Fredrik Lundh wrote:
Jeremy Jones wrote:
#################################
file_dict = {}
a_list = [("a", "a%s" % i) for i in range(2500)]
b_list = [("b", "b%s" % i) for i in range(2500)]
c_list = [("c", "c%s" % i) for i in range(2500)]
d_list = [("d", "d%s" % i) for i in range(2500)]
joined_list = a_list + b_list + c_list + d_list
for key, value in joined_list:
outfile = file_dict.setdefault(key, open("%s.txt" % key, "w"))
you do realize that this opens the file again every time, so you end up having
4x2500 file handles pointing to 4 physical files. that's a bad idea.
That *is* a bad idea, and no, I didn't realize that would be the
result. From the "mapping types" page:
a.setdefault(k[,
x]) |
a[k] if k
in a, else x (also setting it) |
(5) |
- (5)
- setdefault() is like get(), except
that if k is missing, x is both returned and
inserted into
the dictionary as the value of k. x defaults to
None.
I took this to mean that setdefault was a short-circuit and only
created the default object once if the dictionary didn't contain the
specified key. But I guess it's *me* who is passing setdefault a new
file handle a few thousand times :-) Ouch.
if you replace
outfile = file_dict.setdefault(key, open("%s.txt" % key, "w"))
with
outfile = file_dict.get(key)
if outfile is None:
file_dict[key] = outfile = open("%s.txt" % key, "w")
or, if you prefer,
try:
outfile = file_dict[key]
except KeyError:
file_dict[key] = outfile = open("%s.txt" % key, "w")
your code won't depend on any undefined behaviour, and will work properly
on all platforms.
</F>
Thanks (and thanks to you, Duncan, for your reply as well),
Jeremy
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