How to find out if the generator is empty from the very beginning?

Question

Is there an easy way to test if the generator does not have elements like peek, hasNext, isEmpty, something along these lines?

Answer 1

A simple answer to your question: no, there is no easy way. There are many jobs.

There really should not be a simple way that generators are due to: a way to output a sequence of values ​​without storing the sequence in memory. So there is no backtracking.

You could write the has_next function or maybe even tickle it with the generator as a method with a fantastic decorator if you want.

Answer 2

Sentence:

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

Using:

 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. 

Answer 3

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

 iterable = some_generator() _exhausted = object() if next(iterable, _exhausted) == _exhausted: print('generator is empty') 

Edit: Fixed the problem noted in the mehtunguh comment.

Answer 4

I hate to offer a second solution, especially one that I would not use myself, but if you absolutely needed it and did 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 should be used.

Answer 5

The best approach, IMHO, would be to avoid a special test. In most cases, using a generator is a test:

 thing_generated = False # Nothing is lost here. if nothing is generated, # the for block is not executed. Often, that 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 this is not good enough, you can still run an explicit test. At this point, thing will contain the last generated value. If nothing has been created, it will be undefined - if you have not defined a variable already. You can check the value of thing , but it's a little unreliable. Instead, just set the flag inside the block and check it later:

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

Answer 6

next(generator, None) is not None

Or replace None , but no matter what you know, it's not in your generator.

Edit : Yes, this will skip 1 element in the generator. Often, however, I check to see if the generator is empty only for verification purposes, but it is not actually used. Or else I'm doing something like:

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

Answer 7

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

 for item in my_generator: print item 

Now you have found that the generator is empty during its use. Of course, the item will not be displayed if the generator is empty.

This may not match your code, but this is what the generator idiom is for: iteration, so maybe you can change your approach a little or not use generators at all.

Answer 8

I understand that this post is 5 years old, but I found it, looking for an idiomatic way to do this, and did not see how my solution was published. 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 am sure, many commentators will point out that these are hacks and only work in certain limited situations (for example, when the generators are free from side effects). YMMV.

Answer 9

All you have to do to find out if the generator is empty is to try to get the following result. Of course, if you are not ready to use this result, you must save it in order to return it later.

Here's a wrapper class that you can add to an existing iterator to add the __nonzero__ test so you can see that the generator is empty with a simple if . Perhaps it can 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 = self.source.next() self.stored = True except StopIteration: return False return True def next(self): if self.stored: self.stored = False return self.value return self.source.next() 

Here's how you use it:

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

Answer 10

 >>> 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 the StopIteration generator rises, since in your case the end is reached immediately, an exception occurs. But usually you should not check for the following value.

another thing you can do:

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

Answer 11

If you need to know before using the generator, then no, there is no easy way. If you can wait until you use a generator, there is an easy 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() 

Answer 12

Here is my simple approach that I use to keep returning an iterator, checking that something was missing. I just check if the loop is working:

  n = 0 for key, value in iterator: n+=1 yield key, value if n == 0: print ("nothing found in iterator) break 

Answer 13

Just wrap the itertools.chain generator, put something that will represent the end of the iterable as the second iterable, and then just check that.

Example:

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

Now all that remains is to check if this value is added to the end of the iterable, when you read it, it means the end

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

Answer 14

Here is a simple decorator that wraps the generator, so it returns None if it is empty. This can be useful if your code needs to know if the generator will create anything before passing through it.

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

Using:

 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 the template code - i.e. jinja2

 {% if content_generator %} <section> <h4>Section title</h4> {% for item in content_generator %} {{ item }} {% endfor % </section> {% endif %} 

Answer 15

with islice you only need to check the first iteration to see if it is empty.

from itertools import islice

def isempty (iterable):
return list (islice (iterable, 1)) == []

Answer 16

In my case, I needed to know if the array of generators was full before passing it to a function that combined the elements, i.e. zip(...) . The solution is similar, but quite different, from the accepted answer:

Definition:

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

Using:

 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 specific problem has the property that iterators are either empty or have exactly the same number of records.

Answer 17

How about using any ()? I use it with generators and it works fine. There's a guy here explaining a little about it.

Answer 18

I solved this using the sum function. Below is an example that 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 probably won't work for HUGE generators, but should work fine for small lists