One-liner to check whether an iterator yields at least one element?

Question

Currently I\'m doing this:

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

But I would like an expression th

Answer 1

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.

Answer 2

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

Answer 3

__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)).