Tim Williams:
> You could also use a list comprehension for your case
> >>> alist = [1 ,2 ,3]
> >>> alist = [x for x in alist if x != 2]
> >>> alist
> [1, 3]
The list comprehension filtering is the simpler and often the best
solution. For memory-conscious people this is another possible
(un-pythonic) solution, usable in less dynamic languages too (not much
tested), expecially if the good elements are few (Psyco may help too).
I haven't tested its speed, but in Python it may be slower than the
simpler comprehension filtering solution:
array = [1,2,3,1,5,1,6,0]
good = lambda x: x > 2
slow = 0
for fast, item in enumerate(array):
print item,
if good(item):
if slow != fast:
array[slow] = array[fast]
slow += 1
print "\n", array
del array[slow:]
print array
Output:
1 2 3 1 5 1 6 0
[3, 5, 6, 1, 5, 1, 6, 0]
[3, 5, 6]
Bye,
bearophile
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