[Python-Dev] PEP 289 - Generator Expressions

[Jewett, Jim J]

> Try a concrete example?

	t1 = (1,2,3)
	t2 = ("a", "b", "c")
	t3 = ("first", "second", "third")

	(x,y,z for x in t1 for y in t2 for z in t3)

Do What I Mean is the crossproduct, 27 elements long,
which you get with a list comprehension.

This will happen if at least (all but one) generators
are pre-evaluated.  It will also happen if at least
(all but one) generators are restarted after they
raise StopIterator.

Doing strictly what happens today gets only

	(1,'a',"first"), (1,'a',"second"), (1,'a',"third")

because t3 has raised StopIteration by the time t1 or t2
advance.  Effectively, only the last iterator matters.

Keeping the last value of an iterator also adds

	(1, 'b', "third"), (1, 'c', "third"), 
	(2, 'c', "third"), (3, '3', "third")

for seven values -- but perhaps not a well-chosen seven.

Pre-evaluating only the leftmost gets all combinations of
(1,2,3) and ("first", "second", "third"), but still misses
items like (2, 'b', "second").  Since these are in the
middle of the series, they are the most likely to be 
forgotten during testing.

Restarting the second (and later) iterators for each value
of the first iterator does the right thing, without requiring
any magical pre-evaluation.  (Unless the iterator can't be
restarted -- in which case there is no right thing, and the
user should have decided what to cache.)

Thus my suggestion of
	(x,y,z for x in t1 for y in t2 for z in t3)
<==>
	for x in t1.__iter__():
		for y in t2.__iter__():
			for z in t3.__iter__():
				yield x,y,z

For the example data, this creates 3 separate iterators
on the letters and nine iterators on the words - but 
produces the expected result, without having to store
or pre-evaluate anything.

-jJ