On Wed, 30 Jan 2008 06:07:45 -0800, cokofreedom wrote:
Anyone else noticed that the OP has not actually replied to any of the
suggestions...
Sorry. I was just fascinated at the turns it was taking. But the first
answer was fine for me:
for name in apps[:]:
etc.
Thanks all.
--
09 F9 11
Hmm, how does this fare??
for i in range(len(a)):
if a[i]==99: a=a[:i]+a[i+1:]
I like following your guys code noodling. I can come up with something
that does what it appears your doing, sometimes, but as to it's relevant
merits I havent a clue :)
matt.
En Thu, 31 Jan 2008 15:45:42 -0200, [EMAIL PROTECTED]
escribió:
Hmm, how does this fare??
for i in range(len(a)):
if a[i]==99: a=a[:i]+a[i+1:]
I like following your guys code noodling. I can come up with something
that does what it appears your doing, sometimes, but as to it's
On 30 ene, 08:09, Paul Rubin http://[EMAIL PROTECTED] wrote:
Santiago Romero [EMAIL PROTECTED] writes:
li = [1,2,3,4,5]
filter(lambda x: x != 3, li)
[1, 2, 4, 5]
I haven't measured it, but this should be the fast solution in all
the thread ...
li.remove(3) is probably faster.
On Jan 30, 9:50 am, Santiago Romero [EMAIL PROTECTED] wrote:
On 30 ene, 08:09, Paul Rubin http://[EMAIL PROTECTED] wrote:
Santiago Romero [EMAIL PROTECTED] writes:
li = [1,2,3,4,5]
filter(lambda x: x != 3, li)
[1, 2, 4, 5]
I haven't measured it, but this should be the
Santiago Romero [EMAIL PROTECTED] writes:
In a = [1, 2, 3, 3, 3, 4, 3, 3, 2, 3], the filter solution will
efectively remove all items with value == 3 while li.remove(3) will
only remove the first ocurrence.
Hmm, interesting, I didn't realize that (shoulda checked the docs).
Thanks!
--
On Jan 29, 10:59 pm, Paul Hankin [EMAIL PROTECTED] wrote:
If I really had to modify it in place (and the condition wasn't really
x == 99), how about:
bad_indices = [i for i, x in enumerate(a) if x == 99]
for bad_index in reversed(bad_indices):
del a[bad_index]
Or one could use the trick
On Jan 30, 11:57 am, Arnaud Delobelle [EMAIL PROTECTED] wrote:
n = len(a)
for i, x in enumerate(a):
if x == 99: del a[i-n]
Oops. That can't work. Don't know what I was thinking here. I
probably did had one mental refactoring too many...
--
Arnaud
--
Neil Cerutti [EMAIL PROTECTED] writes:
Or one can put on his bellbottoms, horn-rimmed glasses, and wear a mullet:
i = 0
while i len(a):
if a[i] == 99:
del a[i]
else:
i += 1
Quadratic time!! Yowch!! Back to the future:
def rocket_science(xs):
for x in xs:
if x !=
On Jan 30, 2008 6:57 AM, Arnaud Delobelle [EMAIL PROTECTED] wrote:
Or one could use the trick of counting from the right (untested):
n = len(a)
for i, x in enumerate(a):
if x == 99: del a[i-n]
Or one can put on his bellbottoms, horn-rimmed glasses, and wear a mullet:
i = 0
while i
If you don't want to reinvent the wheel all the time you can use this
one:
def inplacefilter(pred, alist):
inplacefilter(pred, alist): filters the given list like
filter(),
but works inplace, minimizing the used memory. It returns None.
pr = lambda x: x 2
l = []
On 30 Jan 2008 05:20:49 -0800, Paul Rubin
http://phr.cx@nospam.invalid wrote:
Neil Cerutti [EMAIL PROTECTED] writes:
Or one can put on his bellbottoms, horn-rimmed glasses, and wear a mullet:
i = 0
while i len(a):
if a[i] == 99:
del a[i]
else:
i += 1
Quadratic
Anyone else noticed that the OP has not actually replied to any of the
suggestions...
--
http://mail.python.org/mailman/listinfo/python-list
Paul Rubin http://[EMAIL PROTECTED] writes:
Quadratic time!! Yowch!! Back to the future:
def rocket_science(xs):
for x in xs:
if x != 99:
yield x
a[:] = list(rocket_science(a))
I call useless use of list!
a[:] = rocket_science(a)
:-)
--
On Jan 29, 8:34 am, William McBrine [EMAIL PROTECTED] wrote:
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
...
1
2
3
5
a
[1, 2, 4, 5]
Sure, the resulting list is correct. But 4 is never printed
On Tue, 29 Jan 2008 16:34:17 GMT, William McBrine [EMAIL PROTECTED] wrote:
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
...
1
2
3
5
a
[1, 2, 4, 5]
You have to iterate over a copy of 'a', so for x in
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
Well ... you could use:
for i in range(len(a)-1, -1, -1):
...print a[i]
...if a[i] == 3: del a[i]
...
5
4
3
2
1
print a
[1, 2, 4, 5]
Bye.
--
On Tue, 29 Jan 2008 09:23:16 -0800 (PST), [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
If you're going to delete elements from
a list while iterating over it, then do
it in reverse order:
Why so hard? Reversing it that way creates a copy, so you might as
well do:
a = [ 98, 99, 100 ]
for i, x
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
...
1
2
3
5
a
[1, 2, 4, 5]
Sure, the resulting list is correct. But 4 is never printed during the
loop!
What I was really trying to do was this:
apps = [name for
On Jan 29, 12:34 pm, William McBrine [EMAIL PROTECTED] wrote:
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
...
1
2
3
5
a
[1, 2, 4, 5]
Sure, the resulting list is correct. But 4 is never printed
Berteun Damman [EMAIL PROTECTED] wrote:
On Tue, 29 Jan 2008 09:23:16 -0800 (PST), [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
If you're going to delete elements from
a list while iterating over it, then do
it in reverse order:
Why so hard? Reversing it that way creates a copy, so you might
On Jan 29, 4:34 pm, William McBrine [EMAIL PROTECTED] wrote:
Look at this -- from Python 2.5.1:
a = [1, 2, 3, 4, 5]
for x in a:
... if x == 3:
... a.remove(x)
... print x
...
1
2
3
5
a
[1, 2, 4, 5]
Sure, the resulting list is correct. But 4 is never printed
On Jan 29, 8:17 pm, Duncan Booth [EMAIL PROTECTED] wrote:
Berteun Damman [EMAIL PROTECTED] wrote:
On Tue, 29 Jan 2008 09:23:16 -0800 (PST), [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
If you're going to delete elements from
a list while iterating over it, then do
it in reverse order:
On Jan 29, 2008 9:23 AM, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
If you're going to delete elements from
a list while iterating over it, then do
it in reverse order:
how about
li = [1,2,3,4,5]
filter(lambda x: x != 3, li)
[1, 2, 4, 5]
--
Santiago Romero [EMAIL PROTECTED] writes:
li = [1,2,3,4,5]
filter(lambda x: x != 3, li)
[1, 2, 4, 5]
I haven't measured it, but this should be the fast solution in all
the thread ...
li.remove(3) is probably faster.
--
http://mail.python.org/mailman/listinfo/python-list
how about
li = [1,2,3,4,5]
filter(lambda x: x != 3, li)
[1, 2, 4, 5]
I haven't measured it, but this should be the fast solution in all
the thread ...
--
http://mail.python.org/mailman/listinfo/python-list
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