"Lawrence D'Oliveiro" <[EMAIL PROTECTED]> wrote in
message
news:[EMAIL PROTECTED]
Why not just build a new list? E.g.
newdevs = []
for dev in devs :
...
if not removing_dev :
newdevs.append(dev)
#end if
#end for
devs = newdevs
En Sun, 09 Sep 2007 22:58:54 -0300, bambam <[EMAIL PROTECTED]> escribi?:
I can try that, but I'm not sure that it will work. The problem
is that devList is just a pointer to a list owned by someone else.
Making devList point to a new list won't work: I need to make
the parent list different. I could do this by adding an extra
level of indirection, but I think at the risk making the call
environment more complex.
Then use [:] to *replace* all the old list items, do not merely rebind
the name. That last statement
should be, instead:
devs[:] = newdevs
(Please don't top-post)
--
Gabriel Genellina
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