In Python, why do function parameters maintain their value beetween function calls

Question

I'm fairly new at Python and i cannot wrap my head around the results im getting Using the code below:

def func(a,b=set()):
    res=list()
    for i in a:
        if i not in b:
            res.append(i)
            b|={i}
    return res

print(func([1,1,2,2,3,4]))
print(func([1,1,2,2,3,4]))

I was getting output:

[1,2,3,4]
[]

I put "print(b)" above "res=list()" and got output:

set()
[1,2,3,4]
{1,2,3,4}
[]

What is going on? Shouldn't "b" be set to "set()" when i call the function? Im using Python 3.6

Have a read of the important warning herehttps://docs.python.org/3/tutorial/controlflow.html#default-argument-values — match, Jan 28 '18 at 09:56

score 4 · Accepted Answer · answered Jan 28 '18 at 10:07

Have a look at the documentation for default parameters:

The default value is evaluated only once. This makes a difference when the default is a mutable object such as a list, dictionary, or instances of most classes.

When you define a function with a default parameter, the default value is only evaluated when the definition is first executed by the interpreter (the actual def statement). This is usually not a problem, except for when a mutable default value is used. That is to say, one that can be modified in place.

In your case, when you modify b in your function the first time you call it, it keeps that value the next time around. To avoid this, you can do like so:

def func(a,b=None):
    if b is None:
        b = set()
    res=list()
    for i in a:
        if i not in b:
            res.append(i)
            b|={i}
    return res

Now b will always have the default value you want.

score 2 · Answer 2 · answered Jan 28 '18 at 10:05

In python functions are objects and arguments are evaluated and executed once. this is a nice explanation : http://effbot.org/zone/default-values.htm

in your example, it could be "fixed" by doing:

def func(a,b=None):
    if b is None:
        b = set()
    res=list()
    for i in a:
        if i not in b:
            res.append(i)
            b|={i}
    return res

score 1 · Answer 3 · edited Jun 20 '20 at 09:12

In First function call b is empty See here

In second function call b is already fill with element

So this condition not run if i not in b and return empty list

Try this

def func(a):
    res=list()
    b=set()
    for i in a:
        if i not in b:
            res.append(i)
            b|={i}
    return res

print(func([1,1,2,2,3,4]))
print(func([1,1,2,2,3,4]))

Output

[1, 2, 3, 4]

[1, 2, 3, 4]

In Python, why do function parameters maintain their value beetween function calls

3 Answers3