How to yield every alternate element from generator?

Question

How do I yield every alternate element from a generator?

The obvious method doesn't seem to work:

>>> generator[::2]
TypeError: 'generator' object is not subscriptable

Sure you could write code that achieves the same purpose:

x = 0
for item in generator:
    if x % 2 == 0:
        do_something(item)
    x += 1

But Python is bad at math in tight loops like this one and I'm eager to find how this could be done better.

`Python is bad at math in tight loops` - could you explain that a bit? `could be done better.` - better how? — wwii, Oct 18 '20 at 21:15
Look at [`itertools.islice`](https://docs.python.org/3.8/library/itertools.html#itertools.islice). You can do `islice(, 0, None, 2)` for example. — Andrej Kesely, Oct 18 '20 at 21:17
When you write your own generator you have complete control over which values are yielded, so I do not see what is the problem you are facing here — sai, Oct 18 '20 at 21:18
Strongly recommend `islice` but you could also consume one element from the generator inside the loop like so: `_ = next(generator)` — Pranav Hosangadi, Oct 18 '20 at 21:25
@PranavHosangadi You don't need to bother assigning the return value. `while True: next(generator); yield next(generator)`. — chepner, Oct 18 '20 at 21:30
@chepner I thought so too, but [my quick test](https://i.stack.imgur.com/XwhDS.png) seemed to [indicate otherwise](https://i.stack.imgur.com/VnLEE.png). I realize now it's prob because it was an interactive terminal :-/ — Pranav Hosangadi, Oct 18 '20 at 21:35

score 2 · Answer 1 · answered Oct 18 '20 at 21:21

This kind of approach can be used to deliver only those values following any kind of repetitive index pattern:

def yield_every_other(g):
    from itertools import compress, cycle
    return compress(g, cycle([True, False]))

Then, e.g.,

>>> for x in yield_every_other(2*i for i in range(20)):
...     print(x)

displays

score 0 · Answer 2 · answered Oct 18 '20 at 21:21

0

You can use enumerate to make the second option cleaner:

for i, item in enumerate(generator):
    if i % 2 == 0:
        do_something(item)

answered Oct 18 '20 at 21:21

wjandrea

2 Answers2