python iterator skip_func

Question

def yield_and_skip(iterable,skip):
    for i in iterable:
        print(skip(i))
        for x in range(skip(i)):
            pass
        yield i

I'm defning a function called skip_yield, which produces a value from the iterable but then it then skips the number of values specified when the function argument is called on the just-produced value. It takes 2 parameters: skip is a small function. However, if I try to do this:

print(''.join([v for v in yield_and_skip('abbabxcabbcaccabb',lambda x : {'a':1,'b':2,'c':3}.get(x,0))]))

it gives me:

abbabcabbcaccabb

but it should give me:

abxccab

Any idea?

What's the point of having a loop with empty body (pass) here? — GingerPlusPlus
– GingerPlusPlus, Commented Apr 28, 2016 at 16:45

vaultah · Accepted Answer · 2016-04-29 15:59:37Z

3

First of all, pass doesn't do anything to iterable. If you need to advance the iterator, call the next function on iterator object skip(i) number of times (either directly or indirectly).

Also, next call may raise StopIteration, thus terminating the loop. yield i before advancing the iterator or provide the second argument, to make next swallow the exception.

Your code would look like

def yield_and_skip(iterable, skip):
    it = iter(iterable)
    for i in it:
        for _ in range(skip(i)):
            next(it, None)
        yield i

edited Apr 29, 2016 at 15:59

answered Apr 28, 2016 at 16:57

vaultah

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1 Comment

Nizam Mohamed Over a year ago

"yield i before advancing the iterator" is not necessary and next(it) should be next(it, None)

Francesco · Accepted Answer · 2016-04-28 16:54:11Z

2

You could define a counter for the number of elements to skip and update it accordingly

def yield_and_skip(iterable,skip):
    num_to_skip = 0
    for i in iterable:
        if num_to_skip == 0:
            num_to_skip = skip(i)
            yield i
        else:
            num_to_skip -= 1

The output of your sample line is

>>> print(''.join([v for v in yield_and_skip('abbabxcabbcaccabb',lambda x : {'a':1,'b':2,'c':3}.get(x,0))]))
abxccab

answered Apr 28, 2016 at 16:54

Francesco

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