On Sun, May 20, 2018 at 11:20 PM, Daniel Foerster
wrote:
> I would guess that a significant amount of the gain is that he doesn't
> have to len() the list every iteration, plus the item unpacking occurs in C.
>
`len(...)` should be constant-time (stored with array), but indeed caching
it in a va
I would guess that a significant amount of the gain is that he doesn't have
to len() the list every iteration, plus the item unpacking occurs in C. I
don't know how much JIT would affect anything unless you ran the tests in
PyPy.
On Sun, May 20, 2018, 23:51 Ian Mallett wrote:
> On Sun, May 20, 2
On Sun, May 20, 2018 at 9:25 PM, Daniel Foerster
wrote:
> The relevance of N exponent versus the discarded coefficients depends on
> how big N may be. With the likely sizes of N in a Pygame class, the
> difference between the algorithms seems probably negligible. A quick test
> with 20% deletion
This is a quality algo improvement. I will note that the del can just be
"del myList[i_write:]" without the manually calculated upper bound.
On Sun, May 20, 2018, 23:39 MrGumm wrote:
> I like this one. I would modify somewhat. If you don't like "not" you
> could change the function name to keep_
I like this one. I would modify somewhat. If you don't like "not" you
could change the function name to keep_item() and reverse the logic.
i_write =0 for itemin myList:
if not needsToBeDeleted(item):
myList[i_write] = item
i_write +=1 del myList[i_write:len(myList)]
On 5/20
Tyler:
That's clever, but still runs into the nasty nasty problems of using
.remove(), which requires iteration over the entire list. In practice, it's
3x slower than either index-tracking algorithm at N=1000, while being
slightly faster at N=100 and 50% faster at N=10— but almost 20x slower at
N=
What about reversing through the list?
for val in reversed(myList):
if check(val):
myList.remove(val)
The way it works is that reversed returns an iterator that doesn't copy
from your list, but handles changes in the list correctly for you. So its
both memory and time efficient (one p
The relevance of N exponent versus the discarded coefficients depends on
how big N may be. With the likely sizes of N in a Pygame class, the
difference between the algorithms seems probably negligible. A quick test
with 20% deletion shows that your algorithm becomes more efficient around
N=7000, bu
On Sun, May 20, 2018 at 8:35 PM, Daniel Foerster
wrote:
> The third, and probably most convenient based on where you seem to be at
> in the curriculum, is to do something like Ian suggested. I think there's a
> simpler way to do it with similar performance though I've not benched to
> find out; I
A couple of options still then.
The simplest on paper is #3 of your original options, but as you mentioned,
list.remove() has problems (and bigger problems if you are working with
potentially duplicate items).
A second option is stay as close to the list comprehension as possible,
using either th
On Sun, May 20, 2018 at 5:55 PM, Irv Kalb wrote:
> Is there a way to do that without the list comprehension? I'm building
> this as part of a class, and I will not have talked about list
> comprehensions up until that point.
>
A list comprehension is the clearest and most-pythonic way to do thi
Thanks very much.
Is there a way to do that without the list comprehension? I'm building this as
part of a class, and I will not have talked about list comprehensions up until
that point.
Thanks,
Irv
> On May 20, 2018, at 2:35 PM, Daniel Foerster wrote:
>
> I think what you're looking for
I think what you're looking for is this:
my_list = [x for x in my_list if not need_to_delete(x)]
On Sun, May 20, 2018, 16:28 Irv Kalb wrote:
> I am building a game where I am keeping track of many objects in a list
> (let's just call it "myList". In every "frame" of my game I need to see if
>
I am building a game where I am keeping track of many objects in a list (let's
just call it "myList". In every "frame" of my game I need to see if any
objects need to be removed from myList.
I know that this code:
for item in myList:
if needsToBeDeleted(item):
myList.remove(item)
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