Continue until all iterators are done Python

Question

I cannot use itertools

So the coding seems pretty simple, but I'm having trouble thinking of the algorithm to keep a generator running until all iterations have been processed fully.

The idea of the function is to take 2 iterables as parameters like this ...

(['a', 'b', 'c', 'd', 'e'], [1,2,5])

And what it does is yield these values ...

a, b, b, c, c, c, c, c

However, in the event that the second iterable runs out of elements first, the function simply iterates the remaining value one time ...

So the remaining values would be iterated like this:

d, e

def iteration(letters, numbers):
    times = 0
    for x,y in zip(letters, numbers):
        try:
            for z in range(y):
                yield x
        except:
            continue

[print(x) for x in iteration(['a', 'b', 'c', 'd'], [1,2,3])]

I'm having difficulty ignoring the first StopIteration and continuing to completion.

Answer 1

Use a default value of 1 for next so you print the letters at least once:

def iteration(letters, numbers): 
     # create iterator from numbers
    it = iter(numbers)
    # get every letter
    for x in letters:
        # either print in range passed or default range of 1
        for z in range(next(it, 1)):
            yield x

Output:

In [60]: for s in iteration(['a', 'b', 'c', 'd', 'e'], [1,2,5]):
   ....:     print(s)
   ....:     
a
b
b
c
c
c
c
c
d
e