Single line to check if iterator gives at least one element?

Question

I am currently doing this:

try: something = iterator.next() # ... except StopIteration: # ... 

But I would like to express an expression that I can place inside a simple if . Is there anything inline that will make this code less awkward?

any() returns False if the iterability is empty, but it will potentially iterate over all elements if it is not. I only need to check the first item.

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Someone asks what I'm trying to do. I wrote a function that executes an SQL query and gives the results. Sometimes, when I call this function, I just want to know if the request returned something and made a decision based on this.

Answer 1

any will not exceed the first element if it is True. If the iterator gives something false, you can write any(True for _ in iterator) .

Answer 2

In Python 2.6+, if the name sentinel bound to a value that the iterator cannot execute,

 if next(iterator, sentinel) is sentinel: print('iterator was empty') 

If you do not know that the iterator can give way, make your own watchdog (for example, at the top of your module) with

 sentinel = object() 

Otherwise, you can use as a gatekeeper any value that you “know” (based on application considerations) that the iterator cannot give.

Answer 3

This is not very clean, but it shows a way to batch convert to a lossless function:

 def has_elements(iter): from itertools import tee iter, any_check = tee(iter) try: any_check.next() return True, iter except StopIteration: return False, iter has_el, iter = has_elements(iter) if has_el: # not empty 

This is not really pythonic, and for some cases there are probably better (but less general) solutions, such as next by default.

 first = next(iter, None) if first: # Do something 

This is not common, because None can be a valid element in many iterations.

Answer 4

you can use:

 if zip([None], iterator): # ... else: # ... 

but it is a little inefficient for the code reader

Answer 5

__length_hint__ evaluates the length of list(it) - this is a private method:

 x = iter( (1, 2, 3) ) help(x.__length_hint__) 1 Help on built-in function __length_hint__: 2 3 __length_hint__(...) 4 Private method returning an estimate of len(list(it)). 

Answer 6

This is the overkill iterator shell, which usually allows you to check if the next element is there (via conversion to logical). Of course, quite inefficient.

 class LookaheadIterator (): def __init__(self, iterator): self.__iterator = iterator try: self.__next = next (iterator) self.__have_next = True except StopIteration: self.__have_next = False def __iter__(self): return self def next (self): if self.__have_next: result = self.__next try: self.__next = next (self.__iterator) self.__have_next = True except StopIteration: self.__have_next = False return result else: raise StopIteration def __nonzero__(self): return self.__have_next x = LookaheadIterator (iter ([])) print bool (x) print list (x) x = LookaheadIterator (iter ([1, 2, 3])) print bool (x) print list (x) 

Output:

 False [] True [1, 2, 3] 

Answer 7

A little late, but ... You can include an iterator in a list, and then work with this list:

 # Create a list of objects but runs out the iterator. l = [_ for _ in iterator] # If the list is not empty then the iterator had elements; else it was empty. if l : pass # Use the elements of the list (ie from the iterator) else : pass # Iterator was empty, thus list is empty.