Hello,
Thanks for this beautiful link. This is amazing.
Thanks,
Arup Rakshit
a...@zeit.io
> On 13-Mar-2019, at 12:11 AM, DL Neil wrote:
>
> Arup,
>
>
> On 13/03/19 3:38 AM, Arup Rakshit wrote:
>> I have questions how nonlocal and global affecting the variable assignment.
>> Also how ea
Arup,
On 13/03/19 3:38 AM, Arup Rakshit wrote:
I have questions how nonlocal and global affecting the variable assignment.
Also how each print statement looking up the values for the spam variable. This
scope thing in python is very confusing too me still. Can anyone help me to
understand th
On 2019-03-12, Arup Rakshit wrote:
> I have questions how nonlocal and global affecting the variable
> assignment. Also how each print statement looking up the values for
> the spam variable. This scope thing in python is very confusing too
> me still. Can anyone help me to understand this code w
I have questions how nonlocal and global affecting the variable assignment.
Also how each print statement looking up the values for the spam variable. This
scope thing in python is very confusing too me still. Can anyone help me to
understand this code w.r.t to scope in Python?
def scope_test()
On Tue, 15 Jun 2010 17:12:47 -0700, Inyeol Lee wrote:
> > "execfile() cannot be used reliably to modify a function’s locals."
[...]
> This is due to CPython's static optimization of local name lookup. Dummy
> 'exec' statement disables this and makes your example work:
>
> def X():
> exec "None"
On Tue, 15 Jun 2010 15:22:17 -0700, Peter wrote:
> I checked help on execfile and could only find the following
> (mystifying) sentence:
>
> "execfile() cannot be used reliably to modify a function’s locals."
What is mystifying about it? It's short and clear -- execfile cannot be
used to reliab
Inyeol Lee wrote:
> On Jun 15, 3:22 pm, Peter wrote:
>> I am puzzled by what appears to be a scope issue - obviously I have
>> something wrong :-)
>>
>> Why does this work:
>>
>> if __name__ == 'main':
>> execfile('test-data.py')
>> print data
>>
>> and yet this doesn't (I get "NameError: global
This one seems to do the trick - thanks! :-)
On Jun 16, 10:12 am, Inyeol Lee wrote:
> On Jun 15, 3:22 pm, Peter wrote:
>
>
>
>
>
> > I am puzzled by what appears to be a scope issue - obviously I have
> > something wrong :-)
>
> > Why does this work:
>
> > if __name__ == 'main':
> > execfile('
On Jun 15, 3:22 pm, Peter wrote:
> I am puzzled by what appears to be a scope issue - obviously I have
> something wrong :-)
>
> Why does this work:
>
> if __name__ == 'main':
> execfile('test-data.py')
> print data
>
> and yet this doesn't (I get "NameError: global name 'data' not
> defined")
Peter wrote:
I am puzzled by what appears to be a scope issue - obviously I have
something wrong :-)
Why does this work:
if __name__ == 'main':
execfile('test-data.py')
print data
and yet this doesn't (I get "NameError: global name 'data' not
defined"):
def X():
execfile('test-data.py')
I am puzzled by what appears to be a scope issue - obviously I have
something wrong :-)
Why does this work:
if __name__ == 'main':
execfile('test-data.py')
print data
and yet this doesn't (I get "NameError: global name 'data' not
defined"):
def X():
execfile('test-data.py')
print data
globalrev wrote:
http://mail.python.org/pipermail/python-list/2003-October/233435.html
why isnt it printing a in the second(second here, last one in OP)
example before complaining?
def run():
a = 1
def run2(b):
a = b
print a
run2(2)
print a
run()
def run():
a = 1
def run2(b
http://mail.python.org/pipermail/python-list/2003-October/233435.html
why isnt it printing a in the second(second here, last one in OP)
example before complaining?
def run():
a = 1
def run2(b):
a = b
print a
run2(2)
print a
run()
def run():
a = 1
def run2(b):
print a
John,
Thanks for writing, and I'm sorry it's taken so long to get back to you. Python
is fun for me -- dinner guests and my boss got in the way.
>> The code ... is the result of noodling around with switches as a learning
>> tool. I've played with python for a few years, but I'm self-taught, so
[EMAIL PROTECTED] wrote:
> The code that follows is the result of noodling around with switches as a
> learning tool. I've played with python for a few years, but I'm self-taught,
> so . . .
>
> Class Switch builds a set of functions. Method switch executes one of them
> given a value of the sw
The code that follows is the result of noodling around with switches as a
learning tool. I've played with python for a few years, but I'm self-taught, so
. . .
Class Switch builds a set of functions. Method switch executes one of them
given a value of the switch variable.
My question is, why a
Steven W. Orr wrote:
> The problem only happens if I try to modify jj.
It only happens if you try to *bind* the name "jj" to an object inside
the function.
> What am I not understanding?
My guess is that you have a C/C++ view of variables and values, where
variables are locations in memory th
On Wednesday, Jan 9th 2008 at 14:01 -, quoth Fredrik Lundh:
=>Steven W. Orr wrote:
=>
=>> So sorry because I know I'm doing something wrong.
=>>
=>> 574 > cat c2.py
=>> #! /usr/local/bin/python2.4
=>>
=>> def inc(jj):
=>> def dummy():
=>> jj = jj + 1
=>> return jj
=>>
On Jan 9, 3:52 pm, Waldemar Osuch <[EMAIL PROTECTED]> wrote:
> On Jan 9, 11:47 am, "Steven W. Orr" <[EMAIL PROTECTED]> wrote:
>
>
>
> > So sorry because I know I'm doing something wrong.
>
> > 574 > cat c2.py
> > #! /usr/local/bin/python2.4
>
> > def inc(jj):
> > def dummy():
> > jj =
On Jan 9, 11:47 am, "Steven W. Orr" <[EMAIL PROTECTED]> wrote:
> So sorry because I know I'm doing something wrong.
>
> 574 > cat c2.py
> #! /usr/local/bin/python2.4
>
> def inc(jj):
> def dummy():
> jj = jj + 1
> return jj
> return dummy
>
> h = inc(33)
> print 'h() = '
Ben Fisher wrote:
> One way to get this to work is:
>
> def inc(jj):
> def dummy(jj = jj):
> jj = jj + 1
> return jj
> return dummy
>
> h = inc(33)
> print h()
>
> It's not very pretty though, especially when you have many variables
> you want to have in
On Jan 9, 8:24 pm, Mike Meyer <[EMAIL PROTECTED]> wrote:
> On Wed, 9 Jan 2008 13:47:30 -0500 (EST) "Steven W. Orr" <[EMAIL PROTECTED]>
> wrote:
>
>
>
> > So sorry because I know I'm doing something wrong.
>
> > 574 > cat c2.py
> > #! /usr/local/bin/python2.4
>
> > def inc(jj):
> > def dummy()
One way to get this to work is:
def inc(jj):
def dummy(jj = jj):
jj = jj + 1
return jj
return dummy
h = inc(33)
print h()
It's not very pretty though, especially when you have many variables
you want to have in the inner scope.
-Ben
On 1/9/08, Mi
On Wed, 9 Jan 2008 13:47:30 -0500 (EST) "Steven W. Orr" <[EMAIL PROTECTED]>
wrote:
> So sorry because I know I'm doing something wrong.
>
> 574 > cat c2.py
> #! /usr/local/bin/python2.4
>
> def inc(jj):
> def dummy():
> jj = jj + 1
> return jj
> return dummy
>
> h =
Steven W. Orr wrote:
> So sorry because I know I'm doing something wrong.
>
> 574 > cat c2.py
> #! /usr/local/bin/python2.4
>
> def inc(jj):
> def dummy():
> jj = jj + 1
> return jj
> return dummy
>
> h = inc(33)
> print 'h() = ', h()
> 575 > c2.py
> h() =
> Tracebac
So sorry because I know I'm doing something wrong.
574 > cat c2.py
#! /usr/local/bin/python2.4
def inc(jj):
def dummy():
jj = jj + 1
return jj
return dummy
h = inc(33)
print 'h() = ', h()
575 > c2.py
h() =
Traceback (most recent call last):
File "./c2.py", line 10,
On 2007-08-06, Nitro <[EMAIL PROTECTED]> wrote:
> Hello,
>
> today I wrote this piece of code and I am wondering why it does
> not work the way I expect it to work. Here's the code:
>
> y = 0
> def func():
> y += 3
> func()
>
> This gives an
>
> UnboundLocalError: local variable 'y' reference
Thanks a lot for clearing this up, Diez!
-Matthias
--
http://mail.python.org/mailman/listinfo/python-list
Nitro wrote:
> Hello,
>
> today I wrote this piece of code and I am wondering why it does not work
> the way I expect it to work. Here's the code:
>
> y = 0
> def func():
> y += 3
> func()
>
> This gives an
>
> UnboundLocalError: local variable 'y' referenced before assignment
>
> If I c
Hello,
today I wrote this piece of code and I am wondering why it does not work
the way I expect it to work. Here's the code:
y = 0
def func():
y += 3
func()
This gives an
UnboundLocalError: local variable 'y' referenced before assignment
If I change the function like this:
y = 0
def f
On Jun 22, 3:53 pm, johnny <[EMAIL PROTECTED]> wrote:
> Scope of ids:
> When I print "ids", it's always empty string '', as I have intialized
> before. That's not what I want. I want the ids to have
> str(r['id']).join(',')
>
> if res:
> ids = ''
> for r in res['key
johnny wrote:
> Scope of ids:
> When I print "ids", it's always empty string '', as I have intialized
> before. That's not what I want. I want the ids to have
> str(r['id']).join(',')
>
> if res:
> ids = ''
> for r in res['key']:
> ids = str(r['
Scope of ids:
When I print "ids", it's always empty string '', as I have intialized
before. That's not what I want. I want the ids to have
str(r['id']).join(',')
if res:
ids = ''
for r in res['key']:
ids = str(r['id']).join(',')
[EMAIL PROTECTED] wrote:
> Ouch! You got me there, I did not copy the code properly. Now I feel
> stupid. Thanks for the enlightment.
>
> I think I am starting to get it.
P.S. The point of the example was to show how nesting isn't necessary
much of the time. The authors wanted to show that it i
> Actually, the code in the book is:
>
> def f1():
> x = 88
> f2(x)
>
> def f2(x):
> print x
>
> f1()
>
> which makes all the difference in the world. Not to mention that this
> particular section of the book is giving an example of how to write the
> code *without* using nested fun
[EMAIL PROTECTED] wrote:
> I do understand (and verified) that if I define f2 within f1, it works
> as expected. But in the "learning pyton 2nd edition" at page 205 it is
> said that "Programs are much simpler if you do not nest defs within
> defs" (juste before the code mentioned in my initial me
Thanks for the answers.
I do understand (and verified) that if I define f2 within f1, it works
as expected. But in the "learning pyton 2nd edition" at page 205 it is
said that "Programs are much simpler if you do not nest defs within
defs" (juste before the code mentioned in my initial message).
[EMAIL PROTECTED] wrote:
[...]
> def f1() :
> x=88
> f2()
> def f2() :
> print 'x=',x
> f1()
>
> that returns an error saying that "NameError: global name 'x' is not
> defined". I expected f2 to "see" the value of x defined in f1 since it
> is nested at runtime.
Ah, no, Python uses "s
[EMAIL PROTECTED] wrote:
> I have a problem understanding the scope of variable in nested
> function. I think I got it nailed to the following example copied from
> Learning Python 2nd edition page 205. Here is the code.
>
> def f1() :
> x=88
> f2()
> def f2() :
> print 'x=',x
> f1()
[EMAIL PROTECTED] wrote:
> I have a problem understanding the scope of variable in nested
> function. I think I got it nailed to the following example copied from
> Learning Python 2nd edition page 205. Here is the code.
>
> def f1() :
> x=88
> f2()
> def f2() :
> print 'x=',x
> f1()
>
I have a problem understanding the scope of variable in nested
function. I think I got it nailed to the following example copied from
Learning Python 2nd edition page 205. Here is the code.
def f1() :
x=88
f2()
def f2() :
print 'x=',x
f1()
that returns an error saying that "NameError:
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