[Tutor] Hey guys!

[Alan Gauld]

On 17/02/15 04:22, Levi Adissi wrote:

Hello. Thanks for your post.
However could you please use a subject that rflects the content. 
Otherwise we wind up with an archive full of "Hey Guys", "Hello chaps", 
"Gday mate", Bonjour" etc  Not very useful for searching.

> So I'm kind of stuck trying to program a function that returns a list of
> tuples.

Note that your function prints the list it does not return the list.
The two are very different.

> The function takes 2 lists containing circles of which it should
> compare list1[0] to list2[0] to see if they intersect. If they intersect or
> touch then I should return them on a list of tuples(in the tuple would be
> both intersecting circles).
>
> I can't get circles_only to work the way I see it I'm comparing h to x only
> if they're both in the same place on the list (hence my "h==x") I know it
> doesn't work because the test returns None so I would really appreciate an
> alternative method if you guys see one.
>
> Here are my functions:
>
>
> def circles_overlap(c1, c2):
>     x=(c2.center.y-c1.center.y)**2
>     y=(c2.center.x-c1.center.x)**2
>     distancemid=math.sqrt(x+y)
>     distancerad=(c1.radius+c2.radius)
>     if distancemid > distancerad:
>         return 1
>     elif distancemid < distancerad:
>         return -1
>     elif distancemid == distancerad:
>         return 0
>

Since this is a test you should probably just return True or False.
Either they overlap or they don't. If you want to use touching as a 
third result you can still give a boolean for the general case:

return
1 for overlap
-1 for touching
0 for not touching

1 and -1 both evaluate to True in a boolean sense
0 evaluates to False

This makes the test code in your second function look like:

         if circles_overlap(c1,c2):
              newlst.append((c1,c2))

and you don't need the else clause

> def circles_only(lst1, lst2):
>     newlst=[]
>     for h in lst1:
>        for x in lst2:
>           if h==x:

Why do you have this test? You only check if they overlap
when they are equal? That seems odd to me.
Also h and x seem an odd choice of name for two circles.
Why not

for c1 in lst1:
   for c2 in lst2:

>              if circles_overlap(lst1[h],lst2[x])== -1:
>                 newlst.append(lst1[h],lst2[x])

Now you are trying to use indexing to extract the circles
but you are (quite rightly) iterating over the lists directly
not using indexes. So h and x are circles, you don't need
to  get them out of the lists.

Also you are trying to store two items in your newlist rather
than a tuple. (See my suggested test above.)

HTH
-- 
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
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