This question already has answers here:
"Least Astonishment" and the Mutable Default Argument
(33 answers)
Closed 1 year ago.
I'm a bit confused about function variable scope, sometimes the function keeps track of its local variables so that if it's called again it recalls the previous values of the variables - and other times it doesn't keep track of the values.
In the following code, the function keeps track of the mylist and appends to it at every call.
def test(mylist=[]):
mylist.append(1)
print (mylist)
test()
test()
test()
The output:
[1]
[1, 1]
[1, 1, 1]
While in the following code, x is set to zero each time the function is called.
def test(x=0):
x+=1
print (x)
test()
test()
test()
The output:
1
1
1
What is the explanation of this behavior?
Also, is there a way to take a look at the current values of variables inside the functions from outside?
[Update]
After comments, I now understand why mylist is updated, but what about x in the second case? why is it not updated? is it because x is immutable while mylist is mutable?
It's not mylist that's kept between invocations of the function, but rather the default value [], which is a single anonymous array that's assigned by reference to mylist each time.
Appending to mylist of course appends to the array that it refers to.
In the first case, x contains a reference to an array, and when you "add to x" you modify that array.
In the second case, x simply contains a number 0, and when you "add to x", you change what x contains.
The key point is that the expression for the default parameter value is evaluated only once, and re-used between calls to the function.
Welcome to Python :-(
Related
This question already has answers here:
How to modify list entries during for loop?
(10 answers)
Closed 5 months ago.
When I try this code:
bar = [1,2,3]
print(bar)
for foo in bar:
print(id(foo))
foo = 0
print(id(foo))
print(bar)
I get this result:
[1, 2, 3]
5169664
5169676
5169652
5169676
5169640
5169676
[1, 2, 3]
I expected the end result to be [0,0,0] and that id would return identical values for each iteration. Why does it behave like this? How can I elegantly assign back to the elements of the list, without using enumerate or range(len(bar))?
See also: How to change variables fed into a for loop in list form
First of all, you cannot reassign a loop variable—well, you can, but that won’t change the list you are iterating over. So setting foo = 0 will not change the list, but only the local variable foo (which happens to contain the value for the iteration at the begin of each iteration).
Next thing, small numbers, like 0 and 1 are internally kept in a pool of small integer objects (This is a CPython implementation detail, doesn’t have to be the case!) That’s why the ID is the same for foo after you assign 0 to it. The id is basically the id of that integer object 0 in the pool.
If you want to change your list while iterating over it, you will unfortunately have to access the elements by index. So if you want to keep the output the same, but have [0, 0, 0] at the end, you will have to iterate over the indexes:
for i in range(len(bar)):
print id(bar[i])
bar[i] = 0
print id(bar[i])
print bar
Otherwise, it’s not really possible, because as soon as you store a list’s element in a variable, you have a separate reference to it that is unlinked to the one stored in the list. And as most of those objects are immutable and you create a new object when assigning a new value to a variable, you won’t get the list’s reference to update.
Yes, the output you got is the ordinary Python behavior. Assigning a new value to foo will change foo's id, and not change the values stored in bar.
If you just want a list of zeroes, you can do:
bar = [0] * len(bar)
If you want to do some more complicated logic, where the new assignment depends on the old value, you can use a list comprehension:
bar = [x * 2 for x in bar]
Or you can use map:
def double(x):
return x * 2
bar = map(double, bar)
you actually didnt change the list at all.
the first thing for loop did was to assign bar[0] to foo(equivalent to foo = bar[0]). foo is just an reference to 1. Then you assign another onject 0 to foo. This changed the reference of foo to 0. But you didnt change bar[0]. Remember, foo as a variable, references bar[0], but assign another value/object to foo doesn't affect bar[0] at all.
bar = [0 for x in bar]
Long answer : foo is just a local name, rebinding does not impact the list. Python variables are really just key:value pairs, not symbolic names for memory locations.
This question already has answers here:
Use value of variable in lambda expression [duplicate]
(3 answers)
Closed 7 years ago.
I've experienced some strange behaviour when storing lambda functions into a dictionary: If you try to pass some default value to a function in a loop, only the last default value is being used.
Here some minimal example:
#!/usr/bin/env python
# coding: utf-8
def myfct(one_value, another_value):
"do something with two int values"
return one_value + another_value
fct_dict = {'add_{}'.format(number): (lambda x: myfct(x, number))
for number in range(10)}
print('add_3(1): {}, id={}'.format(fct_dict['add_3'](1), id(fct_dict['add_3'])))
print('add_5(1): {}, id={}'.format(fct_dict['add_5'](1), id(fct_dict['add_5'])))
print('add_9(1): {}, id={}'.format(fct_dict['add_9'](1), id(fct_dict['add_9'])))
The output reads as follows
add_3(1): 10, id=140421083875280
add_5(1): 10, id=140421083875520
add_9(1): 10, id=140421083876000
You get dissimilar functions (id not identical) but every function uses the same second argument.
Can somebody explain what's going on?
The same holds with python2, python3, pypy...
The fix:
def make_closure(number):
return lambda x: myfct(x, number)
used as
{'add_{}'.format(number): make_closure(number) for number in range(10)}
The reason for this behaviour is, that the variable number (think: named memory location here) is the same during all iterations of the loop (though its actual value changes in each iteration). "Loop" here refers to the dictionary comprehension, which internally is based on a loop. All lambda instances created in the loop will close over the same "location", which retains the value last assigned to it (in the last iteration of the loop).
The following code is not what actually happens underneath. It is merely provided to shed light on the concepts:
# Think of a closure variable (like number) as being an instance
# of the following class
class Cell:
def __init__(self, init=None):
self.value = None
# Pretend, the compiler "desugars" the dictionary comprehension into
# something like this:
hidden_result_dict = {}
hidden_cell_number = Cell()
for number in range(10):
hidden_cell_number.value = number
hidden_result_dictionary['add_{}'.format(number)] = create_lambda_closure(hidden_cell_number)
All lambda closures created by the create_lambda_closure operation share the very same Cell instance and will grab the value attribute at run-time (i.e., when the closure is actually called). By that time, value will refer to last value ever assigned to it.
The value of hidden_result_dict is then answered as the result of the dict comprehension. (Again: this is only meant as be read on a "conceptual" level; it has no relation to the actual code executed by the Python VM).
number is a variable which has different value for each iteration of the dict comprehension. But when you do lambda x: myfct(x, number), it does not use value of number. It just creates a lambda method that will use value of number when it will be called/used. So when you use you add_{} methods, number has value 9 which is used in every call to myfct(x, number).
Suppose I have this function
>>>a=3
>>>def num(a):
a=5
return a
>>>num(a)
5
>>>a
3
Value of a doesnt change.
Now consider this code :
>>> index = [1]
>>> def change(a):
a.append(2)
return a
>>> change(index)
>>> index
>>> [1,2]
In this code the value of index changes.
Could somebody please explain what is happening in these two codes. As per first code, the value of index shouldnt change(ie should remain index=[1]).
You need to understand how python names work. There is a good explanation here, and you can click here for an animation of your case.
If you actually want to operate on a separate list in your function, you need to make a new one, for instance by using
a = a[:]
before anything else. Note that this will only make a new list, but the elements will still be the same.
The value of index doesn't change. index still points to the same object it did before. However, the state of the object index points to has changed. That's just how mutable state works.
Line 3 in the first block of code is assignment and in the second block is mutation, and that's why you are observing that behavior.
The issue you are encountering is:
a = 3
def num(a):
# `a` is a reference to the argument passed, here 3.
a = 5
# Changed the reference to point at 5, and return the reference.
return a
num(a)
The a in the num function is a diffrent object than the a defined globally.
It works in case of the list because a points at the list passed, and you modify the object being referenced to by the variable, not the reference variable itself.
This question already has answers here:
"Least Astonishment" and the Mutable Default Argument
(33 answers)
Closed 6 months ago.
#!/usr/bin/env python3.2
def f1(a, l=[]):
l.append(a)
return(l)
print(f1(1))
print(f1(1))
print(f1(1))
def f2(a, b=1):
b = b + 1
return(a+b)
print(f2(1))
print(f2(1))
print(f2(1))
In f1 the argument l has a default value assignment, and it is only evaluated once, so the three print output 1, 2, and 3. Why f2 doesn't do the similar?
Conclusion:
To make what I learned easier to navigate for future readers of this thread, I summarize as the following:
I found this nice tutorial on the topic.
I made some simple example programs to compare the difference between mutation, rebinding, copying value, and assignment operator.
This is covered in detail in a relatively popular SO question, but I'll try to explain the issue in your particular context.
When your declare your function, the default parameters get evaluated at that moment. It does not refresh every time you call the function.
The reason why your functions behave differently is because you are treating them differently. In f1 you are mutating the object, while in f2 you are creating a new integer object and assigning it into b. You are not modifying b here, you are reassigning it. It is a different object now. In f1, you keep the same object around.
Consider an alternative function:
def f3(a, l= []):
l = l + [a]
return l
This behaves like f2 and doesn't keep appending to the default list. This is because it is creating a new l without ever modifying the object in the default parameter.
Common style in python is to assign the default parameter of None, then assign a new list. This gets around this whole ambiguity.
def f1(a, l = None):
if l is None:
l = []
l.append(a)
return l
Because in f2 the name b is rebound, whereas in f1 the object l is mutated.
This is a slightly tricky case. It makes sense when you have a good understanding of how Python treats names and objects. You should strive to develop this understanding as soon as possible if you're learning Python, because it is central to absolutely everything you do in Python.
Names in Python are things like a, f1, b. They exist only within certain scopes (i.e. you can't use b outside the function that uses it). At runtime a name refers to a value, but can at any time be rebound to a new value with assignment statements like:
a = 5
b = a
a = 7
Values are created at some point in your program, and can be referred to by names, but also by slots in lists or other data structures. In the above the name a is bound to the value 5, and later rebound to the value 7. This has no effect on the value 5, which is always the value 5 no matter how many names are currently bound to it.
The assignment to b on the other hand, makes binds the name b to the value referred to by a at that point in time. Rebinding the name a afterwards has no effect on the value 5, and so has no effect on the name b which is also bound to the value 5.
Assignment always works this way in Python. It never has any effect on values. (Except that some objects contain "names"; rebinding those names obviously effects the object containing the name, but it doesn't affect the values the name referred to before or after the change)
Whenever you see a name on the left side of an assignment statement, you're (re)binding the name. Whenever you see a name in any other context, you're retrieving the (current) value referred to by that name.
With that out of the way, we can see what's going on in your example.
When Python executes a function definition, it evaluates the expressions used for default arguments and remembers them somewhere sneaky off to the side. After this:
def f1(a, l=[]):
l.append(a)
return(l)
l is not anything, because l is only a name within the scope of the function f1, and we're not inside that function. However, the value [] is stored away somewhere.
When Python execution transfers into a call to f1, it binds all the argument names (a and l) to appropriate values - either the values passed in by the caller, or the default values created when the function was defined. So when Python beings executing the call f3(5), the name a will be bound to the value 5 and the name l will be bound to our default list.
When Python executes l.append(a), there's no assignment in sight, so we're referring to the current values of l and a. So if this is to have any effect on l at all, it can only do so by modifying the value that l refers to, and indeed it does. The append method of a list modifies the list by adding an item to the end. So after this our list value, which is still the same value stored to be the default argument of f1, has now had 5 (the current value of a) appended to it, and looks like [5].
Then we return l. But we've modified the default list, so it will affect any future calls. But also, we've returned the default list, so any other modifications to the value we returned will affect any future calls!
Now, consider f2:
def f2(a, b=1):
b = b + 1
return(a+b)
Here, as before, the value 1 is squirreled away somewhere to serve as the default value for b, and when we begin executing f2(5) call the name a will become bound to the argument 5, and the name b will become bound to the default value 1.
But then we execute the assignment statement. b appears on the left side of the assignment statement, so we're rebinding the name b. First Python works out b + 1, which is 6, then binds b to that value. Now b is bound to the value 6. But the default value for the function hasn't been affected: 1 is still 1!
Hopefully that's cleared things up. You really need to be able to think in terms of names which refer to values and can be rebound to point to different values, in order to understand Python.
It's probably also worth pointing out a tricky case. The rule I gave above (about assignment always binding names with no effect on the value, so if anything else affects a name it must do it by altering the value) are true of standard assignment, but not always of the "augmented" assignment operators like +=, -= and *=.
What these do unfortunately depends on what you use them on. In:
x += y
this normally behaves like:
x = x + y
i.e. it calculates a new value with and rebinds x to that value, with no effect on the old value. But if x is a list, then it actually modifies the value that x refers to! So be careful of that case.
In f1 you are storing the value in an array or better yet in Python a list where as in f2 your operating on the values passed. Thats my interpretation on it. I may be wrong
Other answers explain why this is happening, but I think there should be some discussion of what to do if you want to get new objects. Many classes have the method .copy() that allows you create copies. For instance, if we rewrite f1 as
def f1(a, l=[]):
new_l = l.copy()
new_l.append(a)
return(new_l)
then it will continue to return [1]no matter how many times we call it. There is also the library https://docs.python.org/3/library/copy.html for managing copies.
Also, if you're looping through the elements of a container and mutating them one by one, using comprehensions not only is more Pythonic, but can avoid the issue of mutating the original object. For instance, suppose we have the following code:
data = [1,2,3]
scaled_data = data
for i, value in enumerate(scaled_data):
scaled_data[i] = value/sum(data)
This will set scaled_data to [0.16666666666666666, 0.38709677419354843, 0.8441754916792739]; each time you set a value of scaled_data to the scaled version, you also change the value in data. If you instead have
data = [1,2,3]
scaled_data = [x/sum(data) for x in data]
this will set scaled_data to [0.16666666666666666, 0.3333333333333333, 0.5] because you're not mutating the original object but creating a new one.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
“Least Astonishment” in Python: The Mutable Default Argument
I'm kind of confused about how optional parameters work in Python functions/methods.
I have the following code block:
>>> def F(a, b=[]):
... b.append(a)
... return b
...
>>> F(0)
[0]
>>> F(1)
[0, 1]
>>>
Why F(1) returns [0, 1] and not [1]?
I mean, what is happening inside?
Good doc from PyCon a couple years back - Default parameter values explained. But basically, since lists are mutable objects, and keyword arguments are evaluated at function definition time, every time you call the function, you get the same default value.
The right way to do this would be:
def F(a, b=None):
if b is None:
b = []
b.append(a)
return b
Default parameters are, quite intuitively, somewhat like member variables on the function object.
Default parameter values are evaluated when the function definition is executed. This means that the expression is evaluated once, when the function is defined, and that same “pre-computed” value is used for each call. This is especially important to understand when a default parameter is a mutable object, such as a list or a dictionary: if the function modifies the object (e.g. by appending an item to a list), the default value is in effect modified.
http://docs.python.org/reference/compound_stmts.html#function
Lists are a mutable objects; you can change their contents. The correct way to get a default list (or dictionary, or set) is to create it at run time instead, inside the function:
def good_append(new_item, a_list=None):
if a_list is None:
a_list = []
a_list.append(new_item)
return a_list