I found below code at http://pydanny.com/python-dictionary-as-a-class.html. The code works fine and I kind of understand why it does but a similar code, below it gives error.
def newclass(**kwargs):
""" Use kwargs.update() method to handle inheritance """
def set(key, value):
""" Sets key/value to the kwargs.
Replicates self/this clumsily
"""
kwargs[key] = value
kwargs['set'] = set
return kwargs
My trial code:
def closing():
x=1
def closed():
print(x)
x=x+1
return(closed)
a=closing()
a()
Error Message:
Traceback (most recent call last):
File "<pyshell#606>", line 1, in <module>
a()
File "<pyshell#604>", line 4, in closed
print(x)
UnboundLocalError: local variable 'x' referenced before assignment
When I use 'nonlocal x' in the closed function it works but how come the initial code works without 'nonlocal'.
My understanding is that it's a closure and the inside function will keep reference of the outside (free) variable and whenever the inner function is called it will be able to act upon that closed variable but certainly I haven't understood some part of it properly.
Please help me clear the concept I am missing.
Thank you all the folks who answer. SO has been too helpful.
When I use 'nonlocal x' in the closed function it works but how come the initial code works without 'nonlocal'.
In the first snippet, you're mutating the existing kwargs value. You're not assigning a new value to it, or to any other name.
But in the second snippet, you're assigning a new value to x. That forces x to be local (unless you've said otherwise). Therefore, your x+1 is a reference to a local variable that hasn't been assigned yet. Hence the error.
For a more rigorous explanation, read Naming and binding in the documentation. The relevant bits are:
Each occurrence of a name in the program text refers to the binding of that name established in the innermost function block containing the use…
When a name is used in a code block, it is resolved using the nearest enclosing scope…
If a name is bound in a block, it is a local variable of that block, unless declared as nonlocal…
To learn how this works under the covers, you'll also need to read about functions and code objects in The standard type hierarchy. You probably don't need to know that part for understanding, but for debugging, you may want to inspect the __closure__ and other attributes.
Related
I couldn't understand global and local variables in python, especially functions. I look lots of examples, but all of them don't explain in function calls clearly. My example code is below:
def called(d):
del d[0]
b=[1]
return b
def main():
a=[0,1,2,3]
print("first ", a)
c=called(a)
print("second ", a)
main()
Output:
first [0, 1, 2, 3]
second [1, 2, 3]
I expect that "a" is local variable at main. When I call the "called" function a is copyed to d. And d is local at "called". But reality is different. I solve the problem with a "called(a.copy)". But I want to understand the logic.
Python never implicitely copies anything, when you pass an object (and everything in Python is an object) to a function what you have in the function IS the object you passed in, so if you mutate it the change will be seen outside the function.
OTHO, parameters names are local to the function so rebinding a parameter within a function only change what object this name points to within the function, it has no impact on the object passed.
For more in-depth explanation the best reference is Ned Batchelder's article here.
Basically, a global variable is one that can be accessed anywhere, regardless of whether or not it is in a function (or anything else).
A local variable is one that solely exists within the function in question. You cannot declare or access it anywhere else. You could, however, make it global by explicitly including that in the function.
As far as I can see,'d' has not really been defined as a variable, but as a function parameter, as seen in called(d). This could've been changed to called(blabla) and your function would behave the same exact way if you, within the function, also changed d[0] to blabla[0].
What this means is that when you call that function, anything that has 'd' in it would be replaced by what you're calling it with. In this case, the parameter has been changed to the variable a, and the functions are then executed as you've stated.
In order to define a variable, you have to use '='.
This question already has answers here:
UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)
(14 answers)
Why does code like `str = str(...)` cause a TypeError, but only the second time?
(20 answers)
Closed 6 months ago.
Could anyone explain the exception the below code. It only works when I change the var sub in the display() to another name. There is no global variable sub as well. So what happened ?
def sub(a, b):
return a - b
def display():
sub = sub(2,1) // if change to sub1 or sth different to sub, it works
print sub
Any variable you assign to inside a scope is treated as a local variable (unless you declare it global, or, in python3, nonlocal), which means it is not looked up in the surrounding scopes.
A simplified example with the same error:
def a(): pass
def b(): a = a()
Now, consider the different scopes involved here:
The global namespace contains a and b.
The function a contains no local variables.
The function b contains an assignment to a - this means it is interpreted as a local variable and shadows the function a from the outer scope (in this case, the global scope). As a has not been defined inside of b before the call, it is an unbound local variable, hence the UnboundLocalError. This is exactly the same as if you had written this:
def b(): x = x()
The solution to this is simple: choose a different name for the result of the sub call.
It is important to note that the order of use and assignment makes no difference - the error would have still happened if you wrote the function like this:
def display():
value = sub(2,1) #UnboundLocalError here...
print value
sub = "someOtherValue" #because you assign a variable named `sub` here
This is because the list of local variables is generated when the python interpreter creates the function object.
This was originally a comment. The OP found this useful as an answer. Therefore, I am re-posting it as an answer
Initially, sub is a function. Then, it becomes the return value of a function. So when you say print sub, python doesn't know which sub you are referring to.
Edit:
First you define a function sub. Now, python knows what sub is.
When you create a variable and try to assign to it (say x = 2), python evaluates the stuff on the right hand side of the = and assigns the value of the evaluation as the value of the stuff on the left hand side of the =. Thus, everything on the right hand side should actually compute.
So if your statement was x = x+1, then x better have a value assigned to it before that line; and the previously defined x has to be of some type compatible with the addition of 1.
But suppose x is a function, and you make a variable called x in some other function, and try to assign to it, a value computed with function x, then this really starts to confuse python about which x you are referring to. This is really an oversimplification of this answer, which does a much better job of explaining variable scope and shadowing in python functions
For every variable used, Python determines whether it is a local or a nonlocal variable. Referencing a unknown variable marks it as nonlocal. Reusing the same name as a local variable later is considered a programmers mistake.
Consider this example:
def err():
print x # this line references x
x = 3 # this line creates a local variable x
err()
This gives you
Traceback (most recent call last):
File "asd.py", line 5, in <module>
err()
File "asd.py", line 2, in err
print x # this line references x
UnboundLocalError: local variable 'x' referenced before assignment
What happens is basically that Python keeps track of all references to names in code. When it reads the line print x Python knows that x is a variable from a outer scope (upvalue or global). However, in x = 3 the x is used as a local variable. As this is a inconsistency in the code, Python raises an UnboundLocalError to get the Programmers attention.
Python start executing your code and get the function first
def sub(a, b):
return a - b
So after executing this interpreter get the sub as a function. Now when come to next line it found
def display():
sub = sub(2,1) // if change to sub1 or sth different to sub, it works
print sub
so first line sub = sub (2, 1) will convert the sub function to sub variable. From this function you are returning the sub variable. So its create problem.
I saw in a book about language description that says
On the other hand, a name can be bound to no object (a dangling pointer),
one object (the usual case), or several objects (a parameter name in a
recursive function).
How can we bind a name to several objects? Isnt that what we call an array for example where all elements have the same name but with index? For a recursive function like the example here:
x = 0
def f(y):
global x
x += 1
if x < 4 :
y +=100
f(y)
else: return
f(100)
Is the name y binded with multiple values that are created recursively since the nametable has already the y name binded to an initial value which is being reproduced with recursion?
EDITED Just press here Visualizer and see what it generates. :)
No.
A name is bound to one single object . When we are talking about Python - it is either bound to a single object in a given context, or do not exist at all.
What happens, is that the inner workings may have the name defined in several "layers" - but your code will only see one of those.
If a name is a variable in a recursive function, you will only see whatver is bound to it in the current running context - each time there is a function call in Python, the execution frame, which is an object which holds several attributes of the running code, including a reference to the local variables, is frozen. On the called function, a new execuciton frame is created, and there, the variable names are bound again to whatever new values they have in the called context. Your code just "see" this instance.
Then, there is the issue of global variables and builtin objects in Python: if a name is not a local variable in the function execution context, it is searched in the globals variables for the module (again, just one of those will be visible).ANd if the name is not defiend in the globals, them, Python looks for it in globals().__builtins__ that is your last call.
If I understand you correctly, you're asking about what rules Python has for creating variables in different scopes. Python uses lexical scoping on the function level.
It's hard to tell exactly what you're getting at with the code you've written, but, while there may be a different value associated with y in different scopes (with a value of y defined at each level of recursion), your code will only ever be able to see one at a time (the value defined at the scope in which you're operating).
To really understand scoping rules in Python, I would have a look at PEP 227. Also, have a look at this Stack Overflow question.
Finally, to be able to speak intelligently about what a "name" is in Python, I suggest you read about how Python is a "Call-By-Object" language.
At this point, we are capable of understanding that, instead of a "nametable", python uses a dictionary to hold what is accessible in a given scope. See this answer for a little more detail. The implication of this is that you can never have two of the same name in a single scope (for the same reason you can't have two of the same key in a python dictionary). So, while y may exist in a dictionary for a different scope, you have no way of accessing it, since you can only access the variables in the current scope's dictionary.
The key is:
several objects (a parameter name in a recursive function).
The passage is almost certainly not referring to arrays, but simply to the fact that in a recursive function (or any function, but a recursive function is likely to have multiple activations at one time), a parameter may be bound to a different value in each recursive call.
This does not mean that you can access each such object in every stack frame; indeed the point of the technique is to ensure that only one such value is accessible in each stack frame.
Firstly, you should mention in the question that the sentence from the book is not related explicitly to Python (as jsbueno wrote, one name is bound to exactly one object in Python).
Anyway, name bound to no object is a bit inaccurate. Generally, names are related to variables, and name related to a dangling pointer is the name of that pointer variable.
When speaking about the variable scope (i.e. the part of code where the variable is used), one variable name can be used only for a single value at a time. However, there may be other parts of code, independent on the one where we think about that variable. In the other part of code, the same name can be used; however, the two variables with the same name are totally isolated. This is the case of local variables also in the case of function bodies. If the language allows recursion, it must be capable to create another isolated space of local variable even for another call of the same function.
In Python, each function can also access outer variables, but it is more usual to use the inner, local variables. Whenever you assign a name some value, it is created in the local space.
How do I obtain the non-local variables for the current scope? The functions vars, locals, and globals exist, but is there a function to get the nonlocals?
Why aren't the nonlocals listed when calling vars?
Update
My issue is that there's no way to enumerate the variables available in the current scope, as neither vars or globals includes the non-locals AFAICT.
I frequently use vars in code such as the following:
'{meh[0]}/{meh[3]} {a}{b}{c}'.format(**vars())
Which fails if any of these variables are in the scope of a containing function.
From within running code, you can easily get the names of the nonlocal variables - but retriving their content in a way a call to locals gets you a dictionary is a bit trickier.
The used nonlocal variable names are stored in the current running code object, in the co_freevars attribute.
So, getting the nonlocal names is a matter of:
names = inspect.currentframe().f_code.co_freevars
The contents for these variables, however, are stored in the __closure__ attribute (func_closure, in Python 2), of the function object instead. (Not the code object). The problem is that, without "aid from outside", there is no easy way for a running code to get to the function object it is running on. You can get to the frame object, which links to the code object, but there are no links back to the function object. (For a top level defined function one could explicitly use the function known name, as used in the def statement` but for an enclosed function, that is returned to a caller, there is no way of knowing its name either).
So, one has to resort to a trick - getting all the objects that link to the current code object, by using the gc module (garbage collector) - there is a gc.get_referrers call - it will return all the function objects that link to the code object one holds.
So, inside a function with non_local variables one could do:
import inspect, gc
from types import FunctionType
def a(b):
b1 = 2
def c():
nonlocal b1
print (b)
code = inspect.currentframe().f_code
names = code.co_freevars
function = [func for func in gc.get_referrers(code) if isinstance(func, FunctionType)][0]
nonlocals = dict (zip(names, (x.cell_contents for x in function.__closure__ )))
print(nonlocals)
return inspect.currentframe()
return c
c = a(5)
f = c()
And therefore retrieve the names and values of the nonlocals. But this won't work if you have more than one instance of that function around (that is, if the function of interested was created more than once with more than one call to the functin that generates it) - becasue all of those instances will link to the same code object. The example above, assumes there is only one function running with the current code - and would work correctly in this case. Another call to the factrory function would create another function, with potentially other values for the nonlocal variables, but with the same code object - the function = list genrator above would retrieve all of those, and arbitrarily pick the first of those.
The "correct" function is the one on which the current code is executing - I am trying to think of a way of retrieving this information, but can't get to it. If I can, I will complete this answer, but for now, this can't help you to retrieve the nonlocals values values.
(just found out that trying to use "eval" with a nonlocal variable name won't work as well)
It looks like that the only thing linking the current running frame to the function object where the nonlocal variables values are held is created at run time inside the native side of the Python interpreter. I can't think of a way of getting to it short of using the ctypes module to look at interpreters data structures at runtime, which would, of course, be unsuitable for any actual production code.
The bottom line: you can reliably retrieve the nonlocal variable names. But it looks like you can't get their value given their name as a string (nor rebind then).
You could try opening a feature-request for a "nonlocals" call on Python's bug tracker or on Python-ideas mailing list.
I'm currently developing some things in Python and I have a question about variables scope.
This is the code:
a = None
anything = False
if anything:
a = 1
else:
a = 2
print a # prints 2
If I remove the first line (a = None) the code still works as before. However in this case I'd be declaring the variable inside an "if" block, and regarding other languages like Java, that variable would only be visible inside the "if".
How exactly variable scoping works in Python and what's the good way to program in cases like this?
Thanks!
As a rule of thumb, scopes are created in three places:
File-scope - otherwise known as module scope
Class-scope - created inside class blocks
Function-scope - created inside def blocks
(There are a few exceptions to these.)
Assigning to a name reserves it in the scope namespace, marked as unbound until reaching the first assignment. So for a mental model, you are assigning values to names in a scope.
I believe that Python uses function scope for local variables. That is, in any given function, if you assign a value to a local variable, it will be available from that moment onwards within that function until it returns. Therefore, since both branches of your code are guaranteed to assign to a, there is no need to assign None to a initially.
Note that when you can also access variables declared in outer functions -- in other words, Python has closures.
def adder(first):
def add(second):
return first + second
return add
This defines a function called adder. When called with an argument first, it will return a function that adds whatever argument it receives to first and return that value. For instance:
add_two = adder(2)
add_three = adder(3)
add_two(4) # = 6
add_three(4) # = 7
However, although you can read the value from the outer function, you can't change it (unlike in many other languages). For instance, imagine trying to implement an accumulator. You might write code like so:
def accumulator():
total = 0
def add(number):
total += number
return total
return add
Unfortunately, trying to use this code results in an error message:
UnboundLocalError: local variable 'total' referenced before assignment
This is because the line total += number tries to change the value of total, which cannot be done in this way in Python.
There is no problem assigning the variable in the if block.
In this case it is being assigned on both branches, so you can see it will definitely be defined when you come to print it.
If one of the branches did not assign to a then a NameError exception would be raise when you try to print it after that branch
Python doesn't need variables to be declared initially, so you can declare and define at arbitrary points. And yes, the scope is function scope, so it will be visible outside the if.
i'm quite a beginner programmer, but for what i know, in python private variables don't exist. see private variables in the python documentation for a detailed discussion.
useful informations can also be found in the section "scopes and namespaces" on the same page.
personally, i write code like the one you posted pretty much every day, especially when the condition relies in getting input from the user, for example
if len(sys.argv)==2:
f = open(sys.argv[1], 'r')
else:
print ('provide input file')
i do declare variables before using them for structured types, for example i declare an empty list before appending its items within a loop.
hope it helps.