将列表清空到新列表中

Question

I'm trying to make a function that when a list is empty, a second list will empty into the empty one in reverse order. Right now, I tried to do:

a = [1,2,3,4]
b = []
def a_to_b(a, b):
    if not a:
        print('a:',a)        
        print('b:',b)
        for c in b:
            a.append(c)
            b.remove(c)
        print('a:',a)
        print('b:',b)
        return True
    else:
        top = a.pop()
        b.append(top)
        print('a:',a)
        print('b:',b)
        return True

I want it to be after each run:

1) a = [1,2,3]
   b = [4]
2) a = [1,2]
   b = [4,3]
3) a = [1]
   b = [4,3,2]
4) a = []
   b = [4,3,2,1]
5) a = [1,2,3,4]
   b = []

But after this fifth run it is giving me

a = [4,2]
b = [3,1]

And I cannot figure out why it is only applying to every other number in b.

Answer 1

This should work

 a = [1,2,3,4] b = [] for i in range(len(a)): b.append(a[-1]) a.pop() print a,b 

Answer 2

Your problem is caused by you removing elements that you're hopping over in the for loop)

Pointer is going as index 0, 1, 2, 3 but you're already removing the 0th element causing the pointer to go straight to 2 (which is now index 1 in remaining list)

to avoid it, you could change your code to:

for c in b:
    a.append(c)
for c in a:
    b.remove(c)

Answer 3

Here's the reason why you're getting a weird result:

    for c in b:
        a.append(c)
        b.remove(c)

You're changing list b as you're iterating over it. Thus things are not going to come out as you expect them to. Why don't you just do

b.reverse()
a = b
b = []

In the place of what you had before. So it'd be

a = [1,2,3,4]
b = []
def a_to_b(a, b):
    if not a:
        print('a:',a)        
        print('b:',b)
        b.reverse()
        a = b
        b = []
        print('a:',a)
        print('b:',b)
        return True
    else:
        top = a.pop()
        b.append(top)
        print('a:',a)
        print('b:',b)
        return True

Answer 4

def f(a, b):
    if a:
        b.append(a.pop())
    else:
        while b:
            a.append(b.pop())
    print 'a: %s' % a
    print 'b: %s' % b

>>> a = [1, 2, 3, 4]
>>> b = []
>>> f(a, b)
a: [1, 2, 3]
b: [4]

>>> f(a, b)
a: [1, 2]
b: [4, 3]

>>> f(a, b)
a: [1]
b: [4, 3, 2]

>>> f(a, b)
a: []
b: [4, 3, 2, 1]

>>> f(a, b)
a: [1, 2, 3, 4]
b: []

Answer 5

The problem is here:

for c in b:
        a.append(c)
        b.remove(c)

You cant remove objects from the middle of an interable without confusing the loop.

Heres what happens:

You have b=[4,3,2,1] and a=[] . You start the for loop and c is pointing to the first index of b, ie 4. You remove c from the list and put it in a .

Now you have b = [3,2,1] and a=[4] like you expect.

When you try to start the next loop, your index is incremented ( c is now pointing at the 2nd element) but the problem is you've messed with the structure of the iterable. So the loop removes c like its supposed to, but c=2 , not 3 like you are expecting.

Now you have a=[4,2] and b=[1,3] and when the loop checks for index 3, it finds that b is only has 2 elements so it exits.