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How to sort rectangles/track (shapes) in a list: left to right and/or top to bottom?

Sheppie

Hi,

I'm working on code for a pcell, in which I generate an array of rectangles in, for instance M1, and M2. The M1 rectangles are multiple rows and wider than tall, the M2 in multiple columns, and taller than wide.

I have a list for the M1 rectangles (containing the database-IDs of the rectangles) and a separate list for the M2 rectangles (also database-ID's).

I want to sort the M1 list in such a way that the first entry of the list is the M1 rectangle at the top ("northern" most) and the lowest ("southern" most) rectangle as last entry. All other neatly sorted in between of course.

I want to sort the M2 list in such a way that the first entry of the list is the M2 rectangle at the left ("western" most) and the the wan furthes to the right ("eastern" most) rectangle as last entry. All other neatly sorted in between of course.

I think I could come up with a very nice, but elaborate/long/complicated/not-understandable/over-complicated/highly-specific/not-efficient, piece of SKILL code to do this. However, I can imagine that there is a very neat/general piece of code/build-in-procedure that I can use to do this. Does anybody know the existence of this, or has done this before and is willing to share this with me? That would be highly appreciated.

Thanks in advance.

With kind regards,

Sjoerd

  • Hi Sjoerd,

    you can sort any data by creating a custom lambda function to compare the desired properties. For your example it would look like this:

    M1shapes = sort(M1shapes lambda((shape1 shape2) topEdge(shape1->bBox) > topEdge(shape2->bBox)))
    M2shapes = sort(M2shapes lambda((shape1 shape2) leftEdge(shape1->bBox) < leftEdge(shape2->bBox)))

    You can put arbitrarily complex expressions within the lambda functions and use them to re-order your data.

    Regards,

    Petar

  • Like Petar said - customizing sort() is the solution. Below is a chunk of code that orders figures (an instance is a figure)  left to right and top to bottom:

    (sort M1
       (lambda (s1 s2)
          (let  ( (x1 (xCoord (car s1->bBox)))
                   (x2 (xCoord (car s2->bBox)))
                   (y1 (yCoord (car s1->bBox)))
                   (y2 (yCoord (car s2->bBox))))
             (cond
                (y1==y2
                   (if x1<=x2 t nil))
                (y1<=y2 t nil)))))

  • in reply to p94todorov

    Hi Petar, Thank you very much for your quick response. I'll test it later today and will report back. I haven't really had a use for lambda before, so I will be most interested in getting this to work for the first time. Once again, thank you. Kind regards, Sjoerd

  • in reply to mbracht

    Hi Max, thank you too of course. Like I wrote in my response to Petar, I'll give it a try and will report back. Much appreciated. With kind regards, Sjoerd

  • in reply to mbracht

    Hi, It works perfectly! Thank a lot. This is something I wouldn't have come up with myself. Very nice solution. With kind regards, Sjoerd

  • in reply to p94todorov

    Hi, although I ended up using the solution of Max, I first tested it with your version, and that worked flawlessly. Thanks a lot. Now that I have seen how the lambda function can be used, I'll go over some of my recent code to try to apply something similar to other pieces of code. So far, the setof() function was my favorite (very powerful to select anything), but this may be better. With kind regards, Sjoerd

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