Qfyilyi

Consider the following Python code:

b = [1,2,3,4,5,6,7]

a = iter(b)

for x in a :

    if (x % 2) == 0 :

        print(next(a))

Which will print 3, 5, and 7. Is the use of next on the variable being iterated on a reliable construct (you may assume that a StopIteration exception is not a concern or will be handled), or does the modification of the iterator being looped over inside the loop constitute a violation of some principle?

There's nothing wrong here protocol-wise or in theory that would stop you from writing such code. An exhausted iterator it will throw StopIteration on every subsequent call to it.__next__, so the for loop technically won't mind if you exhaust the iterator with a next/__next__ call in the loop body.

I advise against writing such code because the program will be very hard to reason about. If the scenario gets a little more complex than what you are showing here, at least I would have to go through some inputs with pen and paper and work out what's happening.

In fact, your code snippet possibly does not even behave like you think it behaves, assuming you want to print every number that is preceded by an even number.

>>> b = [1, 2, 4, 7, 8]                                              

>>> a = iter(b)                                                      

>>> for x in a: 

...:    if x%2 == 0: 

...:        print(next(a, 'stop'))                                   

4

stop

Why is 7 skipped although it's preceded by the even number 4?

>>>> a = iter(b)                                                      

>>>> for x in a: 

...:     print('for loop assigned x={}'.format(x)) 

...:     if x%2 == 0: 

...:         nxt = next(a, 'stop') 

...:         print('if popped nxt={} from iterator'.format(nxt)) 

...:         print(nxt)

...:                                               

for loop assigned x=1

for loop assigned x=2

if popped nxt=4 from iterator

4

for loop assigned x=7

for loop assigned x=8

if popped nxt=stop from iterator

stop

Turns out x = 4 is never assigned by the for loop because the explicit next call popped that element from the iterator before the for loop had a chance to look at the iterator again.

That's something I'd hate to work out the details of when reading code.

It depends what you mean by 'safe', as others have commented, it is okay, but you can imagine some contrived situations that might catch you out, for example consider this code snippet:

b = [1,2,3,4,5,6,7]

a = iter(b)

def yield_stuff():

    for item in a:

        print(item)

        print(next(a))

    yield 1



list(yield_stuff())

On Python <= 3.6 it runs and outputs:

1

2

3

4

5

6

7

But on Python 3.7 it raises RuntimeError: generator raised StopIteration. Of course this is expected if you read PEP 479 and if you're thinking about handling StopIteration anyway you might never encounter it, but I guess the use cases for calling next() inside a for loop are rare and there are normally clearer ways of re-factoring the code.

If you modify you code to see what happens to iterator a:

b = [1,2,3,4,5,6,7]

a = iter(b)



for x in a :

    print 'x', x

    if (x % 2) == 0 :

        print 'a', next(a)

You will see the printout:

x 1

x 2

a 3

x 4

a 5

x 6

a 7

It means that when you are doing next(a) you are moving forward your iterator. If you need (or will need in the future) to do something else with iterator a, you will have problems. For complete safety use various recipes from itertools module. For example:

from itertools import tee, izip



def pairwise(iterable):

    "s -> (s0,s1), (s1,s2), (s2, s3), ..."

    a, b = tee(iterable)

    next(b, None)

    return izip(a, b) 



b = [1,2,3,4,5,6,7]

a = iter(b)

c = pairwise(a)



for x, next_x in c:

    if x % 2 == 0:

        print next_x

Not that here you have full control in any place of the cycle either on current iterator element, or next one.