We Keep Coding

Python: `locals()` as a default function argument

Suppose I have a module PyFoo.py that has a function bar. I want bar to print all of the local variables associated with the namespace that called it.
For example:
#! /usr/bin/env python
import PyFoo as pf
var1 = 'hi'
print locals()
pf.bar()
The two last lines would give the same output. So far I've tried defining bar as such:
def bar(x=locals):
print x()
def bar(x=locals()):
print x
But neither works. The first ends up being what's local to bar's namespace (which I guess is because that's when it's evaluated), and the second is as if I passed in globals (which I assume is because it's evaluated during import).
Is there a way I can have the default value of argument x of bar be all variables in the namespace which called bar?
EDIT 2018-07-29:
As has been pointed out, what was given was an XY Problem; as such, I'll give the specifics.
The module I'm putting together will allow the user to create various objects that represent different aspects of a numerical problem (e.x. various topology definitions, boundary conditions, constitutive models, ect.) and define how any given object interacts with any other object(s). The idea is for the user to import the module, define the various model entities that they need, and then call a function which will take all objects passed to it, make needed adjustments to ensure capability between them, and then write out a file that represents the entire numerical problem as a text file.
The module has a function generate that accepts each of the various types of aspects of the numerical problem. The default value for all arguments is an empty list. If a non-empty list is passed, then generate will use those instances for generating the completed numerical problem. If an argument is an empty list, then I'd like it to take in all instances in the namespace that called generate (which I will then parse out the appropriate instances for the argument).
EDIT 2018-07-29:
Sorry for any lack of understanding on my part (I'm not that strong of a programmer), but I think I might understand what you're saying with respect to an instance being declared or registered.
From my limited understanding, could this be done by creating some sort of registry dataset (like a list or dict) in the module that will be created when the module is imported, and that all module classes take this registry object in by default. During class initialization self can be appended to said dataset, and then the genereate function will take the registry as a default value for one of the arguments?

There's no way you can do what you want directly.
locals just returns the local variables in whatever namespace it's called in. As you've seen, you have access to the namespace the function is defined in at the time of definition, and you have access to the namespace of the function itself from within the function, but you don't have access to any other namespaces.
You can do what you want indirectly… but it's almost certainly a bad idea. At least this smells like an XY problem, and whatever it is you're actually trying to do, there's probably a better way to do it.
But occasionally it is necessary, so in case you have one of those cases:
The main good reason to want to know the locals of your caller is for some kind of debugging or other introspection function. And the way to do introspection is almost always through the inspect library.
In this case, what you want to inspect is the interpreter call stack. The calling function will be the first frame on the call stack behind your function's own frame.
You can get the raw stack frame:
inspect.currentframe().f_back
… or you can get a FrameInfo representing it:
inspect.stack()[1]
As explained at the top of the inspect docs, a frame object's local namespace is available as:
frame.f_locals
Note that this has all the same caveats that apply to getting your own locals with locals: what you get isn't the live namespace, but a mapping that, even if it is mutable, can't be used to modify the namespace (or, worse in 2.x, one that may or may not modify the namespace, unpredictably), and that has all cell and free variables flattened into their values rather than their cell references.
Also, see the big warning in the docs about not keeping frame objects alive unnecessarily (or calling their clear method if you need to keep a snapshot but not all of the references, but I think that only exists in 3.x).

How is the for-loop able to use a variable that isn't defined yet

I'm new to coding and I'm a little confused. How/why can a for loop use a variable that isn't defined yet?
For example:
demond = {'green':'grass', 'red':'fire', 'yellow':'sun'}
for i in demond:
print(i)
Output:
green
yellow
red

In python, you don't need to declare variables. In C/C++/JAVA etc. you will have to declare them first and then use them.
Variables are nothing but reserved memory locations to store values.Based on the data type of a variable, the interpreter allocates memory and decides what can be stored in the reserved memory.Python variables do not need explicit declaration to reserve memory space. The declaration happens automatically when you assign a value to a variable.

There are two things that you need to keep in mind:
because Python is a weakly-typed language, you do not need to explicitly declare any variable to a certain object type. This is something you already know, and why you can assign things without having to state what type they will be.
For loop constructs do a lot of things in the background that you don't explicitly see. This means that although it doesnt LOOK like anything is being defined, it is.
With that in mind, I dont want to really explain how for loops work, because there are already answers available for that but the main point is that a for loop in python is the same as the following pseudo code.
#set up your iterable
demond = SOME_ITERABLE_OBJECT (this can be a list, string, dict, etc)
#this
for i in demond:
do_something(i)
#is the same as this
i = demond[0] # the first item in demond
do_something(i)
i = demond[1] # the second item in demond
do_something(i)
i = demond[2]
...
...
..
i = demond[n] # the last item in demond
do_something(i)
Now your follow up question may be this: what makes it so that, in your code, for i in demond sets i to equal to it's keys? Well that is just part of the design of python, specifically how dicts work. What the for loop is ACTUALLY doing is calling an iterables next() function until the iterable generator is done. Each iterable can have a different result from a for loop (see the first link).
NOTE:
In my code example, I am setting i = demond[some_index]. This looks like a list index grab but it is really meant to just show that is iterating through the list in some sort of order. IT IS PSUEDO CODE. Just keep that in mind.

How to assign the same value to multiple variables in Python

I have a question about Python, which I am kinda new to. Let's assume I want to assign a 5x5 matrix to 10 different variables. I searched across the board, and what I found was this:
a, b, c, d, e = myMatrix
That is all good, but in Python, this means that when I change a, I also change the values of the other variables, because they all come down to the same memory adress if I got this correctly.
My question: Is there a fast way of assigning myMatrix to multiple Variables and giving each of them a unique memory adress? So that I can change myMatrix without changing a, b or c. I do explicitly search for some kind of multi-assignment.
Thanks in advance!

use the [copy] module
>>> import copy
>>> new_matrix = copy.deepcopy(myMatrix)

As Burhan Khalid and juanchopanza have pointed out, what happens in your example will be different in, for example,
the case where "myMatrix" is actually an array of 5 values (in which case "a" will get the first value and "e" will get the last value), and
the case where "myMatrix" is an instance of an Object (in which case "a" through "e" will each refer to the same object).
It sounds like you're thinking of case 2, and hoping for something like a macro which will automatically expand your single assignment statement (with a single Right Hand Side Value, whether Deep Copied or not) into 5 assignment statements, each with its own Left Hand Side, Right Hand Side, and Deep Copy.
I don't know of any way to do this, and I would point out that:
When most OO languages encounter an assignment operation like yours with an Object on the Right Hand Side, the compiler/interpreter looks for a "copy constructor" for the class of the RHS Object, and uses it (if found) to generate the value (an Object reference) which is actually assigned to the LHS. Can you even imagine what the syntax could look like for what you're describing, where the copy constructor is supposed to be called 5 times to yield 5 different Objects on the RHS, references to which are then assigned to five different variables on the LHS? What could you possibly write in a single assignment statement that would make this intent clear?
If you're writing code where Deep vs. Shallow copies will actually have an effect on behavior then IMHO you owe it to yourself and anyone else who has to read and maintain your code to make this obvious and explicit - like the answer from wong2, repeated 5 times (once for each of the 5 variables).

memory management with objects and lists in python

I am trying to understand how exactly assignment operators, constructors and parameters passed in functions work in python specifically with lists and objects. I have a class with a list as a parameter. I want to initialize it to an empty list and then want to populate it using the constructor. I am not quite sure how to do it.
Lets say my class is --
class A:
List = [] # Point 1
def __init1__(self, begin=[]): # Point 2
for item in begin:
self.List.append(item)
def __init2__(self, begin): # Point 3
List = begin
def __init3__(self, begin=[]): # Point 4
List = list()
for item in begin:
self.List.append(item)
listObj = A()
del(listObj)
b = listObj
I have the following questions. It will be awesome if someone could clarify what happens in each case --
Is declaring an empty like in Point 1 valid? What is created? A variable pointing to NULL?
Which of Point 2 and Point 3 are valid constructors? In Point 3 I am guessing that a new copy of the list passed in (begin) is not made and instead the variable List will be pointing to the pointer "begin". Is a new copy of the list made if I use the constructor as in Point 2?
What happens when I delete the object using del? Is the list deleted as well or do I have to call del on the List before calling del on the containing object? I know Python uses GC but if I am concerned about cleaning unused memory even before GC kicks in is it worth it?
Also assigning an object of type A to another only makes the second one point to the first right? If so how do I do a deep copy? Is there a feature to overload operators? I know python is probably much simpler than this and hence the question.
EDIT:
5. I just realized that using Point 2 and Point 3 does not make a difference. The items from the list begin are only copied by reference and a new copy is not made. To do that I have to create a new list using list(). This makes sense after I see it I guess.
Thanks!

In order:
using this form is simply syntactic sugar for calling the list constructor - i.e. you are creating a new (empty) list. This will be bound to the class itself (is a static field) and will be the same for all instances.
apart from the constructor name which must always be init, both are valid forms, but mean different things.
The first constructor can be called with a list as argument or without. If it is called without arguments, the empty list passed as default is used within (this empty list is created once during class definition, and not once per constructor call), so no items are added to the static list.
The second must be called with a list parameter, or python will complain with an error, but using it without the self. prefix like you are doing, it would just create a new local variable name List, accessible only within the constructor, and leave the static A.List variable unchanged.
Deleting will only unlink a reference to the object, without actually deleting anything. Once all references are removed, however, the garbage collector is free to clear the memory as needed.
It is usually a bad idea to try to control the garbage collector. instead. just make sure you don't hold references to objects you no longer need and let it make its work.
Assigning a variable with an object will only create a new reference to the same object, yes. To create a deep copy use the related functions or write your own.
Operator overloading (use with care, it can make things more confusing instead of clearer if misused) can be done by overriding some special methods in the class definition.
About your edit: like i pointed above, when writing List=list() inside the constructor, without the self. (or better, since the variable is static, A.) prefix, you are just creating an empty variable, and not overriding the one you defined in the class body.
For reference, the usual way to handle a list as default argument is by using a None placeholder:
class A(object):
def __init__(self, arg=None):
self.startvalue = list(arg) if arg is not None else list()
# making a defensive copy of arg to keep the original intact
As an aside, do take a look at the python tutorial. It is very well written and easy to follow and understand.

"It will be awesome if someone could clarify what happens in each case" isn't that the purpose of the dis module ?
http://docs.python.org/2/library/dis.html

How are closures implemented?

"Learning Python, 4th Ed." mentions that:
the enclosing scope variable is looked up when the nested functions
are later called..
However, I thought that when a function exits, all of its local references disappear.
def makeActions():
acts = []
for i in range(5): # Tries to remember each i
acts.append(lambda x: i ** x) # All remember same last i!
return acts
makeActions()[n] is the same for every n because the variable i is somehow looked up at call time. How does Python look up this variable? Shouldn't it not exist at all because makeActions has already exited? Why doesn't Python do what the code intuitively suggests, and define each function by replacing i with its current value within the for loop as the loop is running?

I think it's pretty obvious what happens when you think of i as a name not some sort of value. Your lambda function does something like "take x: look up the value of i, calculate i**x" ... so when you actually run the function, it looks up i just then so i is 4.
You can also use the current number, but you have to make Python bind it to another name:
def makeActions():
def make_lambda( j ):
return lambda x: j * x # the j here is still a name, but now it wont change anymore
acts = []
for i in range(5):
# now you're pushing the current i as a value to another scope and
# bind it there, under a new name
acts.append(make_lambda(i))
return acts
It might seem confusing, because you often get taught that a variable and it's value are the same thing -- which is true, but only in languages that actually use variables. Python has no variables, but names instead.
About your comment, actually i can illustrate the point a bit better:
i = 5
myList = [i, i, i]
i = 6
print(myList) # myList is still [5, 5, 5].
You said you changed i to 6, that is not what actually happend: i=6 means "i have a value, 6 and i want to name it i". The fact that you already used i as a name matters nothing to Python, it will just reassign the name, not change it's value (that only works with variables).
You could say that in myList = [i, i, i], whatever value i currently points to (the number 5) gets three new names: mylist[0], mylist[1], mylist[2]. That's the same thing that happens when you call a function: The arguments are given new names. But that is probably going against any intuition about lists ...
This can explain the behavior in the example: You assign mylist[0]=5, mylist[1]=5, mylist[2]=5 - no wonder they don't change when you reassign the i. If i was something muteable, for example a list, then changing i would reflect on all entries in myList too, because you just have different names for the same value!
The simple fact that you can use mylist[0] on the left hand of a = proves that it is indeed a name. I like to call = the assign name operator: It takes a name on the left, and a expression on the right, then evaluates the expression (call function, look up the values behind names) until it has a value and finally gives the name to the value. It does not change anything.
For Marks comment about compiling functions:
Well, references (and pointers) only make sense when we have some sort of addressable memory. The values are stored somewhere in memory and references lead you that place. Using a reference means going to that place in memory and doing something with it. The problem is that none of these concepts are used by Python!
The Python VM has no concept of memory - values float somewhere in space and names are little tags connected to them (by a little red string). Names and values exist in separate worlds!
This makes a big difference when you compile a function. If you have references, you know the memory location of the object you refer to. Then you can simply replace then reference with this location.
Names on the other hand have no location, so what you have to do (during runtime) is follow that little red string and use whatever is on the other end. That is the way Python compiles functions: Where
ever there is a name in the code, it adds a instruction that will figure out what that name stands for.
So basically Python does fully compile functions, but names are compiled as lookups in the nesting namespaces, not as some sort of reference to memory.
When you use a name, the Python compiler will try to figure out where to which namespace it belongs to. This results in a instruction to load that name from the namespace it found.
Which brings you back to your original problem: In lambda x:x**i, the i is compiled as a lookup in the makeActions namespace (because i was used there). Python has no idea, nor does it care about the value behind it (it does not even have to be a valid name). One that code runs the i gets looked up in it's original namespace and gives the more or less expected value.

What happens when you create a closure:
The closure is constructed with a pointer to the frame (or roughly, block) that it was created in: in this case, the for block.
The closure actually assumes shared ownership of that frame, by incrementing the frame's ref count and stashing the pointer to that frame in the closure. That frame, in turn, keeps around references to the frames it was enclosed in, for variables that were captured further up the stack.
The value of i in that frame keeps changing as long as the for loop is running – each assignment to i updates the binding of i in that frame.
Once the for loop exits, the frame is popped off the stack, but it isn't thrown away as it might usually be! Instead, it's kept around because the closure's reference to the frame is still active. At this point, though, the value of i is no longer updated.
When the closure is invoked, it picks up whatever value of i is in the parent frame at the time of invocation. Since in the for loop you create closures, but don't actually invoke them, the value of i upon invocation will be the last value it had after all the looping was done.
Future calls to makeActions will create different frames. You won't reuse the for loop's previous frame, or update that previous frame's i value, in that case.
In short: frames are garbage-collected just like other Python objects, and in this case, an extra reference is kept around to the frame corresponding to the for block so it doesn't get destroyed when the for loop goes out of scope.
To get the effect you want, you need to have a new frame created for each value of i you want to capture, and each lambda needs to be created with a reference to that new frame. You won't get that from the for block itself, but you could get that from a call to a helper function which will establish the new frame. See THC4k's answer for one possible solution along these lines.

The local references persist because they're contained in the local scope, which the closure keeps a reference to.

I thought that when a function exits, all of its local references disappear.
Except for those locals which are closed over in a closure. Those do not disappear, even when the function to which they are local has returned.

Intuitively one might think i would be captured in its current state but that is not the case. Think of each layer as a dictionary of name value pairs.
Level 1:
acts
i
Level 2:
x
Every time you create a closure for the inner lambda you are capturing a reference to level one. I can only assume that the run-time will perform a look-up of the variable i, starting in level 2 and making its way to level 1. Since you are not executing these functions immediately they will all use the final value of i.
Experts?