [41]: f[1]()
Out[41]: 9
In [42]: ff[0]()
Out[42]: 0
In [43]: ff[1]()
Out[43]: 1

I don't understand why in the first case f[for all i in 0..9]==9

In the first case, i is a free variable. That means that Python will
get it from other place (the global namespace, likely [surely?])

>>f=[lambda:i for i in range(10)]
f[0]()

9

>>i=123
f[0]()

123

>>print f[0].func_closure

None

In the second case, the inner i is a free variable, but local to its
enclosing scope (outer lambda). It's a closure:

>>ff = map(lambda i: lambda : i, range(10))
ff[4]()

4

>>ff[4].func_closure

(<cell at 0x00ACEA90: int object at 0x00995344>,)

>>i=321
ff[4]()

4

what is different from (more usefull)

In [44]: f = ["%i" % i for i in range(10)]
In [45]: f
Out[45]: ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

This is a simple expression evaluated at each iteration

doing it like this works again

In [54]: def d(x):
....: return lambda:x
....:

x inside lambda is a free variable, but since it's local to d, this
becomes a closure:

>>d(1)

<function <lambdaat 0x00A35EF0>

>>d(1).func_closure

(<cell at 0x00A2A4F0: int object at 0x00995338>,)

In [55]: f = [d(i) for i in range(10)]
In [56]: f[0]()
Out[56]: 0
In [57]: f[1]()
Out[57]: 1

This is similar to the ff example above

in a C programmer sence I would say there seems to be no "sequence
point" which would separate "now" from "next"

No, the problem is that C has no way to express a closure; this is a
functional concept absolutely extraneous to the C language.

--
Gabriel Genellina