Efficient way to remove objects from a list

[=?GB2312?B?0rvK18qr?=]

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?

[Peter Otten]

Chris Rebert wrote:

On Mon, Nov 3, 2008 at 1:40 AM, 一首诗 <ne******@gmail.comwrote:

>Hi all,

Today I wrote some code like this:

Build a new list as you go, then overwrite the old list with it.

unfinished = []

> for m in self.messages:
if not m.finished:

unfinished.append(m)

> continue

#process the message

Remove the following code

> fini = [m for m in self.messages if m.finished]
for m in fini:
self.messages.remove(m)

self.messages[:] = unfinished

Just

self.messages = unfinished

if also OK unless you have multiple references to the self.messages list.

This way you aren't calling .remove() multiple times and only iterate
through the list once.

[=?GB2312?B?0rvK18qr?=]

Thanks! That's a more clear way!

On Nov 3, 9:38 pm, Peter Otten <[e-mail]:

Chris Rebert wrote:

On Mon, Nov 3, 2008 at 1:40 AM, Ò»Ê×Ê« <[e-mail]>wrote:

Hi all,

Today I wrote some code like this:

Build a new list as you go, then overwrite the old list with it.

unfinished = []

for m in self.messages:
if not m.finished:

unfinished.append(m)

continue

#process the message

Remove the following code

fini = [m for m in self.messages if m.finished]
for m in fini:
self.messages.remove(m)

self.messages[:] = unfinished

Just

self.messages = unfinished

if also OK unless you have multiple references to the self.messages list.

This way you aren't calling .remove() multiple times and only iterate
through the list once.