Harald Luessen a écrit :
On Thu, 18 Sep 2008 Bruno Desthuilliers wrote:
# Harald : uncomment this and run test_results. As far as I can tell, it
# doesn't yields the expected results
## IN7 = IN[:]
## def sortk7(n):
## return n.coordinates[0]
## def doubles7():
## "is ordering better ? - Nope Sir, it's broken..."
## IN7.sort(key=sortk)
## SN = []
## sn_append = SN.append
## in_len = len(IN)
## for i in xrange(in_len):
## node_i = IN[i]
## coords_i = node_i.coordinates
## for j in xrange(i+1, in_len):
## if coords_i[0] == IN[j].coordinates[0]:
## if coords_i[1] == IN[j].coordinates[1]:
## sn_append(node_i)
## else:
## break
## return SN
...
Results here (py2.5, gentoo linux, athlonxp1800, 512 ram):
test_results()
True
test_times()
doubles0 : 1.55667901039
doubles1 : 0.719144105911
doubles2 : 0.703393936157
doubles3 : 0.700654983521
doubles4 : 0.706257104874
doubles5 : 0.528184890747
doubles6 : 0.461633205414
doubles8 : 0.0134379863739
doubles9 : 0.0108540058136
When you change my code then do it right. :-)
You forgot to change the IN to IN7 at _every_ place.
And the sortk should be sortk7 in _both_ places.
doh :(
Sorry Harald, my fault, you're right...
I never let the code run before myself. I just wrote it
in the newsreader. But now i did and I have a second
version as bonus.
IN7 = IN[:]
def sortk7(n):
return n.coordinates[0], n.coordinates[1]
def doubles7():
IN7.sort(key=sortk7)
SN = []
sn_append = SN.append
in_len = len(IN7)
for i in xrange(in_len):
node_i = IN7[i]
coords_i = node_i.coordinates
for j in xrange(i+1, in_len):
if coords_i[0] == IN7[j].coordinates[0]:
if coords_i[1] == IN7[j].coordinates[1]:
sn_append(node_i)
else:
break
return SN
def comp7( x, y ):
return cmp( x.coordinates, y.coordinates )
def doubles7a():
IN7.sort( comp7 )
SN = []
sn_append = SN.append
in_len = len(IN7)
for i in xrange(in_len):
node_i = IN7[i]
for j in xrange(i+1, in_len):
node_j = IN7[j]
if comp7( node_i, node_j ) == 0:
sn_append(node_i)
else:
break
return SN
Here are the results. (py2.5, WindowsXP, Pentium4, 2.6GHz, 1.5GB):
My version is not so bad.
doubles0 : 1.03830598582
doubles1 : 0.47943719104
doubles2 : 0.487412506338
doubles3 : 0.475924733451
doubles4 : 0.466548681466
doubles5 : 0.340487967046
doubles6 : 0.278480365521
doubles7 : 0.0953190978183
doubles7a : 0.0784233750379
doubles8 : 0.010236496538
doubles9 : 0.00742803903848
Point taken. Now there's one thing I find questionnable with your
solution (which is probably why I didn't bother double-checking it when
it *appeared* to be broken) : you assume that it's ok to loose original
ordering, which I strongly suspect is not the case for the OP use case,
and given the OP use case list's size, working on a copy might not be an
acceptable solution.
But anyway: this is not an excuse for me having broken your code. My
apologies.
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