Efficient way to remove objects from a list

[Tino Wildenhain]

一首诗 wrote:
> Hi all,
> 
> Today I wrote some code like this:
> 
>         for m in self.messages:
>             if not m.finished:
>                 continue
> 
>             #process the message
> 
>         fini = [m for m in self.messages if m.finished]
>         for m in fini:
>             self.messages.remove(m)
> 
> As you can, I want to find these finished messages in
> "self.messages",
> process them, and then remove them from the list.
> 
> Because a list can not be modified while iterating it,  I have to use
> a list "fini" to accomplish the target.
> 
> I found a smell of bad performance here.
> Is there any faster ways?

I'm not getting what your code really wants to do -
you are iterating over messages, and if its
"not" finished, skip over it and process all
which are finished. (If they are finished, why
process them?) Wouln't it make more sense
to call the process where you process the messages?
And you could just pop() the list or use some
of the queue implementations for better performance.

e.g.

     def enqueue_messages(self,msg):
         self.messages.append(msg)
     def process_messages(self):
         if self.messages:
             msg=self.messages.pop(0)
             process(msg)

if you want some post processing,
its really like you are acting in
different stages, so you could just
have two lists for your messages:
one for the messages in phase1,
one for the messages passed phase1 and
now in phase2 and work over them in
the way shown above.

Regards
Tino

> --
> http://mail.python.org/mailman/listinfo/python-list


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