Home Forums Nazca Identify Cells already created

This topic contains 4 replies, has 2 voices, and was last updated by  Ronald 2 months, 1 week ago.

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  • #5550

    Daneel
    Member

    I’m doing a design where many rectangles with different sizes will be created as cells. The cells will be names according to their length and width:

    def rectangles(length=80, width=80):
        with nd.Cell('rectangle_' + str(length) + 'x' + str(width)) as rectangle:
            rect = geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
        return rectangle

    In a second cell many of these rectangle cells will be created and used with different sizes.

    I do not want to create a cell with the same dimensions twice, instead I want to use the cell which already exists.

    Is there a way to determine if a rectangle cell with specific dimensions does already exist (maybe by checking if the name already exist)?

    Thank you very much for your help.

    #5552

    Ronald
    Keymaster

    Dear Daneel,

    Using the @hashme decorator should work for you as discussed in this topic

    import nazca as nd
    
    @nd.bb_util.hashme('rectangle', 'length', 'width') # gives default cell name 'rectangle_80_80'
    def rectangles(length=80, width=80):
        with nd.Cell(hashme=True) as rectangle:
            rect = geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
        return rectangle

    In case you still want to know which cells have been defined you can try the following code to print all cell names:

    import nazca as nd
    
    for name in nd.cfg.cellnames.keys():
        print(name)

    Ronald

    • This reply was modified 8 months, 2 weeks ago by  Ronald. Reason: Added the missing @
    #5553

    Daneel
    Member

    Thank you so much for the help.

    #5920

    Daneel
    Member

    Hello Ronald,

    I have another follow-up question on that topic.

    I have a function which defines a cell which itself consists of several subcells. Is it possible to use the hasme = True statement with the last cell, which is returned by the function? If I do it like that I receive an exceptio: Exception: Detected a call nd.Cell(hashme=True) without first calling the @hashme(‘cellname’) decorator with a non-empty ‘cellname’.

    Does that mean the hashme only works with one cell per function, does the cell which is hashed has to be follow the function definition directly or do i have to hash each cell in the function?

    import nazca as nd
    import nazca.geometries as geom
    
    @nd.bb_util.hashme('rectangle', 'length', 'width') # gives default cell name 'rectangle_80_80'
    def rectangles(length=80, width=80):
        with nd.Cell('Text') as text:
            nd.text(text = 'Rectangle' + str(length) +' ,' + str(width),
            height = 30, layer = 'Label', align='cc').put(0,0)
    
        with nd.Cell(hashme=True) as rectangle:
            rect = geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
            text.put(100,0)
        
        return rectangle

    rectangles().put()
    rectangles().put()

    nd.export_gds()

    • This reply was modified 2 months, 1 week ago by  Ronald. Reason: Updated the code to make it run and export
    #5922

    Ronald
    Keymaster

    Dear Daneel,

    The first cell you create after evoking the @hashme will get the name created by hashme.

    You can simply nest your ‘text’ cell inside the ‘rectangle’ cell to get what you need:

    import nazca as nd
    import nazca.geometries as geom
    
    @nd.bb_util.hashme('rectangle', 'length', 'width')
    def rectangles(length=80, width=80):
        with nd.Cell() as rectangle:
            with nd.Cell() as text:
                nd.text('text', height=30, layer='Label', align='cc').put(0)
            rect=geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
            text.put(100, 0)
        return rectangle
    
    rectangles().put(0)
    rectangles().put(0)
    
    nd.export_gds()

    If all you ultimately want to do here is to get access the cell name to print it you have other options too. You can use the Cell attribute cell_name and get rid of the second explicit cell altogether:

    @nd.bb_util.hashme('rectangle', 'length', 'width')
    def rectangles(length=80, width=80):
        with nd.Cell() as rectangle:
            text = nd.text(text=rectangle.cell_name[:-6], height=30,
                layer='Label', align='cc')
            rect=geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
            text.put(100, 0)
        return rectangle

    A last alternative is to use a bounding box:

    @nd.bb_util.hashme('rectangle', 'length', 'width')
    def rectangles(length=80, width=80):
        with nd.Cell() as rectangle:
            rectangle.autobbox = True        
            rect=geom.box(length=length, width=width)
            nd.Polygon(points=rect).put(0, 0)
        return rectangle

    Ronald

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