Python passing list as argument

If i were to run this code:

def function(y):
    y.append('yes')
 开发者_如何学JAVA   return y

example = list()
function(example)
print(example)

Why would it return ['yes'] even though i am not directly changing the variable 'example', and how could I modify the code so that 'example' is not effected by the function?

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Everything is a reference in Python. If you wish to avoid that behavior you would have to create a new copy of the original with list(). If the list contains more references, you'd need to use deepcopy()

def modify(l):
 l.append('HI')
 return l

def preserve(l):
 t = list(l)
 t.append('HI')
 return t

example = list()
modify(example)
print(example)

example = list()
preserve(example)
print(example)

outputs

['HI']
[]

---

The easiest way to modify the code will be add the [:] to the function call.

def function(y):
    y.append('yes')
    return y



example = list()
function(example[:])
print(example)

---

"Why would it return ['yes']"

Because you modified the list, example.

"even though i am not directly changing the variable 'example'."

But you are, you provided the object named by the variable example to the function. The function modified the object using the object's append method.

As discussed elsewhere on SO, append does not create anything new. It modifies an object in place.

See Why does list.append evaluate to false?, Python append() vs. + operator on lists, why do these give different results?, Python lists append return value.

and how could I modify the code so that 'example' is not effected by the function?

What do you mean by that? If you don't want example to be updated by the function, don't pass it to the function.

If you want the function to create a new list, then write the function to create a new list.

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Its because you called the function before printing the list. If you print the list then call the function and then print the list again, you would get an empty list followed by the appended version. Its in the order of your code.