How to pick just one item from a generator?

I have a generator function like the following:

def myfunct():
  ...
  yield result

The usual way to call this function would be:

for r in myfunct():
  dostuff(r)

My question, is there a way to get just one element from the generator whenever I like? For example, I'd like to do something like:

while True:
  ...
  if something:
      my_element = pick_just_one_element(myfunct())
      dostuff(my_element)
  ...

Create a generator using

g = myfunct()

Everytime you would like an item, use

next(g)

(or g.next() in Python 2.5 or below).

If the generator exits, it will raise StopIteration. You can either catch this exception if necessary, or use the default argument to next():

next(g, default_value)

For picking just one element of a generator use break in a for statement, or list(itertools.islice(gen, 1))

According to your example (literally) you can do something like:

while True:
  ...
  if something:
      for my_element in myfunct():
          dostuff(my_element)
          break
      else:
          do_generator_empty()

If you want "get just one element from the [once generated] generator whenever I like" (I suppose 50% thats the original intention, and the most common intention) then:

gen = myfunct()
while True:
  ...
  if something:
      for my_element in gen:
          dostuff(my_element)
          break
      else:
          do_generator_empty()

This way explicit use of generator.next() can be avoided, and end-of-input handling doesn't require (cryptic) StopIteration exception handling or extra default value comparisons.

The else: of for statement section is only needed if you want do something special in case of end-of-generator.

Note on next() / .next():

In Python3 the .next() method was renamed to .__next__() for good reason: its considered low-level (PEP 3114). Before Python 2.6 the builtin function next() did not exist. And it was even discussed to move next() to the operator module (which would have been wise), because of its rare need and questionable inflation of builtin names.

Using next() without default is still very low-level practice - throwing the cryptic StopIteration like a bolt out of the blue in normal application code openly. And using next() with default sentinel - which best should be the only option for a next() directly in builtins - is limited and often gives reason to odd non-pythonic logic/readablity.

Bottom line: Using next() should be very rare - like using functions of operator module. Using for x in iterator , islice, list(iterator) and other functions accepting an iterator seamlessly is the natural way of using iterators on application level - and quite always possible. next() is low-level, an extra concept, unobvious - as the question of this thread shows. While e.g. using break in for is conventional.

Generator is a function that produces an iterator. Therefore, once you have iterator instance, use next() to fetch the next item from the iterator. As an example, use next() function to fetch the first item, and later use for in to process remaining items:

# create new instance of iterator by calling a generator function
items = generator_function()

# fetch and print first item
first = next(items)
print('first item:', first)

# process remaining items:
for item in items:
    print('next item:', item)

You can pick specific items using destructuring, e.g.:

>>> first, *middle, last = range(10)
>>> first
0
>>> middle
[1, 2, 3, 4, 5, 6, 7, 8]
>>> last
9

Note that this is going to consume your generator, so while highly readable, it is less efficient than something like next(), and ruinous on infinite generators:

>>> first, *rest = itertools.count()
🔥🔥🔥

I don't believe there's a convenient way to retrieve an arbitrary value from a generator. The generator will provide a next() method to traverse itself, but the full sequence is not produced immediately to save memory. That's the functional difference between a generator and a list.

generator = myfunct()
while True:
   my_element = generator.next()

make sure to catch the exception thrown after the last element is taken

For those of you scanning through these answers for a complete working example for Python3... well here ya go:

def numgen():
    x = 1000
    while True:
        x += 1
        yield x

nums = numgen() # because it must be the _same_ generator

for n in range(3):
    numnext = next(nums)
    print(numnext)

This outputs:

1001
1002
1003

I believe the only way is to get a list from the iterator then get the element you want from that list.

l = list(myfunct())
l[4]