a break for comprehensions

Nick Perkins nperkins7 at home.com
Thu Jul 26 18:55:57 CEST 2001


"Tom Jenkins" <tjenkins at nospiced.ham.devis.com> wrote in message
news:3B6031C3.6010707 at nospiced.ham.devis.com...
> Kevin Lacker wrote:
>
> > Can you do this:
> >
> > answer = []
> > for x in my_list:
> >     if not is_good(x):
> >         break
> >     answer.append(process(x))
> >
> > with a list comprehension somehow? I'm starting to dislike explicit for
> > loops just for the purposes of constructing another list from a current
one.
>
>
> do you mean like this?
>  >>> def even(x):
> ... return int(divmod(x,2)[1]) == 0
> ...
>  >>> a = [1, 2, 3, 4, 5, 6, 7]
>  >>> b = [z for z in a if not even(z)]
>  >>> print b
> [1, 3, 5, 7]
>  >>> c = [z for z in a if even(z)]
>  >>> print c
> [2, 4, 6]
>  >>>
>
> oops sorry just saw that you had process(x) in there... lets try it:
>
>  >>> def process(x):
> ... return str(x)
> ...
>  >>> process(1)
> '1'
>  >>> c = [process(z) for z in a if even(z)]
>  >>> print c
> ['2', '4', '6']
>  >>>
>
> hmmm, seems to work for me (python 2.1)
>


This misses the point of the original post.
Notice the 'break' in the original for-loop version.
The idea is to return consecutive elements,
up to, but not including, the first element which
does not meet the condition.

ie.
lst = [0,2,4,5,6]
evens = [0,2,4]

I don't think list comprehensions can do this.


Here's a generator approach to the problem:



from __future__ import generators

def conditional_iter(lst,cond):
    for item in lst:
        if cond(item):
            yield item
        else:
            raise StopIteration


mylist = [0,2,4,5,6]
def even(num): return not (num%2)

for num in conditional_iter(mylist, even)
    print num








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