Python "for in" loop to print the last item in a list

Question

Recently, I found out about lists and loops, as well as a command .pop()that points and removes the last item in a list.

So, I tried to write code to remove the last items in the list one by one until it was left with only one item.

The code:

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

for i in list_A:
    print(list_A.pop())
    if 'c' not in list_A:
        break

print("job done.")

The result of python 3.6 gives me the following:

/Library/Frameworks/Python.framework/Versions/3.6/bin/python3.6
j
i
h
g
f
job done.

As you can see, it really worked, but half that?

I expected:

j
i
h
g
f
e
d
c
job done

I mean, it will be more convenient for me if he returns some error, this means that the code is incorrect. But why did it work, but not completely?

Answer 1

.

while:

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

while 'c' in list_A:
    print(list_A.pop())

print('job done')

:

j
i
h
g
f
e
d
c
job done

( ):

try:
    pos = list_A.index('c')
    list_A[:] = list_A[:pos]
    # del list_A[pos:]           # more efficient alternative suggested by @ShadowRanger
except ValueError as e:
    pass

Answer 2

for..in Python .

:

• , a
• pop() , j
• 5 . , .
• e f
• a - e, e,

, , , - . while, , . :

while list_A:
    print(list_A.pop())
    if "c" not in list_A:
        break

, , , c.

Answer 3

, : , .

( ) , :

class CustomIterator(object):
    def __init__(self, seq):
        self.seq = seq
        self.idx = 0

    def __iter__(self):
        return self

    def __next__(self):
        print('give next element:', self.idx)
        for idx, item in enumerate(self.seq):
            if idx == self.idx:
                print(idx, '--->', item)
            else:
                print(idx, '    ', item)
        try:
            nxtitem = self.seq[self.idx]
        except IndexError:
            raise StopIteration
        self.idx += 1
        return nxtitem

    next = __next__  # py2 compat

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

for i in CustomIterator(list_A):
    print(list_A.pop())
    if 'c' not in list_A:
        break

:

give next element: 0
0 ---> a
1      b
2      c
3      d
4      e
5      f
6      g
7      h
8      i
9      j
j
give next element: 1
0      a
1 ---> b
2      c
3      d
4      e
5      f
6      g
7      h
8      i
i
give next element: 2
0      a
1      b
2 ---> c
3      d
4      e
5      f
6      g
7      h
h
give next element: 3
0      a
1      b
2      c
3 ---> d
4      e
5      f
6      g
g
give next element: 4
0      a
1      b
2      c
3      d
4 ---> e
5      f
f
give next element: 5
0      a
1      b
2      c
3      d
4      e

, - break, , ( : !).

'c' not in listA O(n), O(n**2). 'c' :

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

try:
    c_index = list_A.index('c')
except ValueError:
    # no 'c' in the list, probably should do something more useful here ...
    pass
else:
    for item in reversed(list_A[c_index:]):  # print the items
        print(item)
    del list_A[c_index:]  # remove the items from the list

( ):

j
i
h
g
f
e
d
c

Answer 4

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']

while list_A: # while list_A has elements, in case 'c' wasn't present
    el = list_A.pop() # save the last element
    print(el)
    if 'c'==el: # if 'c' was popped (reached)
        break
print("job done.")

, 'c' , , . 'c' , .

Based on @MSeifert's comment, if the loop should not stop at the first c, then instead it stops whenever there is no list c, a small modification of the above code leads to:

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'c', 'h', 'i', 'j']

while list_A:
    print(list_A.pop())
    if 'c' not in list_A:
        break
print("job done.")

We could go faster, but I don’t know how you learned to cut the list and the compromise, but here is the best and quickest solution:

list_A = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
try:
  p=list_A.index('c')
  r='\n'.join(list_A[list_A.index('c'):][::-1])
except ValueError:
  r='\n'.join(list_A[::-1])
print(r)
print('job done.')