Delete all items in the list

Hi. This must be a simple command but I just can't find it in the
Phthon manual. How do I delete all items with a certain condition from
a list? For instance:

L=['a', 'b', 'c', 'a']

I want to delete all 'a's from the list.
But if L.remove('a') only deletes the first 'a'.

How do you delete all 'a's?
I would really appreciate your help.

Thanks.

Clarendon

On Thu, Feb 26, 2009 at 3:05 AM, Clarendon <> wrote:
> Hi. This must be a simple command but I just can't find it in the
> Phthon manual. How do I delete all items with a certain condition from
> a list? For instance:
>
> L=['a', 'b', 'c', 'a']
>
> I want to delete all 'a's from the list.
> But if L.remove('a') only deletes the first 'a'.
>
> How do you delete all 'a's?

There are several ways. I'd go with a list comprehension:

L = [i for i in L if i != 'a']

Or to modify the list in-place:

L[:] = [i for i in L if i != 'a']

Cheers,
Chris

--
Follow the path of the Iguana...
http://rebertia.com

Chris Rebert

Clarendon a écrit :
> Hi. This must be a simple command but I just can't find it in the
> Phthon manual. How do I delete all items with a certain condition from
> a list? For instance:
>
> L=['a', 'b', 'c', 'a']
>
> I want to delete all 'a's from the list.
> But if L.remove('a') only deletes the first 'a'.
>
> How do you delete all 'a's?

L[:] = [item for item in L if item != 'a']

Bruno Desthuilliers

Chris Rebert wrote:
> On Thu, Feb 26, 2009 at 3:05 AM, Clarendon <> wrote:
....
>> L=['a', 'b', 'c', 'a']
>>
>> I want to delete all 'a's from the list.
>> But if L.remove('a') only deletes the first 'a'.
>>
>> How do you delete all 'a's?
>
> There are several ways. I'd go with a list comprehension:

and for a couple other ways

while 'a' in L : L.remove('a')

L = filter('a'.__ne__,L)

Boris Borcic

En Thu, 26 Feb 2009 11:00:30 -0200, Boris Borcic <>
escribió:

> Chris Rebert wrote:
>> On Thu, Feb 26, 2009 at 3:05 AM, Clarendon <> wrote:
> ...
>>> L=['a', 'b', 'c', 'a']
>>>
>>> I want to delete all 'a's from the list.
>>> But if L.remove('a') only deletes the first 'a'.
>>>
>>> How do you delete all 'a's?
>> There are several ways. I'd go with a list comprehension:
>
> and for a couple other ways
>
> while 'a' in L : L.remove('a')

This is probably the worst way, takes cuadratic time (and two searches per
loop).

> L = filter('a'.__ne__,L)

And this is probably the fastest. But not all types define __ne__ so a
more generic answer would be:

from functools import partial
from operator import ne
L = filter(partial(ne, 'a'), L)

Of course, this is only relevant if L is large and there are many
duplicates. In other cases I'd stick with the list comprehension as posted
by Chris Rebert.

--
Gabriel Genellina

Gabriel Genellina

>>
>> > L = filter('a'.__ne__,L)
>>
>> And this is probably the fastest. But not all types define
>> __ne__ so a
>> more generic answer would be:
>>
>> from functools import partial
>> from operator import ne
>> L = filter(partial(ne, 'a'), L)
>>

And don't forget this "traditional" solution:

>>> L=['a', 'b', 'c', 'a']
>>> filter(lambda arg: arg != 'a', L)
['b', 'c']

-John





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John Posner

>>
>> from functools import partial
>> from operator import ne
>> L = filter(partial(ne, 'a'), L)
>>

I learned about functools.partial only recently (on this list). functools is
implemented in C, not Python, so I couldn't answer this question for myself:

Is functools.partial completely equivalent to a manually-coded closure?

Help, please.

-John





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John Posner

Gabriel Genellina wrote:

>
>> L = filter('a'.__ne__,L)
>
> And this is probably the fastest. But not all types define __ne__

In Py3, all classes inherit .__ne__ from 'object'.

Terry Reedy

Terry Reedy wrote:
> Gabriel Genellina wrote:
>
>>
>>> L = filter('a'.__ne__,L)
>>
>> And this is probably the fastest. But not all types define __ne__
>
> In Py3, all classes inherit .__ne__ from 'object'.
>
Isn't that inherited method just an identity comparison?

However in this particular case the main point is that str *does*
implement __ne__. So as long as that built-in method doesn't call the
other operand's __ne__ there should be no problem.

regards
Steve
--
Steve Holden +1 571 484 6266 +1 800 494 3119
Holden Web LLC http://www.holdenweb.com/

Steve Holden

Steve Holden wrote:
> Terry Reedy wrote:
>> Gabriel Genellina wrote:
>>
>>>> L = filter('a'.__ne__,L)
>>> And this is probably the fastest. But not all types define __ne__
>> In Py3, all classes inherit .__ne__ from 'object'.
>>
> Isn't that inherited method just an identity comparison?

Yes. And so is, by default, the snipped slower proposed alternative of
lambda x: a!= x.

> However in this particular case the main point is that str *does*
> implement __ne__. So as long as that built-in method doesn't call the
> other operand's __ne__ there should be no problem.
>
> regards
> Steve

Terry Reedy