How to pick just one item from a generator?

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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)

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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.

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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.

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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

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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)

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Comments
  • Note, it will only raise StopIteration when you try to use g.next() after the last item in g has been provided.
  • next(gen, default) may also be used to avoid the StopIteration exception. For example next(g, None) for a generator of strings will either yield a string or None after the iteration was finished.
  • in Python 3000, next() is __next__()
  • @JonathanBaldwin: Your comment is somewhat misleading. In Python 3, you would use the second syntax given in my answer, next(g). This will internally call g.__next__(), but you don't really have to worry about that, just as you usually don't care that len(a) internally calls a.__len__().
  • I should have been more clear. g.next() is g.__next__() in py3k. The builtin next(iterator) has been around since Python 2.6, and is what should be used in all new Python code, and it's trivial to backimplement if you need to support py <= 2.5.
  • This is way too much work for just getting the first element of a list result. Often enough I do not need it to be lazy but do not have a choice in py3. Is there not something akin to mySeq.head ?
  • Not valid for Python 3, see the excellent answer by kxr.
  • Just replace "generator.next()" with "next(generator)" for Python 3.
  • Sven's answer is probably better, but I'll just leave this here incase it's more inline with your needs.
  • Make sure that you have a finite generator before doing this.
  • Sorry, this has complexity the length of the iterator, while the problem is obviously O(1).
  • Wasting too many memory and process to suck from generator! Also, as @Seth mentioned before, generators are not guaranteed for, when to stop generating.
  • This is obviously not the only way (or the best way if myfunct() generates a large number of values) since you can use the built-in function next to get the next generated value.