When I come across closure in python, from the below code, I don't understand how the values for the arguments assigned for x and y when calling the function. In Point 1, we passing argument value 5 for x. It is then assigned to x in the function. In Point 2, we passing 7 to inc5, My doubt arises here, again the value 7 should assign to x, but instead how come it is assigning to y.
def makeInc(x):
def inc(y):
return y + x
return inc
inc5 = makeInc(5) #Point 1
inc5(7) #Point 2
I can't understand why you think 7 should be assigned to x. The outer function makeInc returns the inner function inc, and it's that which you assign to inc5. So inc5 accepts the argument y, and that's what is bound to the value 7.
At Point 1 you are passing a varialbe inc5 the returning result of a makeInc function, which will be another function inc with a local variable x set to 5. Then you call for this new function and pass a 7 as its one and only parameter y.
Related
This question already has answers here:
Why can a function modify some arguments as perceived by the caller, but not others?
(13 answers)
Closed 7 months ago.
Once I define a list in main() and passed it as a function parameter, I expected that its content will remain the same once we exit the function. I have the following code snippet:
def test(a):
a.append(1000)
return a
def main():
a = [1,2,3]
print(a, test(a))
main()
Output is clear.
# Output: [1,2,3,1000] [1,2,3,1000]
Even though the initial list contains 3 elements, function test() seems to modify its content. This thing does not happen to variables:
def test(a):
a = a + 10
return a
def main():
a = 5
print(a, test(a))
main()
# Output: 5 15
Why is that? Why does Python modify the initial lists and not variables?
a.append()
mutates the a which was passed by reference
a = a + 10
assigns a value to a new local variable in the function called a
Integers are immutable, but if there were some method that could be called on that a (on the right hand side) to mutate it, it would also be reflected in the caller.
Some objects in python (like lists, dicts and sets) are mutable, whereas others (like tuples, frozensets, strings and numbers) are immutable.
Operations on mutable objects modify the object themselves. Many immutable objects also allow operations, which in this case return a new object. (Internally this is implemented with the dunder "magic methods", which in this case just return a new object. x OPERATOR y actually calls a method on x(and/or y).)
Since in python x = y binds the name x to point to y, if y is mutable and you modify it, x will point to the modified y. But if y is immutable (but allows operations), operations on y will create a new y, to which x no longer points:
x = [] # mutable
y = x # y and x both point to a particular '[]'
x += ["hi"]
y == ["hi"] # True
x = 0
y = x # y and x both point to 0
x += 7 # x now points to 7
y == 7 # False
Exactly the same thing goes on when you parse an argument to a function---you bind a name in the function's scope pointing to that object. What then happens to the object pointed to by the name depends upon the type of object and what you do to it. If it's mutable and you mutate the object pointed to by the name in the function, you will mutate the object itself, and any other names bound to that object ouside the function will see the changed object.
This is the reason that we don't generally do this:
def f(x=[]):
...
Since f is evaluated once, when the module is first loaded, x will point to the same list for the duration of the function. Thus if the function does anything to x, x will be in that state next time it's called (which is rarely what you want).
def f():
print (x)
def g():
print (x)
x = 1
x = 3
f()
x = 3
g()
I am learning functions now, why are empty arguments used in these functions ?
No arguments are being used (as none are being passed).
Inside f, x is a free variable. There is no local variable by that name, so its value is looked up in the next enclosing scope, which in this case is the global scope.
Inside g, x is a local variable (by virtue of a value being assigned to it), but it is not yet defined when you call print(x). If you were to reverse the order of the assignment and the call to print, you would see 1 as the output, since the global variable by the same name is ignored.
I have come across this example from Python hitchhikers guide:
def create_multipliers():
return [lambda x, i=i : i * x for i in range(5)]
The example above is the solution to some issues caused with late binding, where variables used in closures are looked up at the time the inner function is called.
What does the i=i mean and why is it making such difference?
It's actually not just for lambdas; any function that takes default parameters will use the same syntax. For example
def my_range(start, end, increment=1):
ans = []
while start < end:
ans.append(start)
start += increment
return ans
(This is not actually how range works, I just thought it would be a simple example to understand). In this case, you can call my_range(5,10) and you will get [5,6,7,8,9]. But you can also call my_range(5,10,increment=2), which will give you [5, 7, 9].
You can get some surprising results with default arguments. As this excellent post describes, the argument is bound at function definition, not at function invocation as you might expect. That causes some strange behavior, but it actually helps us here. Consider the incorrect code provided in your link:
def create_multipliers():
return [lambda x : i * x for i in range(5)]
for multiplier in create_multipliers():
print multiplier(2)
When you call multiplier(2), what is it actually doing? It's taking your input parameter, 2, and returning i * 2. But what is i? The function doesn't have any variable called i in its own scope, so it checks the surrounding scope. In the surrounding scope, the value of i is just whatever value you left it -- in this case 4. So every function gives you 8.
On the other hand, if you provide a default parameter, the function has a variable called i in its own scope. What's the value of i? Well, you didn't provide one, so it uses its default value, which was bound when the function was defined. And when the function was defined, i had a different value for each of the functions in your list!
It is a bit confusing that they've used the same name for the parameter variable as they did for the iterating variable. I suspect you could get the same result with greater readability with
def create_multipliers():
return [(lambda x, y=i: y*x) for i in range(5)]
In that case, each number in the range, will be assigned to the optional parameters of each lambda function:
def create_multipliers():
return [lambda x, i=i : i * x for i in range(5)]
lambda x, i=0
lambda x, i=1
lambda x, i=2
lambda x, i=3
lambda x, i=4
So, you can call the functions now with one parameter (because they already have the default)
for f in create_multipliers():
print(f(3))
0
3
6
9
12
Or you can call the function and give the parameter you want, that's why is optional
for f in create_multipliers():
print(f(3,2))
6
6
6
6
6
There are examples where optional parameter are needed, such as recursion
For example, square in terms of square:
square = lambda n, m=0: 0 if n==m else n+square(n,m+1)
Look that the optional parameter there is used as accumulator
This question already has answers here:
What are the differences between swap in C++ and Python?
(4 answers)
Closed 7 years ago.
I think the output should be x = 5, y = 3. But, when I tried executing it in jes it shows that x=3 and y=5.
My swap function is as follows:
def swap (x,y):
temp=x
x=y
y=temp
And from my driver function I call swap():
def driver():
x=3
y=5
swap(x,y)
print x
print y
I want to know why isn't the output as expected?
Well this is not a big issue in python you can return multiple values like try this snippet it might help you.
def swap(a,b):
return (b,a)
def driver(x,y):
print "Previous x=",x,"y=",y
x,y = swap(x,y)
print "Now x=", x, "y=",y
driver(3,5)
As other answers have suggested, this really doesn't require functions. Since you have decided to use them though, you might as well try and understand a bit more about scopes.
In order to retain the swapped values you need to return them or else they get lost after the execution of swap():
def swap (x,y):
temp=x
x=y
y=temp
return x, y # must return the swapped values.
Then, in the function where you call swap() you assign the returned values to the variables that you swapped:
def driver():
x=3
y=5
x, y = swap(x,y) # re-assign values
print x
print y
Running driver() now will give you the swapped value:
5
3
This happens because in the function swap() the variables x and y you pass as arguments are treated as local to that function, only swap can see their value. If you don't return this value back, it is lost and forgotten.
Integers are immutable in python. You are setting y to x and x to y in the local scope, but are not modifying the reference to the initial integers passed in. In python, the best you can do if you want to encapsulate a swap function that modifies in-place instead of by return is pass some kind of container object that is mutable and contain references to the thing you want to swap:
def swap(container):
container[0], container[1] = container[1], container[0]
And then call it like so:
x = 3
y = 5
container = [x,y]
swap(container)
FWIW, swap in python can be implemented in one line as simply as:
x, y = y, x
Which is probably syntactically clearer in most cases anyway.
See also Python passing an integer by reference
In python assignment changes the identity of an object rather than its value (unless you are mutating the content). This is worth noting that, as per Python Data Model
Every object has an identity, a type and a value.
So interchanging the identities inside swap would not change the identity inside the driver. Moreover, python does not allow you to change the value of an immutable types so, there is no other possible ways to swap the values inside the swap method.
Id like to define a list of functions by a list of parameters, to actually compose a potential from a finite number of monopoles and dipoles.
Functions in python list, however, for some reason do not behave as expected (by me).
The following example:
def function(coefficients):
phi = []
i = 0
for y in coefficients:
f = lambda x: y*(x**i)
i += 1
phi.append(f)
print f(2)
return phi
> phi = function([1,2,3])
2
8
24
> k0 = phi[0]
> k1 = phi[1]
> k2 = phi[2]
> print k0(2), k1(2), k2(2)
24 24 24
Always the last function in the list is retrieved independent of which function is picked from the list.
Any ideas?
Thanks in advance.
The lambda function holds a reference to the variable y, because of the closure property. So, all the functions have the reference to the same object and they will get the last value of y during the invocation. To fix this,
f = lambda x,y=y,i=i: y*(x**i)
Now, we have created another parameter y and that has the default value y. It makes the function depend on a local reference now. So, even if the value of y changes in the loop, the lambda function no longer refers to that y.
Note: The same explanation is applicable to the variable i as well.
To make things very clear, I would recommend using different variable names
f = lambda x, z=y, j=i: z * (x ** j)
Note: Your original output was 2 8 24. That was because, when you invoke the function, i got incremented and I don't think that is what you really wanted. The actual output should have been 1 4 12.