Identify Cells already created

[Daneel]

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.

[Ronald]

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

[Daneel]

Thank you so much for the help.

[Daneel]

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()

[Ronald]

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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