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When defining a lambda function with a variable inside, how to prevent changes in function when I change the variable?

For example: in this code, function changes when the variable changes. I would like to learn how to prevent the change in function behaviour when i change the variable. Is there some way to only get the value of the variable instead of the variable itself? Also are there any sources where I can learn more about problems like this?
a = 5
adder = lambda x: x + a
print(adder(5)) # prints 10
a = 50
print(adder(5)) # prints 55

Just like the equivalent function defined by a def statement (which is what you should be using, rather than assigning the result of a lambda expression to a name explicitly)
def adder(x):
return x + a
the name a isn't looked up until the function is called.
One way to make a function that specifically computes x + 5 when a == 5 at definition time is use a default argument value:
def adder(x, a=a):
return x + a
where the left-hand a is a new parameter (which isn't intended to be set explicitly), and the right-hand a is the value of the a in the current scope.
A better idea, though, is to define a closure, so that your function doesn't have a "hidden" parameter that can be abused.
# This is where a lambda expression does make sense: you want
# a function, but don't need to give it a name.
def make_adder(a):
return lambda x: x + a
adder = make_adder(5)
a in adder is still a free variable, but now it refers to a variable in a scope which you don't "outside" access to after make_adder returns.

how to assign a new value to a variable inside a function? [duplicate]

Sorry if this is a dumb question, but I've looked for a while and not really found the answer.
If I'm writing a python function, for example:
def function(in1, in2):
in1=in1+1
in2=in2+1
How do I make these changes stick?
I know why they dont, this has been addressed in many answers, but I couldn't find an answer to the question of how to actually make them do so. Without returning values or making some sort of class, is there really no way for a function to operate on its arguments in a global sense?
I also want these variables to not be global themselves, as in I want to be able to do this:
a=1
b=2
c=3
d=4
function(a,b)
function(c,d)
Is this just wishful thinking?

It can be done but I'm warning you - it won't be pretty! What you can do is to capture the caller frame in your function, then pick up the call line, parse it and extract the arguments passed, then compare them with your function signature and create an argument map, then call your function and once your function finishes compare the changes in the local stack and update the caller frame with the mapped changes. If you want to see how silly it can get, here's a demonstration:
# HERE BE DRAGONS
# No, really, here be dragons, this is strictly for demonstration purposes!!!
# Whenever you use this in code a sweet little pixie is brutally killed!
import ast
import inspect
import sys
def here_be_dragons(funct): # create a decorator so we can, hm, enhance 'any' function
def wrapper(*args, **kwargs):
caller = inspect.getouterframes(inspect.currentframe())[1] # pick up the caller
parsed = ast.parse(caller[4][0], mode="single") # parse the calling line
arg_map = {} # a map for our tracked args to establish global <=> local link
for node in ast.walk(parsed): # traverse the parsed code...
# and look for a call to our wrapped function
if isinstance(node, ast.Call) and node.func.id == funct.__name__:
# loop through all positional arguments of the wrapped function
for pos, var in enumerate(funct.func_code.co_varnames):
try: # and try to find them in the captured call
if isinstance(node.args[pos], ast.Name): # named argument!
arg_map[var] = node.args[pos].id # add to our map
except IndexError:
break # no more passed arguments
break # no need for further walking through the ast tree
def trace(frame, evt, arg): # a function to capture the wrapped locals
if evt == "return": # we're only interested in our function return
for arg in arg_map: # time to update our caller frame
caller[0].f_locals[arg_map[arg]] = frame.f_locals.get(arg, None)
profile = sys.getprofile() # in case something else is doing profiling
sys.setprofile(trace) # turn on profiling of the wrapped function
try:
return funct(*args, **kwargs)
finally:
sys.setprofile(profile) # reset our profiling
return wrapper
And now you can easily decorate your function to enable it to perform this ungodly travesty:
# Zap, there goes a pixie... Poor, poor, pixie. It will be missed.
#here_be_dragons
def your_function(in1, in2):
in1 = in1 + 1
in2 = in2 + 1
And now, demonstration:
a = 1
b = 2
c = 3
d = 4
# Now is the time to play and sing along: Queen - A Kind Of Magic...
your_function(a, b) # bam, two pixies down... don't you have mercy?
your_function(c, d) # now you're turning into a serial pixie killer...
print(a, b, c, d) # Woooo! You made it! At the expense of only three pixie lives. Savage!
# prints: (2, 3, 4, 5)
This, obviously, works only for non-nested functions with positional arguments, and only if you pass simple local arguments, feel free to go down the rabbit hole of handling keyword arguments, different stacks, returned/wrapped/chained calls, and other shenanigans if that's what you fancy.
Or, you know, you can use structures invented for this, like globals, classes, or even enclosed mutable objects. And stop murdering pixies.

If you are looking to modify the value of the variables you could have your code be
def func(a,b):
int1 = a + 2
int2 = b + 3
return int1,int2
a = 2
b = 3
a,b = func(a,b)
This allows you to actually change the values of the a and b variables with the function.

you can do:
def function(in1, in2):
return in1 + 1 , in2 + 1
a, b = function(a,b)
c, d = function(c,d)
python functions are closed -> when function(a,b) s called, a and b get reassigned to a local (to the function) references/pointers in1 and in2, which are not accessible outside of the function. provide references to those new values w/o using globals, you will need to pass that back through return.
When you pass an array or non primitive object into a function, you can modify the object's attributes and have those modifications be visible to other references for that object outside, because the object itself contain the pointers to those values, making the visible to anything else holding a pointer to that object.

Python global keyword [duplicate]

This question already has answers here:
Is it possible to modify a variable in python that is in an outer (enclosing), but not global, scope?
(9 answers)
Closed 5 months ago.
I am confused with the global keyword behavior in below code snippet, I was expecting 30, 30, 30 in all 3 prints.
def outer_function():
#global a ###commented intentionally
a = 20
def inner_function():
global a
a = 30
print('a =',a)
inner_function()
print('a =',a)
a = 10
outer_function()
print('a =',a)
#Output:
#30
#20 #Expecting 30 here
#30
All the confusion coming from "global a" after outer function definition. As my understanding at this point of time is " All the reference and assignment to variable become globally reflected on declaration of global keyword on that variable". If I am uncommenting that first global statement I am getting expected output 30,30,30.
Why global declaration inside inner_function and value change does not reflect on 2nd print i:e to outer_function(or outer scope), whereas got reflected in global namespace.

A common acronym to be familiar with is LEGB:
Local
Enclosed
Global
Built-in
This is the order in which Python will search the namespaces to find variable assignments.
Local
The local namespace is everything that happens within the current code block. Function definitions contain local variables that are the first thing that is found when Python looks for a variable reference. Here, Python will look in the local scope of foo first, find x with the assignment of 2 and print that. All of this happens despite x also being defined in the global namespace.
x = 1
def foo():
x = 2
print(x)
foo()
# prints:
2
When Python compiles a function, it decides whether each of the variables within the definition code block are local or global variables. Why is this important? Let's take a look at the same definition of foo, but flip the two lines inside of it. The result can be surprising
x = 1
def foo():
print(x)
x = 2
foo()
# raises:
UnboundLocalError: local variable 'x' referenced before assignment
This error occurs because Python compiles x as a local variable within foo due to the assignment of x = 2.
What you need to remember is that local variables can only access what is inside of their own scope.
Enclosed
When defining a multi-layered function, variables that are not compiled as local will search for their values in the next highest namespace. Here is a simple example.
x = 0
def outer_0():
x = 1
def outer_1():
def inner():
print(x)
inner()
outer_1()
outer_0()
# print:
1
When inner() is compiled, Python sets x as a global variable, meaning it will try to access other assignments of x outside of the local scope. The order in which Python searches for a value of x in moving upward through the enclosing namespaces. x is not contained in the namespace of outer_1, so it checks outer_0, finds a values and uses that assignment for the x within inner.
x --> inner --> outer_1 --> outer_0 [ --> global, not reached in this example]
You can force a variable to not be local using the keywords nonlocal and global (note: nonlocal is only available in Python 3). These are directives to the compiler about the variable scope.
nonlocal
Using the nonlocal keyword tells python to assign the variable to first instance found as it moves upward through the namespaces. Any changes made to the variable will be made in the variable's original namespace as well. In the example below, when 2 is assigned x, it is setting the value of x in the scope of outer_0 as well.
x = 0
def outer_0():
x = 1
def outer_1():
def inner():
nonlocal x
print('inner :', x)
x = 2
inner()
outer_1()
print('outer_0:', x)
outer_0()
# prints:
inner : 1
outer_0: 2
Global
The global namespace is the highest level namespace that you program is running in. It is also the highest enclosing namespace for all function definitions. In general it is not good practice to pass values in and out of variables in the global namespace as unexpected side effects can occur.
global
Using the global keyword is similar to non-local, but instead of moving upward through the namespace layers, it only searches in the global namespace for the variable reference. Using the same example from above, but in this case declaring global x tells Python to use the assignment of x in the global namespace. Here the global namespace has x = 0:
x = 0
def outer_0():
x = 1
def outer_1():
def inner():
global x
print('inner :', x)
inner()
outer_1()
outer_0()
# prints:
0
Similarly, if a variable is not yet defined in the global namespace, it will raise an error.
def foo():
z = 1
def bar():
global z
print(z)
bar()
foo()
# raises:
NameError: name 'z' is not defined
Built-in
Last of all, Python will check for built-in keywords. Native Python functions such as list and int are the final reference Python checks for AFTER checking for variables. You can overload native Python functions (but please don't do this, it is a bad idea).
Here is an example of something you SHOULD NOT DO. In dumb we overload the the native Python list function by assigning it to 0 in the scope of dumb. In the even_dumber, when we try to split the string into a list of letters using list, Python will find the reference to list in the enclosing namespace of dumb and try to use that, raising an error.
def dumb():
list = 0
def even_dumber():
x = list('abc')
print(x)
even_dumber()
dumb()
# raises:
TypeError: 'int' object is not callable
You can get back the original behavior by referencing the global definition of list using:
def dumb():
list = [1]
def even_dumber():
global list
x = list('abc')
print(x)
even_dumber()
dumb()
# returns:
['a', 'b', 'c']
But again, DO NOT DO THIS, it is bad coding practice.
I hope this helps bring to light some of how the namespaces work in Python. If you want more information, chapter 7 of Fluent Python by Luciano Ramalho has a wonderful in-depth walkthrough of namespaces and closures in Python.

From the documentation:
The global statement is a declaration which holds for the entire
current code block. It means that the listed identifiers are to be
interpreted as globals.
Note it only applies to current code block. So the global in inner_function only applies within inner_function. Outside of it, the identifier is not global.
Note how “identifier” is not the same as “variable”. So what it tells the interpreter is “when I use identifier a within this code block, do not apply normal scope resolution, I actually mean the module-level variable, ”.

Just uncomment your global command in the outer_function, otherwise you're declaring a local variable with value 20, changing a global variable then printing that same local variable.

It's not a good idea use global variabilities. If you want only reset the value of a variable, you just use this lines:
def outer_function():
a = 20
def inner_function():
a = 30
print('a =',a)
return a
a = inner_function()
print('a =',a)
return a
a = 10
a = outer_function()
print('a =',a)

How to make two functions share the same non global variable (Python)

Is there a way to make function B to be able to access a non global variable that was declared in only in function A, without return statements from function A.
As asked, the question:
Define two functions:
p: prints the value of a variable
q: increments the variable
such that
Initial value of the variable is 0. You can't define the variable in the global
enviroment.
Variable is not located in the global environment and the only way to change it is by invoking q().
The global enviroment should know only p() and q().
Tip: 1) In python, a function can return more than 1 value. 2) A function can be
assigned to a variable.
# Example:
>>> p()
0
>>> q()
>>> q()
>>> p()
2

The question says the global enviroment should know only p and q.
So, taking that literally, it could be done inline using a single function scope:
>>> p, q = (lambda x=[0]: (lambda: print(x[0]), lambda: x.__setitem__(0, x[0] + 1)))()
>>> p()
0
>>> q()
>>> q()
>>> p()
2

Using the tips provided as clues, it could be done something like this:
def make_p_and_q():
context = {'local_var': 0}
def p():
print('{}'.format(context['local_var']))
def q():
context['local_var'] += 1
return p, q
p, q = make_p_and_q()
p() # --> 0
q()
q()
p() # --> 2

The collection of things that functions can access is generally called its scope. One interpretation of your question is whether B can access a "local variable" of A; that is, one that is defined normally as
def A():
x = 1
The answer here is "not easily": Python lets you do a lot, but local variables are one of the things that are not meant to be accessed inside a function.
I suspect what your teacher is getting at is that A can modify things outside of its scope, in order to send information out without sending it through the return value. (Whether this is good coding practise is another matter.) For example, functions are themselves Python objects, and you can assign arbitrary properties to Python objects, so you can actually store values on the function object and read them from outside it.
def a():
a.key = "value"
a()
print a.key
Introspection and hacking with function objects
In fact, you can sort of get at the constant values defined in A by looking at the compiled Python object generated when you define a function. For example, in the example above, "value" is a constant, and constants are stored on the code object:
In [9]: a.func_code.co_consts
Out[9]: (None, 'value')
This is probably not what you meant.

Firstly, it's bad practise to do so. Such variables make debugging difficult and are easy to lose track of, especially in complex code.
Having said that, you can accomplish what you want by declaring a variable as global:
def funcA():
global foo
foo = 3
def funcB():
print foo # output is 3

That's one weird homework assignment; especially the tips make me suspect that you've misunderstood or left out something.
Anyway, here's a simpler solution than the accepted answer: Since calls to q increment the value of the variable, it must be a persistent ("static") variable of some sort. Store it somewhere other than the global namespace, and tell p about it. The obvious place to store it is as an attribute of q:
def q():
q.x += 1
q.x = 0 # Initialize
def p():
print(q.x)

Can you explain closures (as they relate to Python)?

I've been reading a lot about closures and I think I understand them, but without clouding the picture for myself and others, I am hoping someone can explain closures as succinctly and clearly as possible. I'm looking for a simple explanation that might help me understand where and why I would want to use them.

Closure on closures
Objects are data with methods
attached, closures are functions with
data attached.
def make_counter():
i = 0
def counter(): # counter() is a closure
nonlocal i
i += 1
return i
return counter
c1 = make_counter()
c2 = make_counter()
print (c1(), c1(), c2(), c2())
# -> 1 2 1 2

It's simple: A function that references variables from a containing scope, potentially after flow-of-control has left that scope. That last bit is very useful:
>>> def makeConstantAdder(x):
... constant = x
... def adder(y):
... return y + constant
... return adder
...
>>> f = makeConstantAdder(12)
>>> f(3)
15
>>> g = makeConstantAdder(4)
>>> g(3)
7
Note that 12 and 4 have "disappeared" inside f and g, respectively, this feature is what make f and g proper closures.

To be honest, I understand closures perfectly well except I've never been clear about what exactly is the thing which is the "closure" and what's so "closure" about it. I recommend you give up looking for any logic behind the choice of term.
Anyway, here's my explanation:
def foo():
x = 3
def bar():
print x
x = 5
return bar
bar = foo()
bar() # print 5
A key idea here is that the function object returned from foo retains a hook to the local var 'x' even though 'x' has gone out of scope and should be defunct. This hook is to the var itself, not just the value that var had at the time, so when bar is called, it prints 5, not 3.
Also be clear that Python 2.x has limited closure: there's no way I can modify 'x' inside 'bar' because writing 'x = bla' would declare a local 'x' in bar, not assign to 'x' of foo. This is a side-effect of Python's assignment=declaration. To get around this, Python 3.0 introduces the nonlocal keyword:
def foo():
x = 3
def bar():
print x
def ack():
nonlocal x
x = 7
x = 5
return (bar, ack)
bar, ack = foo()
ack() # modify x of the call to foo
bar() # print 7

I like this rough, succinct definition:
A function that can refer to environments that are no longer active.
I'd add
A closure allows you to bind variables into a function without passing them as parameters.
Decorators which accept parameters are a common use for closures. Closures are a common implementation mechanism for that sort of "function factory". I frequently choose to use closures in the Strategy Pattern when the strategy is modified by data at run-time.
In a language that allows anonymous block definition -- e.g., Ruby, C# -- closures can be used to implement (what amount to) novel new control structures. The lack of anonymous blocks is among the limitations of closures in Python.

I've never heard of transactions being used in the same context as explaining what a closure is and there really aren't any transaction semantics here.
It's called a closure because it "closes over" the outside variable (constant)--i.e., it's not just a function but an enclosure of the environment where the function was created.
In the following example, calling the closure g after changing x will also change the value of x within g, since g closes over x:
x = 0
def f():
def g():
return x * 2
return g
closure = f()
print(closure()) # 0
x = 2
print(closure()) # 4

# A Closure is a function object that remembers values in enclosing scopes even if they are not present in memory.
# Defining a closure
# This is an outer function.
def outer_function(message):
# This is an inner nested function.
def inner_function():
print(message)
return inner_function
# Now lets call the outer function and return value bound to name 'temp'
temp = outer_function("Hello")
# On calling temp, 'message' will be still be remembered although we had finished executing outer_function()
temp()
# Technique by which some data('message') that remembers values in enclosing scopes
# even if they are not present in memory is called closures
# Output: Hello
Criteria to met by Closures are:
We must have nested function.
Nested function must refer to the value defined in the enclosing function.
Enclosing function must return the nested function.
# Example 2
def make_multiplier_of(n): # Outer function
def multiplier(x): # Inner nested function
return x * n
return multiplier
# Multiplier of 3
times3 = make_multiplier_of(3)
# Multiplier of 5
times5 = make_multiplier_of(5)
print(times5(3)) # 15
print(times3(2)) # 6

Here's a typical use case for closures - callbacks for GUI elements (this would be an alternative to subclassing the button class). For example, you can construct a function that will be called in response to a button press, and "close" over the relevant variables in the parent scope that are necessary for processing the click. This way you can wire up pretty complicated interfaces from the same initialization function, building all the dependencies into the closure.

In Python, a closure is an instance of a function that has variables bound to it immutably.
In fact, the data model explains this in its description of functions' __closure__ attribute:
None or a tuple of cells that contain bindings for the function’s free variables. Read-only
To demonstrate this:
def enclosure(foo):
def closure(bar):
print(foo, bar)
return closure
closure_instance = enclosure('foo')
Clearly, we know that we now have a function pointed at from the variable name closure_instance. Ostensibly, if we call it with an object, bar, it should print the string, 'foo' and whatever the string representation of bar is.
In fact, the string 'foo' is bound to the instance of the function, and we can directly read it here, by accessing the cell_contents attribute of the first (and only) cell in the tuple of the __closure__ attribute:
>>> closure_instance.__closure__[0].cell_contents
'foo'
As an aside, cell objects are described in the C API documentation:
"Cell" objects are used to implement variables referenced by multiple
scopes
And we can demonstrate our closure's usage, noting that 'foo' is stuck in the function and doesn't change:
>>> closure_instance('bar')
foo bar
>>> closure_instance('baz')
foo baz
>>> closure_instance('quux')
foo quux
And nothing can change it:
>>> closure_instance.__closure__ = None
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: readonly attribute
Partial Functions
The example given uses the closure as a partial function, but if this is our only goal, the same goal can be accomplished with functools.partial
>>> from __future__ import print_function # use this if you're in Python 2.
>>> partial_function = functools.partial(print, 'foo')
>>> partial_function('bar')
foo bar
>>> partial_function('baz')
foo baz
>>> partial_function('quux')
foo quux
There are more complicated closures as well that would not fit the partial function example, and I'll demonstrate them further as time allows.

Here is an example of Python3 closures
def closure(x):
def counter():
nonlocal x
x += 1
return x
return counter;
counter1 = closure(100);
counter2 = closure(200);
print("i from closure 1 " + str(counter1()))
print("i from closure 1 " + str(counter1()))
print("i from closure 2 " + str(counter2()))
print("i from closure 1 " + str(counter1()))
print("i from closure 1 " + str(counter1()))
print("i from closure 1 " + str(counter1()))
print("i from closure 2 " + str(counter2()))
# result
i from closure 1 101
i from closure 1 102
i from closure 2 201
i from closure 1 103
i from closure 1 104
i from closure 1 105
i from closure 2 202

we all have used Decorators in python. They are nice examples to show what are closure functions in python.
class Test():
def decorator(func):
def wrapper(*args):
b = args[1] + 5
return func(b)
return wrapper
#decorator
def foo(val):
print val + 2
obj = Test()
obj.foo(5)
here final value is 12
Here, the wrapper function is able to access func object because wrapper is "lexical closure", it can access it's parent attributes.
That is why, it is able to access func object.

I would like to share my example and an explanation about closures. I made a python example, and two figures to demonstrate stack states.
def maker(a, b, n):
margin_top = 2
padding = 4
def message(msg):
print('\n’ * margin_top, a * n,
' ‘ * padding, msg, ' ‘ * padding, b * n)
return message
f = maker('*', '#', 5)
g = maker('', '♥’, 3)
…
f('hello')
g(‘good bye!')
The output of this code would be as follows:
***** hello #####
 good bye! ♥♥♥
Here are two figures to show stacks and the closure attached to the function object.
when the function is returned from maker
when the function is called later
When the function is called through a parameter or a nonlocal variable, the code needs local variable bindings such as margin_top, padding as well as a, b, n. In order to ensure the function code to work, the stack frame of the maker function which was gone away long ago should be accessible, which is backed up in the closure we can find along with the 'message's function object.

For me, "closures" are functions which are capable to remember the environment they were created. This functionality, allows you to use variables or methods within the closure wich, in other way,you wouldn't be able to use either because they don't exist anymore or they are out of reach due to scope. Let's look at this code in ruby:
def makefunction (x)
def multiply (a,b)
puts a*b
end
return lambda {|n| multiply(n,x)} # => returning a closure
end
func = makefunction(2) # => we capture the closure
func.call(6) # => Result equal "12"
it works even when both, "multiply" method and "x" variable,not longer exist. All because the closure capability to remember.

The best explanation I ever saw of a closure was to explain the mechanism. It went something like this:
Imagine your program stack as a degenerate tree where each node has only one child and the single leaf node is the context of your currently executing procedure.
Now relax the constraint that each node can have only one child.
If you do this, you can have a construct ('yield') that can return from a procedure without discarding the local context (i.e. it doesn't pop it off the stack when you return). The next time the procedure is invoked, the invocation picks up the old stack (tree) frame and continues executing where it left off.