[Tutor] filtering within a function.

[Michael P. Reilly]

> On 12 Mar 2001, Michael P. Reilly replied:
> > That is probably because unstray relies on 
> > side-effects of the function parameter, which
> > is usually a bad design choice (but not always).
> > Here the problem is that unstray returns nothing
> > and modifies listinput.  
> 
> In this case, returning nothing and modifying the
> list was a conscious choice, not just how I got
> it to work--no need to retain the original list.
> 
> unstray is a crock for the same reason as:
> 
> foo = [1, 2, 1, 3, 1, 4]
> for x in foo:
>     if x >= 2:
> 	foo.remove(x)
> 
> That looks like it works, but foo[1] is removed,
> foo[2] falls into foo[1], placing it behind the
> iteration, as foo = [1, 2, 3, 4] would reveal.

Yup, if you're going to do this, then it is generally better to use
indices, which just makes life more interesting. ;)  The remove()
method removes the first element that matches.

>>> i = 0
>>> while i < len(foo):
...   if foo[i] >= 2:
...     del foo[i] # do NOT increment i
...   else:
...     i = i + 1
>>>

> > What you can do is overwrite the list's elements.
> >
> > How about:
> > uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
> > def unstray(listinput, f=lambda x: x in uppercase):
> >  l = filter(f, listinput)
> >  # replace the contents of listinput with the result of the filtering
> >  listinput[:] = l
> >
> > But then, maybe I'm missing the point. :)
> 
> Not that I can see.  Is this bad form?:
> 
> def unstray(listinput):
>   f=lambda x: x in uppercase
>   l = filter(f, listinput)
>   listinput[:] = l

About the only thing is that putting the lambda in the function
declaration does not recreate the lambda function every time and also
keeps it out of the global namespace.

  -Arcege

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