Problem with iterating generator functions

Question

Let's say we have this generator function:

def count_up_to(max):
    count = 1
    while count <= max:
        yield count
        count += 1

Now if I want to call the next() function, I will have to assign the count_up_to function to a variable, otherwise the output will always be "1".

The code below works fine and iterates through the numbers:

counter = count_up_to(10)
print(next(counter))
print(next(counter))
print(next(counter))
.
.
.

But this one does not work and keeps printing "1".

print(next(count_up_to(10)))
print(next(count_up_to(10)))
print(next(count_up_to(10)))
.
.
.

But why? is print(next(counter)) any different from print(next(count_up_to(10))) ?!

On your last snippet, you're initialising your generator anew for every print line, so every next() call gets a fresh new iterator. — Talon
– Talon, Commented Apr 28, 2020 at 20:32
of course it's different. In the first case, you create one generator object and keep re-using it. In the second case, you create a new generator object each time you use it. You seem to have already discovered this? What exactly do you not understand? — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Apr 28, 2020 at 20:35

Rafael · Accepted Answer · 2020-04-28 20:33:13Z

3

In your last snippet, you always make a new generator, that's why it always prints "1".

In the second one, you assign the generator returned by count_up_to to the variable count, not the function. And with every call of next(count), you tell the generator to yield the next value.

I hope that this helps :)

answered Apr 28, 2020 at 20:33

Rafael

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5 Comments

Sherafati Over a year ago

But using a for loop with the function name works: for num in count_up_to(10): print(num). as far as i know the for loop keeps calling the next() function so why is it working fine in this case but now when we call the next() function manually?

Dmitry Egerev Over a year ago

@Sherafati it's because for loop does not call function on every iteration. It call your function once and then uses generator to get new values.

Rafael Over a year ago

Python then makes a temporary variable and uses the same generator for every iteration. I don't exactly know how Python handles this internally, but you can work with it like the generator can be used as a list.

Rafael Over a year ago

@Sherafati you could also do this for example: list(count_up_tp(10))

Sherafati Over a year ago

So when we use the for loop it calls the function only once and assigns it to a variable(that is of a "generator" type) and keeps calling the next() function on that variable. correct?

Dmitry Egerev · Accepted Answer · 2020-04-28 20:39:00Z

It's completely expected behavior. Lets me describe why.

Every single call of count_up_to will return you a new generator, which starts from 1.

Simple check in Python REPL:

>>> count_up_to(100)
<generator object count_up_to at 0x10e28beb0>
>>> count_up_to(100)
<generator object count_up_to at 0x10e28bf00>
>>> count_up_to(100)
<generator object count_up_to at 0x10e28beb0>
>>> count_up_to(100)
<generator object count_up_to at 0x10e28bf00>

Look at the address. It is always different, there are 4 different generators in memory.

So, count_up_to function is not a generator itself, but it returns a generator when you call it.

And when you stores result of count_up_to call in a variable, you store generator. Further, you can get values from the generator, stored in the variable.

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