How do I know if a generator is empty from the start?


Is there a simple way of testing if the generator has no items, like peek, hasNext, isEmpty, something along those lines?

Asked By: Dan



>>> gen = (i for i in [])
>>> next(gen)
Traceback (most recent call last):
  File "<pyshell#43>", line 1, in <module>

At the end of generator StopIteration is raised, since in your case end is reached immediately, exception is raised. But normally you shouldn’t check for existence of next value.

another thing you can do is:

>>> gen = (i for i in [])
>>> if not list(gen):
    print('empty generator')
Answered By: SilentGhost

Sorry for the obvious approach, but the best way would be to do:

for item in my_generator:
     print item

Now you have detected that the generator is empty while you are using it. Of course, item will never be displayed if the generator is empty.

This may not exactly fit in with your code, but this is what the idiom of the generator is for: iterating, so perhaps you might change your approach slightly, or not use generators at all.

Answered By: Ali Afshar

I hate to offer a second solution, especially one that I would not use myself, but, if you absolutely had to do this and to not consume the generator, as in other answers:

def do_something_with_item(item):
    print item

empty_marker = object()

     first_item =     
except StopIteration:
     print 'The generator was empty'
     first_item = empty_marker

if first_item is not empty_marker:
    for item in my_generator:

Now I really don’t like this solution, because I believe that this is not how generators are to be used.

Answered By: Ali Afshar

The best approach, IMHO, would be to avoid a special test. Most times, use of a generator is the test:

thing_generated = False

# Nothing is lost here. if nothing is generated, 
# the for block is not executed. Often, that's the only check
# you need to do. This can be done in the course of doing
# the work you wanted to do anyway on the generated output.
for thing in my_generator():
    thing_generated = True

If that’s not good enough, you can still perform an explicit test. At this point, thing will contain the last value generated. If nothing was generated, it will be undefined – unless you’ve already defined the variable. You could check the value of thing, but that’s a bit unreliable. Instead, just set a flag within the block and check it afterward:

if not thing_generated:
    print "Avast, ye scurvy dog!"
Answered By: vezult

The simple answer to your question: no, there is no simple way. There are a whole lot of work-arounds.

There really shouldn’t be a simple way, because of what generators are: a way to output a sequence of values without holding the sequence in memory. So there’s no backward traversal.

You could write a has_next function or maybe even slap it on to a generator as a method with a fancy decorator if you wanted to.

Answered By: David Berger


def peek(iterable):
        first = next(iterable)
    except StopIteration:
        return None
    return first, itertools.chain([first], iterable)


res = peek(mysequence)
if res is None:
    # sequence is empty.  Do stuff.
    first, mysequence = res
    # Do something with first, maybe?
    # Then iterate over the sequence:
    for element in mysequence:
        # etc.
Answered By: John Fouhy

If you need to know before you use the generator, then no, there is no simple way. If you can wait until after you have used the generator, there is a simple way:

was_empty = True

for some_item in some_generator:
    was_empty = False

if was_empty:
Answered By: Ethan Furman

A simple way is to use the optional parameter for next() which is used if the generator is exhausted (or empty). For example:

_exhausted  = object()

if next(some_generator, _exhausted) is _exhausted:
    print('generator is empty')
Answered By: Mikko Koho

I realize that this post is 5 years old at this point, but I found it while looking for an idiomatic way of doing this, and did not see my solution posted. So for posterity:

import itertools

def get_generator():
    Returns (bool, generator) where bool is true iff the generator is not empty.
    gen = (i for i in [0, 1, 2, 3, 4])
    a, b = itertools.tee(gen)
    except StopIteration:
        return (False, b)
    return (True, b)

Of course, as I’m sure many commentators will point out, this is hacky and only works at all in certain limited situations (where the generators are side-effect free, for example). YMMV.

Answered By: Real John Connor

All you need to do to see if a generator is empty is to try to get the next result. Of course if you’re not ready to use that result then you have to store it to return it again later.

Here’s a wrapper class that can be added to an existing iterator to add an __nonzero__ test, so you can see if the generator is empty with a simple if. It can probably also be turned into a decorator.

class GenWrapper:
    def __init__(self, iter):
        self.source = iter
        self.stored = False

    def __iter__(self):
        return self

    def __nonzero__(self):
        if self.stored:
            return True
            self.value = next(self.source)
            self.stored = True
        except StopIteration:
            return False
        return True

    def __next__(self):  # use "next" (without underscores) for Python 2.x
        if self.stored:
            self.stored = False
            return self.value
        return next(self.source)

Here’s how you’d use it:

with open(filename, 'r') as f:
    f = GenWrapper(f)
    if f:
        print 'Not empty'
        print 'Empty'

Note that you can check for emptiness at any time, not just at the start of the iteration.

Answered By: Mark Ransom

Quick-dirty solution:

next(my_generator(), None) is not None

Or replace None by whatever value you know it’s not in your generator.

Edit: Yes, this will skip 1 item in the generator. Sometimes, however, I check whether a generator is empty only for validation purposes, then don’t really use it. Otherwise, I do something like:

def foo(self):
    if next(self.my_generator(), None) is None:
        raise Exception("Not initiated")

    for x in self.my_generator():

That is, this works if your generator comes from a function, as in my_generator().

Answered By: juanmirocks

I solved it by using the sum function. See below for an example I used with glob.iglob (which returns a generator).

def isEmpty():
    files = glob.iglob(search)
    if sum(1 for _ in files):
        return True
    return False

*This will probably not work for HUGE generators but should perform nicely for smaller lists

Simply wrap the generator with itertools.chain, put something that will represent the end of the iterable as the second iterable, then simply check for that.


import itertools

g = some_iterable
eog = object()
wrap_g = itertools.chain(g, [eog])

Now all that’s left is to check for that value we appended to the end of the iterable, when you read it then that will signify the end

for value in wrap_g:
    if value == eog: # DING DING! We just found the last element of the iterable
        pass # Do something
Answered By: smac89

Here’s a simple decorator which wraps the generator, so it returns None if empty. This can be useful if your code needs to know whether the generator will produce anything before looping through it.

def generator_or_none(func):
    """Wrap a generator function, returning None if it's empty. """

    def inner(*args, **kwargs):
        # peek at the first item; return None if it doesn't exist
            next(func(*args, **kwargs))
        except StopIteration:
            return None

        # return original generator otherwise first item will be missing
        return func(*args, **kwargs)

    return inner


import random

def random_length_generator():
    for i in range(random.randint(0, 10)):
        yield i

gen = random_length_generator()
if gen is None:
    print('Generator is empty')

One example where this is useful is in templating code – i.e. jinja2

{% if content_generator %}
    <h4>Section title</h4>
    {% for item in content_generator %}
      {{ item }}
    {% endfor %
{% endif %}
Answered By: Greg

In my case I needed to know if a host of generators was populated before I passed it on to a function, which merged the items, i.e., zip(...). The solution is similar, but different enough, from the accepted answer:


def has_items(iterable):
        return True, itertools.chain([next(iterable)], iterable)
    except StopIteration:
        return False, []


def filter_empty(iterables):
    for iterable in iterables:
        itr_has_items, iterable = has_items(iterable)
        if itr_has_items:
            yield iterable

def merge_iterables(iterables):
    populated_iterables = filter_empty(iterables)
    for items in zip(*populated_iterables):
        # Use items for each "slice"

My particular problem has the property that the iterables are either empty or has exactly the same number of entries.

Answered By: André C. Andersen

Use the peek function in cytoolz.

from cytoolz import peek
from typing import Tuple, Iterable

def is_empty_iterator(g: Iterable) -> Tuple[Iterable, bool]:
        _, g = peek(g)
        return g, False
    except StopIteration:
        return g, True

The iterator returned by this function will be equivalent to the original one passed in as an argument.

Answered By: W.P. McNeill

Prompted by Mark Ransom, here’s a class that you can use to wrap any iterator so that you can peek ahead, push values back onto the stream and check for empty. It’s a simple idea with a simple implementation that I’ve found very handy in the past.

class Pushable:

    def __init__(self, iter):
        self.source = iter
        self.stored = []

    def __iter__(self):
        return self

    def __bool__(self):
        if self.stored:
            return True
        except StopIteration:
            return False
        return True

    def push(self, value):

    def peek(self):
        if self.stored:
            return self.stored[-1]
        value = next(self.source)
        return value

    def __next__(self):
        if self.stored:
            return self.stored.pop()
        return next(self.source)
Answered By: sfkleach

Just fell on this thread and realized that a very simple and easy to read answer was missing:

def is_empty(generator):
    for item in generator:
        return False
    return True

If we are not suppose to consume any item then we need to re-inject the first item into the generator:

def is_empty_no_side_effects(generator):
        item = next(generator)
        def my_generator():
            yield item
            yield from generator
        return my_generator(), False
    except StopIteration:
        return (_ for _ in []), True


>>> g=(i for i in [])
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
>>> g=(i for i in range(10))
>>> g,empty=is_empty_no_side_effects(g)
>>> empty
>>> list(g)
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Answered By: Romaric

I found only this solution as working for empty iterations as well.

def is_generator_empty(generator):
    a, b = itertools.tee(generator)
    except StopIteration:
        return True, b
    return False, b

is_empty, generator = is_generator_empty(generator)

Or if you do not want to use exception for this try to use

def is_generator_empty(generator):
    a, b = itertools.tee(generator)
    for item in a:
        return False, b
    return True, b

is_empty, generator = is_generator_empty(generator)

In the marked solution you are not able to use it for empty generators like

def get_empty_generator():
    while False:
        yield None 

generator = get_empty_generator()
Answered By: y0j0

This is an old and answered question, but as no one has shown it before, here it goes:

for _ in generator:

You can read more here

Answered By: Paulo Alves

There’s a very simple solution: if next(generator,-1) == -1 then the generator is empty!

Answered By: Apostolos

Just to try to help with my "2 cents", I am going to describe my experience:

I have a generator that I need slicing it using itertools.islice into small generators. Then to check if my sub generators are empty or not, I just convert/consume them to a small list and I check if the list is empty or not.

For example:

from itertools import islice

def generator(max_yield=10):
    a = 0

    while True:
        a += 1

        if a > max_yield:
            raise StopIteration()

        yield a

tg = generator()

label = 1

while True:
    itg = list(islice(tg, 3))

    if not itg:  # <-- I check if the list is empty or not

    for i in itg:
        print(f'#{label} - {i}')

    label += 1


#1 - 1
#1 - 2
#1 - 3
#2 - 4
#2 - 5
#2 - 6
#3 - 7
#3 - 8
#3 - 9
#4 - 10

Maybe this is not the best approach, mainly because it consumes the generator, however it works to me.

Answered By: rmmariano

bool(generator) will return the correct result

Answered By: zhuomingzhe

Inspecting the generator before iterating over it conforms to the LBYL coding style. Another approach (EAFP) would be to iterate over it and then check whether it was empty or not.

is_empty = True

for item in generator:
    is_empty = False

if is_empty:
    print('Generator is empty')

This approach also handles well infinite generators.

Answered By: Jeyekomon

peekable from more-itertools allows checking whether it’s exhausted by checking its truth value. Demo with one empty and one non-empty iterator:

from more_itertools import peekable

for source in '', 'foobar':
    it = iter(source)
    if it := peekable(it):
        print('values:', *it)


values: f o o b a r
Answered By: Kelly Bundy
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