在 python 的 for 循环中列出 append 的可变性

Question

I'm new to the concept of mutability of lists in the context of for loops.

Can someone explain why the output for the below code is an empty list?

I would expect L3 = [3,4] .

Thanks!

L1 = [1,2,3,4]
L2 = [1,2,5,6]

def no_dups(L1, L2):
    L3 = []
    for e in L1:
        if e not in L2:
            L3.append(e)
    return(L3)

print(L3)

Answer 1

The last line print(L3) does not even call no_dups function. It throws an expection because L3 is not defined outside of function.

Probably you meant this:

L3 = no_dups(L1, L2)
print(L3)

In which case you do inded get the expected result.

Answer 2

This works:

>>> L1 = [1,2,3,4]
>>> L2 = [1,2,5,6]
>>> 
>>> def no_dups(L1, L2):
...     L3 = []
...     for e in L1:
...         if e not in L2:
...             L3.append(e)
...     return(L3)
... 
>>> print(no_dups(L1,L2))
[3, 4]
>>> 

L3 does not actually exists outside of no_dups() , so you cannot print it or whatever