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 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.
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