One-liner to check whether an iterator yields at l

问题:

Currently I'm doing this:

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

But I would like an expression that I can place inside a simple if statement. Is there anything built-in which would make this code look less clumsy?

any() returns False if an iterable is empty, but it will potentially iterate over all the items if it's not. I only need it to check the first item.

---

Someone asks what I'm trying to do. I have written a function which executes an SQL query and yields its results. Sometimes when I call this function I just want to know if the query returned anything and make a decision based on that.

回答1:

any won't go beyond the first element if it's True. In case the iterator yields something false-ish you can write any(True for _ in iterator).

回答2:

In Python 2.6+, if name sentinel is bound to a value which the iterator can't possibly yield,

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

If you have no idea of what the iterator might possibly yield, make your own sentinel (e.g. at the top of your module) with

sentinel = object()

Otherwise, you could use, in the sentinel role, any value which you "know" (based on application considerations) that the iterator can't possibly yield.

回答3:

This isn't really cleaner, but it shows a way to package it in a function losslessly:

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 isn't really pythonic, and for particular cases, there are probably better (but less general) solutions, like the next default.

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

This isn't general because None can be a valid element in many iterables.

回答4:

you can use:

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

but it's a bit nonexplanatory for the code reader

回答5:

What about:

In [1]: i=iter([])

In [2]: bool(next(i,False))
Out[2]: False

In [3]: i=iter([1])

In [4]: bool(next(i,False))
Out[4]: True

回答6:

__length_hint__ estimates the length of list(it) - it's private method, though:

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)).

回答7:

This is an overkill iterator wrapper that generally allows to check whether there's a next item (via conversion to boolean). Of course pretty 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]

回答8:

A little late, but... You could turn the iterator into a list and then work with that 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 (i.e. from the iterator)
else :
    pass # Iterator was empty, thus list is empty.