[Python-Dev] Fwd: summing a bunch of numbers (or "whatevers")
Guido van Rossum
guido@python.org
Mon, 21 Apr 2003 08:48:58 -0400
OK, let me summarize and pronounce.
sum(sequence_of_strings) is out. *If* "".join() is really too ugly (I
still think it's a matter of getting used to, like indentation), we
could add join(seq, delim) as a built-in. VB has one. :-)
sum([]) could either return 0 or raise ValueError. I lean towards 0
because that is occasionally useful and reinforces the numeric
intention. I think making it return 0 will prevent end-case bugs
where a newbie sums a list that is occasionally empty. If we made it
an error, I expect that in 99% of the cases the response to that error
would be to change the program to make it return 0 if the list is
empty, and I can't imagine many bugs caused by choosing 0 over some
other numerical zero. Having to teach the idiom sum(S or [0]) is
ugly, and this doesn't work if S is an iterator.
I appreciate Tim's point of wanting to sum "number-like" objects that
can't be added to 0. OTOH if we provide *any* way of providing a
different starting point, some creative newbie is going to use
sum(list_of_strings, "") instead of "".join(), and be hurt by the
performance months later.
If we add an optional argument for Tim's use case, it could be used in
two different ways: (1) only when the sequence is empty, (2) always
used as a starting point. IMO (2) is more useful and more consistent.
Here's one suggestion to deal with the sequence_of_strings issue
(though maybe too pedantic): explicitly check whether the second
argument is a string or unicode object, and in that case raise a
TypeError indicating that a numeric value is required and suggesting
to use "".join() for summing a sequence of strings.
So here's a strawman implementation:
def sum(seq, start=0):
if isinstance(start, basestring):
raise TypeError, "can't sum strings; use ''.join(seq) instead"
return reduce(operator.add, seq, start)
Alex, go ahead and implement this!
--Guido van Rossum (home page: http://www.python.org/~guido/)