Jacob,

Please reduce the problem to a small code-set which reproduces the
problem. If we can reproduce same, then that tells us something. At the
very least, we can experiment without having to expend amounts of time
in a (likely faulty) bid to reproduce the same environment.

Also, code is the ultimate description!


Perhaps start with a small experiment:

- after l_servers is created, print its id()
- after the global statement, print its id()
- after the clear/reassignment, print its id()

Is Python always working with the same list?
Please advise...


On 6/03/24 07:13, Jacob Kruger via Python-list wrote:

Hi there

Working with python 3.11, and, issue that confused me for a little
while, trying to figure out what was occurring - unless am completely confused, or missing something - was that, for example, when having pre-defined a variable, and then included it in the global statement
inside a function, that function was still referring to a completely
local instance, without manipulating outside variable object at all
unless I first executed a form of referral to it, before then possibly assigning a new value to it.

Now, this does not seem to occur consistently if, for example, I just
run bare-bones test code inside the python interpreter, but consistently occurs inside my actual testing script.

Basically, in a file with python code in that am using for a form of
testing at the moment, at the top of the file, under all the import statements, I initiate the existence of a list variable to make use of

later:

# code snippet

l_servers = []

# end of first code snippet

Then, lower down, inside a couple of different functions, the first line inside the functions includes the following:
# code snippet
    global l_servers
# end code snippet

That should, in theory, mean that if I assign a value to that variable
inside one of the functions, it should reflect globally?

However, it seems like that, while inside those functions, it can be
assigned a new list of values, but if I then return to the scope outside

the functions, it has reverted back to being an empty list = []?

The issue seems to specifically (or not) occur when I make a call to one function, and, in the steps it's executing in one context, while it's
not doing anything to the list directly, it's then making a call to the second function, which is then meant to repopulate the list with a brand
new set of values.

Now, what almost seems to be occurring, is that while just manipulating
the contents of a referenced variable is fine in this context, the
moment I try to reassign it, that's where the issue is occurring .

Here are relevant excerpts from the file:-

# start code

# original assignation in main part of file

l_servers = []

# function wich is initially being executed

def interact():
    global l_servers
    # extra code inbetween choosing what to carry out

    # ...

    # end of other code

    bl_response, o_out = list_servers()

    if bl_response: # just make sure other function call was successful

        l_servers.clear() # first make reference to global variable

        for srv in o_out: l_servers.append(srv) # now re-populate items

    # end code snippet from inside interact function

# end of interact function

# end of code snippet

That other function being called from within, list_servers() was
initially just trying to populate the values inside the global list
variable itself, but was ending up in a similar fashion - reverting to initial empty value, but, the above now seems to work, as long as I
first make reference to/manipulate/work with global variable instead of
just trying to reassign it a brand new value/set of items?

So, am I missing something obvious, have I forgotten about something
else - yes, know that if was working from within an embedded function, I might need/want to then use the nonlocal statement against that variable name, but, honestly, just not sure how this can be occurring, and, it's
not just with this one list variable, etc.?

If I try simple test code from within the python interpreter, using
different types of variables, this does also not seem to be the same all
the time, but, don't think it can relate to an iterable like a list, or
else, just in case, here is the code snippet with all the import
statements from the top of that file, in case something could be
overriding standard behaviour - not likely in this context, but, really
not sure what's occurring:

# import code snippet

import requests, time
from requests.auth import HTTPBasicAuth
import psutil as psu
import pytz
import bcrypt
from copy import copy
from datetime import datetime, timedelta, timezone
from dateutil.parser import parse

# end of import snippet

Thanks if you have any ideas/thoughts on the matter

Jacob Kruger
+2782 413 4791
"Resistance is futile!...Acceptance is versatile..."

--
Regards,
=dn

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On 05Mar2024 20:13, Jacob Kruger <jacob.kruger.work@gmail.com> wrote:

Now, what almost seems to be occurring, is that while just manipulating
the contents of a referenced variable is fine in this context, the
moment I try to reassign it, that's where the issue is occurring .

Because there are no variable definitions in Python, when you write a
function Python does a static analysis of it to decide which variables
are local and which are not. If there's an assignment to a variable, it
is a local variable. _Regardless_ of whether that assignment has been executed, or gets executed at all (eg in an if-statement branch which
doesn't fire).

You can use `global` or `nonlocal` to change where Python looks for a particular name.

In the code below, `f1` has no local variables and `f2` has an `x` and
`l1` local variable.

x = 1
l1 = [1, 2, 3]

def f1():
print("f1 ...")
l1[1] = 5 # _not_ an assignment to "l1"
print("in f1, x =", x, "l1 =", l1)

def f2():
print("f2 ...")
x = 3
l1 = [6, 7, 9] # assignment to "l1"
print("in f2, x =", x, "l1 =", l1)

print("outside, x =", x, "l1 =", l1)
f1()
print("outside after f1, x =", x, "l1 =", l1)
f2()
print("outside after f2, x =", x, "l1 =", l1)

Cheers,
Cameron Simpson <cs@cskk.id.au>

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On 3/6/24 05:55, Jacob Kruger via Python-list wrote:

Ok, simpler version - all the code in a simpler test file, and working
with two separate variables to explain exactly what am talking about:

If you import the contents of that file into the python interpreter, dt_expiry will start off as "1970-01-01 00:00", and, if you execute
do_it function, it will print out the new value assigned to the
dt_expiry variable inside that function, but if you then again check the value of the dt_expiry variable afterwards, it's reverted to the 1970... value?

If I take out the line that removes values from l_test # l_test.clear()
# before appending new value to it, then it will also not retain it's new/additional child items after the function exits, and will just
revert back to [1, 2, 3] each and every time.

In other words, with some of the variable/object types, if you use a
function that manipulates the contents of a variable, before then re-assigning it a new value, it seems like it might then actually update/manipulate the global variable, but, either just calling purely content retrieval functions against said objects, or assigning them new values from scratch seems to then ignore the global scope specified in
the first line inside the function?

Hope this makes more sense

No, it doesn't. Your code is working as one would expect. For example,
adding prints for the l_test variable, and removing the .clear() which
you claim makes it not work, shows me:

before: l_test=[1, 2, 3], id(l_test)=140153285385856
leaving do_it: l_test=[1, 2, 3, 1, 2, 3, 99], id(l_test)=140153285385856
after: l_test=[1, 2, 3, 1, 2, 3, 99], id(l_test)=140153285385856

It's the same list object, as you can see by the id values. And the list
is updating as expected.

And... you don't need the global statement for l_test. As it's mutable,
you can mutate it in the function; the global only acts on assignment.
Using "global" for that may make your intent more clear to readers
though, although static checkers will grumble at you.

You must be doing something additional that you're not telling us about.

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On 3/6/2024 5:59 AM, Alan Gauld via Python-list wrote:

On 05/03/2024 22:46, Grant Edwards via Python-list wrote:

Unfortunately (presumably thanks to SEO) the enshittification of
Google has reached the point where searching for info on things like
Python name scope, the first page of links are to worthless sites like
geeksforgeeks.

And not just Google, I just tried bing, yahoo and duckduckgo
and they are all the same. Not a one listed anything from
python.org on the first page... In fact it didn't even appear
in the first 100 listings, although wikipedia did manage an
entry, eventually.

I don't know ... I just searched for "python local vs global variables"
and a python.org page on it was the second hit. I usually use StartPage
- who knows where they aggregate from - but the same search on Google
and Bing also popped up the python.org link as the second hit. As usual
Bing was a nasty experience, though.

Still, if your search phrase isn't as well focused as that or you are
less lucky, for sure you'll get all sorts of junk.

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On 3/6/2024 7:55 AM, Jacob Kruger via Python-list wrote:

Ok, simpler version - all the code in a simpler test file, and working
with two separate variables to explain exactly what am talking about:

# start code

from datetime import datetime, timezone, timedelta

from copy import copy

# initialise original values

dt_expiry = datetime.strptime("1970-01-01 00:00", "%Y-%m-%d %H:%M").replace(tzinfo=timezone.utc)

l_test = [1, 2, 3]

def do_it():
    global dt_expiry, l_test # asked python to refer to global
variables for both

    # assign new value immediately

    dt_expiry = datetime.now()+timedelta(minutes=5)
    print(dt_expiry.strftime("%Y-%m-%d %H:%M")) # just to show new
value has been assigned
    # grab copy of list for re-use of items
    l_temp = copy(l_test)
    # following line means l_test will later on retain value in global scope because it was manipulated inside function instead of just
assigned new value
    l_test.clear()
    # replace original set of values
    for i in l_temp: l_test.append(i)
    # add new item
    l_test.append(99)
# end of do_it function

# end code

If you import the contents of that file into the python interpreter, dt_expiry will start off as "1970-01-01 00:00", and, if you execute
do_it function, it will print out the new value assigned to the
dt_expiry variable inside that function, but if you then again check the value of the dt_expiry variable afterwards, it's reverted to the 1970... value?

Not when I run your code. With a little annotation added to the print statements I get (I added the import statements to make it run, and I
used the same date-time formatting for all three print statements):

List before: [1, 2, 3]
start: 1970-01-01 00:00
inside after reassignment: 2024-03-06 08:57
outside after: 2024-03-06 08:57
List after: [1, 2, 3, 99]

As an aside, you have gone to some trouble to copy, clear, and
reconstruct l_test. It would be simpler like this (and you wouldn't
have to import the "copy" library):

l_temp = l_test[:]
l_test = []

Instead of those lines and then this:

for i in l_temp: l_test.append(i)

you could achieve the same thing with this single statement:

l_test = l_test[:]

If I take out the line that removes values from l_test # l_test.clear()
# before appending new value to it, then it will also not retain it's new/additional child items after the function exits, and will just
revert back to [1, 2, 3] each and every time.

In other words, with some of the variable/object types, if you use a
function that manipulates the contents of a variable, before then re-assigning it a new value, it seems like it might then actually update/manipulate the global variable, but, either just calling purely content retrieval functions against said objects, or assigning them new values from scratch seems to then ignore the global scope specified in
the first line inside the function?

Hope this makes more sense

Jacob Kruger
+2782 413 4791
"Resistance is futile!...Acceptance is versatile..."

On 2024/03/05 20:23, dn via Python-list wrote:

Jacob,

Please reduce the problem to a small code-set which reproduces the
problem. If we can reproduce same, then that tells us something. At
the very least, we can experiment without having to expend amounts of
time in a (likely faulty) bid to reproduce the same environment.

Also, code is the ultimate description!


Perhaps start with a small experiment:

- after l_servers is created, print its id()
- after the global statement, print its id()
- after the clear/reassignment, print its id()

Is Python always working with the same list?
Please advise...


On 6/03/24 07:13, Jacob Kruger via Python-list wrote:

Hi there


Working with python 3.11, and, issue that confused me for a little
while, trying to figure out what was occurring - unless am completely
confused, or missing something - was that, for example, when having
pre-defined a variable, and then included it in the global statement
inside a function, that function was still referring to a completely
local instance, without manipulating outside variable object at all
unless I first executed a form of referral to it, before then
possibly assigning a new value to it.


Now, this does not seem to occur consistently if, for example, I just
run bare-bones test code inside the python interpreter, but
consistently occurs inside my actual testing script.


Basically, in a file with python code in that am using for a form of
testing at the moment, at the top of the file, under all the import
statements, I initiate the existence of a list variable to make use of

later:


# code snippet

l_servers = []

# end of first code snippet


Then, lower down, inside a couple of different functions, the first line >>> inside the functions includes the following:
# code snippet
     global l_servers
# end code snippet

That should, in theory, mean that if I assign a value to that variable
inside one of the functions, it should reflect globally?

However, it seems like that, while inside those functions, it can be
assigned a new list of values, but if I then return to the scope outside >>>
the functions, it has reverted back to being an empty list = []?


The issue seems to specifically (or not) occur when I make a call to
one function, and, in the steps it's executing in one context, while
it's not doing anything to the list directly, it's then making a call
to the second function, which is then meant to repopulate the list
with a brand new set of values.


Now, what almost seems to be occurring, is that while just
manipulating the contents of a referenced variable is fine in this
context, the moment I try to reassign it, that's where the issue is
occurring .


Here are relevant excerpts from the file:-


# start code

# original assignation in main part of file

l_servers = []


# function wich is initially being executed

def interact():
     global l_servers
     # extra code inbetween choosing what to carry out

     # ...

     # end of other code

     bl_response, o_out = list_servers()

     if bl_response: # just make sure other function call was successful

         l_servers.clear() # first make reference to global variable

         for srv in o_out: l_servers.append(srv) # now re-populate items

     # end code snippet from inside interact function

# end of interact function

# end of code snippet


That other function being called from within, list_servers() was
initially just trying to populate the values inside the global list
variable itself, but was ending up in a similar fashion - reverting
to initial empty value, but, the above now seems to work, as long as
I first make reference to/manipulate/work with global variable
instead of just trying to reassign it a brand new value/set of items?


So, am I missing something obvious, have I forgotten about something
else - yes, know that if was working from within an embedded
function, I might need/want to then use the nonlocal statement
against that variable name, but, honestly, just not sure how this can
be occurring, and, it's not just with this one list variable, etc.?


If I try simple test code from within the python interpreter, using
different types of variables, this does also not seem to be the same
all the time, but, don't think it can relate to an iterable like a
list, or else, just in case, here is the code snippet with all the
import statements from the top of that file, in case something could
be overriding standard behaviour - not likely in this context, but,
really not sure what's occurring:

# import code snippet

import requests, time
from requests.auth import HTTPBasicAuth
import psutil as psu
import pytz
import bcrypt
from copy import copy
from datetime import datetime, timedelta, timezone
from dateutil.parser import parse

# end of import snippet


Thanks if you have any ideas/thoughts on the matter


Jacob Kruger
+2782 413 4791
"Resistance is futile!...Acceptance is versatile..."

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Op 6/03/2024 om 13:55 schreef Jacob Kruger via Python-list:

If you import the contents of that file into the python interpreter, [...]

What exactly to you mean by "import the contents of that file into the
python interpreter"? Other people have put your code in a script,
executed it, and saw it working as expected. I pasted in IPython, and
likewise saw it working as expected, and the same with IDLE. It seems to
me you must be doing something different from us; maybe the way you
execute that code might be the key to this whole confusion.

--
"Il semble que la perfection soit atteinte non quand il n'y a plus rien à ajouter, mais quand il n'y a plus rien à retrancher."
"Perfectie is niet bereikt als er niets meer toe te voegen is, maar als er niets meer weg te nemen is."
-- Antoine de Saint-Exupéry

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Op 6/03/2024 om 16:39 schreef Roel Schroeven via Python-list:

Op 6/03/2024 om 13:55 schreef Jacob Kruger via Python-list:

If you import the contents of that file into the python interpreter,
[...]

What exactly to you mean by "import the contents of that file into the
python interpreter"? Other people have put your code in a script,
executed it, and saw it working as expected. I pasted in IPython, and likewise saw it working as expected, and the same with IDLE. It seems
to me you must be doing something different from us; maybe the way you execute that code might be the key to this whole confusion.

(As an aside, I made a type; "What exactly _do_ you mean" is what I
meant to type)

If you want, you can play with the code online here: https://tio.run/##pVPbitswEH3XVwwJITaNQ9KllATy0EKhL/2AUsqiWKNkWlkykrzZ7NJvT0e@JWT7VmPsEZozc87RqD7Ho7MPl8sUQpQ@QukUCqG9q0DJiJEqBKpqx1vDegHp@@JsHyk0UfaY0tXnIT/FQogpkKVI0lBAcJ4OZKWBJ2kaDEKo@IjPNfkz7MYGyxB9nYJsst58XBWrNb@
wWm1Xq8kCJrPvxawqZgpmX7ezb5N86bE2ssQsvpDVbjewWzaxzIUwjxFD5PI/1gt4v4CHn0xKoQblHilm@VYAPwfj9kxrpLOAHjcFGX6jgrp1CqIDjxp9CnrMk/Qk9wYDaOdh7@JRtCUTMtDBgsVTp5edYbuIZZpzl/NP@dadsvzdaG1WkW2Yy@
5D3mJqTzZmIzK5pTu37p3JmcOvhkUw2XB0pxsmRxlgj2h7jqh6ygcv990pOg18ZLEV5bFo0ulpoIhVaHOTP1XNvFN21rmV91W0M8adyB64hEWoUNowGHoiY8CwVg/sp8coyQ722MHTwEWRCZYy9SVGMd9KWqqbBFWcGkgh6MaWkbgOryNKlSi2igdZV8kj6RCXpUHps4FtPz/X4QwYU6F@
RlNSMoESv071digk6xqtymgoJVXXMdl027DP22xyvg8cpfLt/I0KRL@RLiDwhHanPP@M3Brn@bAu2mdc6/k4B7vXMfxzs98R2L12/@tOPrb48owlz1fH575TMTbsr8sb@CfNR3kH@x9@l8tf

(sorry for the long URL; that's where tio.run puts the code)

Again it works as expected, but with or without the l_test.clear() line.

--
"Il semble que la perfection soit atteinte non quand il n'y a plus rien à ajouter, mais quand il n'y a plus rien à retrancher."
"Perfectie is niet bereikt als er niets meer toe te voegen is, maar als er niets meer weg te nemen is."
-- Antoine de Saint-Exupéry

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On 3/6/24 08:28, Jacob Kruger via Python-list wrote:

C:\temp\py_try>python
Python 3.11.7 (tags/v3.11.7:fa7a6f2, Dec 4 2023, 19:24:49) [MSC v.1937 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.

from scoping2 import *

And it becomes clear: only do `from ... import *` when the module has been specifically designed to support that.

If you were to also do `import scoping2` and, after calling `do_it()`, `print(scoping2.dt_expiry)`, you would see that
it had changed.

I know there are good explanations for how variables and names work in Python, but I couldn't find any at the moment.
Sorry.

--
~Ethan~

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Op 6/03/2024 om 17:40 schreef Jacob Kruger via Python-list:

from scoping2 import *

Ah yes, that explains what's happening. After that statement, the name dt_expiry in the current namespace is bound to the same object that the
name dt_expiry in the namespace of module scoping2 is bound to. Function
do_it re-binds that last one to a new one, with the new value; name
dt_expiry in the current namespace is still bound to the old object. (If
all of that sounds like gibberish, have a look at "Facts and myths about
Python names and values" (text:
https://nedbatchelder.com/text/names.html; slides and video: https://nedbatchelder.com/text/names1.html)

I would advice not to use 'import *', if at all possible, for multiple
reasons, one of which is to prevent problems like this.

I would also advice not to use global variables from other modules
directly, and in fact would advice to minimize the use of globals in
general as much as possible. If you need to keep state between methods,
it might be better to use a class.

--
"There is a theory which states that if ever anyone discovers exactly what the Universe is for and why it is here, it will instantly disappear and be
replaced by something even more bizarre and inexplicable.
There is another theory which states that this has already happened."
-- Douglas Adams, The Restaurant at the End of the Universe

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Grant Edwards via Python-list schreef op 6/03/2024 om 18:59:

On 2024-03-06, Roel Schroeven via Python-list <python-list@python.org>
wrote:

Op 6/03/2024 om 17:40 schreef Jacob Kruger via Python-list:

from scoping2 import *

[...]

I would advice not to use 'import *', if at all possible, for

multiple > reasons, one of which is to prevent problems like this.

Unfortunately, many (most?) tutorials for particular modules (and even example code in the Python documentation itself) are all written
assuming that you do "from <module> import *". It saves the tutorial
write a few keystrokes, but causes untold trouble for people who learn incorrectly that "from <module> import *" is the proper way to do
things.

I know ... it's really irritating.

--
"There is a theory which states that if ever anyone discovers exactly what the Universe is for and why it is here, it will instantly disappear and be
replaced by something even more bizarre and inexplicable.
There is another theory which states that this has already happened."
-- Douglas Adams, The Restaurant at the End of the Universe

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On 7/03/24 05:28, Jacob Kruger via Python-list wrote:
...

So, yes, know this comes across like some form of a scam/joke, or list-garbage, since it doesn't make any sense to me at all, but still
just wondering if missing something, or should I shift over to 3.12 to
see if if works differently, or just try reinstalling 3.11 from scratch,
or should I retry the above in something like the VS code console, or a different python console, etc.?

Some of the facts, such as HOW the code was being executed were missing
(see earlier request for a cut-down scenario, AND reports from others
saying 'but it works for me').

The idea of importing a module into the REPL and then (repeatedly)
manually entering the code to set-up and execute is unusual (surely type
such into a script (once), and run that (repeatedly). As you say, most
of us would be working from an IDE and hitting 'Run'. Am wondering why
you weren't - but it's not important.

That said, the REPL is the perfect place to learn, experiment, and
prototype - particularly when compared with the facilities of other
language's eco-systems. The entirety of the on-line Tutorial cannot be
wrong! (although, after a quick review, I've failed to see where the
Tutorial mentions the usual development mode, apart from two very brief
asides (the most useful is almost at the very end(?)) - but then (as
they say) the objective is to show the language!

The lesson-learned is that there are different 'environments' and
different ways of building the environment in which the code will run.
That's a valuable lesson, and full of subtlety!

Glad to see that comparing id()s was useful - for diagnosis but not
solution. Other tools might include the locals() and globals()
functions. You may also have detected that many of us try to avoid
globals and the implicit assumptions about the behavior of mutable
collections (eg lists) when treated as 'global'. Then there are
"closures", the "LEGB" rule, namespaces, scope, and ...

--
--
Regards,
=dn

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On 06Mar2024 15:12, Jacob Kruger <jacob.kruger.work@gmail.com> wrote:

So, this does not make sense to me in terms of the following snippet
from the official python docs page: >https://docs.python.org/3/faq/programming.html

"In Python, variables that are only referenced inside a function are >implicitly global. If a variable is assigned a value anywhere within
the function’s body, it’s assumed to be a local unless explicitly >declared as global."

So, I would then assume that if I explicitly include a variable name
inside the global statement, then even just assigning it a new value
should update the variable in the global context, outside the
function?

Yes. Note that the "global" namespace is the module in which the
function is defined.

x = 1

def f(n):
global x
x += n

This updates the `x` global variable in the module where `f` was
defined.

If you import `f` and use it in another module it will _still_ update
`x` in the original module namespace.

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