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

Suggestion:

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

Usage:

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

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') 

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().

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 do_work(thing) 

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!" 

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

Example:

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

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 try: self.stored.append(next(self.source)) except StopIteration: return False return True def push(self, value): self.stored.append(value) def peek(self): if self.stored: return self.stored[-1] value = next(self.source) self.stored.append(value) return value def __next__(self): if self.stored: return self.stored.pop() return next(self.source) 

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() try: first_item = my_generator.next() except StopIteration: print 'The generator was empty' first_item = empty_marker if first_item is not empty_marker: do_something_with_item(first_item) for item in my_generator: do_something_with_item(item) 

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

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 try: 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' else: print 'Empty' 

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

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.

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) try: a.next() 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.

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

def is_generator_empty(generator): a, b = itertools.tee(generator) try: next(a) 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() 

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

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') 

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 do_something_with(some_item) if was_empty: handle_already_empty_generator_case() 

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.

Ex:

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 

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:

Definition:

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

Usage:

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.

Use the peek function in cytoolz.

from cytoolz import peek from typing import Tuple, Iterable def is_empty_iterator(g: Iterable) -> Tuple[Iterable, bool]: try: _, 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.

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 break for i in itg: print(f'#{label} - {i}') label += 1 

Output:

#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.

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 do_something(item) if is_empty: print('Generator is empty') 

This approach also handles well infinite generators.

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 try: next(func(*args, **kwargs)) except StopIteration: return None # return original generator otherwise first item will be missing return func(*args, **kwargs) return inner 

Usage:

import random @generator_or_none 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 %} <section> <h4>Section title</h4> {% for item in content_generator %} {{ item }} {% endfor % </section> {% endif %} 

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) else: print('empty') 

Output:

empty values: f o o b a r 

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

for _ in generator: break else: print('Empty') 

You can read more here

Existing answers use custom solutions to address this problem, but an off the shelf solution is available via more_itertools.peekable.

This allows the user to retain the memory efficiency of the iterator, whilst also:

• being able to check the next available element without consuming it
• check that the generator isn't empty

 from more_itertools import peekable def gen_fun(n: int): for i in range(n): yield i generator_object = gen_fun(5) peekable_generator_object = peekable(generator_object) if not peekable_generator_object: print("Empty Generator") 

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