jython-users

[Jython-users] List problem

[Kevin M. <kev...@er...>]

Hi,

why does the following not work:

a = [1, 2]
for i in a:
	a.remove(i)


The loop exits after the first iteration leaving a = [2]

How do I get around this problem?

Any help appreciated,
Kevin


_________________________________________________
Name/Title : Kevin McNamee, Software Consultant
Phone      : +46 13 32 1165

Re: [Jython-users] List problem

[Kevin B. <kb...@ca...>]

That's cool!  :-)

It looks like what is happening is that the 'for i in a:' tracks the index it is working on, so if you remove items, you get every other one:

>>> a = range( 10 )
>>> for i in a:
...   print i
...   a.remove( i )
... 
0
2
4
6
8
>>> 

You can make a copy of the list, and iterate over that copy:

>>> a = range( 10 )
>>> for i in tuple(a):
...   print i
...   a.remove( i )
... 
0
1
2
3
4
5
6
7
8
9
>>> a
[]
>>> 

kb

Kevin McNamee wrote:
> 
> Hi,
> 
> why does the following not work:
> 
> a = [1, 2]
> for i in a:
>         a.remove(i)
> 
> The loop exits after the first iteration leaving a = [2]
> 
> How do I get around this problem?
> 
> Any help appreciated,
> Kevin
> 
> _________________________________________________
> Name/Title : Kevin McNamee, Software Consultant
> Phone      : +46 13 32 1165
> 
> _______________________________________________
> Jython-users mailing list
> Jyt...@li...
> https://lists.sourceforge.net/lists/listinfo/jython-users

Re: [Jython-users] List problem

[Ype K. <yk...@xs...>]

>Hi,
>
>why does the following not work:
>
>a = [1, 2]
>for i in a:
>	a.remove(i)
>
>
>The loop exits after the first iteration leaving a = [2]
>
>How do I get around this problem?

Iterating over a list that is being modified results
in undefined behaviour. A better way is:

    for i in a[:]: # iterate over a copy
         a.remove(i) # remove from original

The problem with this is that it still needs to access each element
in the list (at least twice).

When you need to remove some of the elements in a list you could
also use list comprehensions, but these always create a new list:

    a = [i for i in a if not someRemoveCondition(i)]


To empty a list in a single operation you can also:

    a[:] = [] # modify in place to become an empty list. (A)

or

    a = [] # change to refer to a new empty list object. (B)

or, preferably:

    del a[:] # modify in place deleting all entries (C)

The last variant does not create a new object, which can make a measurable
difference for garbage collection in case it avoids creating a lot of them.
This technique is also known as reusing existing objects.
I have never measured the difference between (A) and the other
two, but (C) is definitely faster than (B) in Jython.
It is probably faster in CPython, too.

>
>Any help appreciated,
>Kevin

My pleasure,
Ype