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Comment by orangecat | Hacker News Reader

That fails for sets and other non-list iterables. It would have to be something like:

  def get_single(l):
      i = iter(l)
      val = i.next()
      try:
          i.next() # expected to throw exception for one-element iterable
      except StopIteration:
          return val
      raise AssertionError('More than one object')

Eww.

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You don't need to do that for a set. sets support `len()`. Also set supports `pop()`, so the code is very similar to the list one:

    assert s and len(s)==1
    return s.pop()

Or if you want to stay in the immutable land:

    assert s and len(s)==1
    return tuple(s)[0]

Both of those still fail for other non-list iterables, like generators.

They also have the side effect of removing an item from the collection.

A property which is exposed by the article's solution too:

   >>> x = (lambda: (yield 1))() # generator with one step
   >>> y = tuple(x)[0]
   >>> y
   1
   >>> list(x) # exhausted
   []

   >>> x = (lambda: (yield 1))()
   >>> y, = x
   >>> y
   1
   >>> list(x) # exhausted, too!
   []

Because that's just what you inevitably need to do to fetch a value from a generator. There is no peeking action or some such.

Personally, I'm willing to restrict this to just lists to keep things simple. I rarely ever need a way to extract one value from a list, set, and iterable.

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