Is it possible to filter an instance on a circular shape ?

Comments

• tagger5896

  April 2021

  @ahmedo

  This will find all boxes outside of a layer/circle
  Run the script and you will see a circle with boxes inside,
  click undo and you can see what has been removed

  
  # Enter your Python code here
  import pya
  import array
  import sys
  import math
  import time
  
  mw = pya.Application.instance().main_window()
  layout = pya.Application.instance().main_window().current_view().active_cellview().layout() 
  topcell = pya.Application.instance().main_window().current_view().active_cellview().cell
  lv = pya.Application.instance().main_window().current_view()
  
  if layout == None:
    raise Exception("No layout")
  
  cv = pya.Application.instance().main_window().current_view().active_cellview()
  
  dbu =layout.dbu
  
  cell = pya.Application.instance().main_window().current_view().active_cellview().cell
  if cell == None:
    raise Exception("No cell")
  
  n =   2000 # number of points
  r =   2*2  # radius
  da = 2 * math.pi / n
  
  
  def coords(path):
      result = []
      for p in path:
        result.append(DPoint.new(p[0]/dbu,p[1]/dbu))
      return result
  
  def create_polygon(crd,lay):
    layer_info = LayerInfo.new(lay,0)
    layer_index = layout.layer(layer_info)
    cv.cell.shapes(layer_index).insert(Polygon.new(crd))
  
  def create_box(crd,lay):
    layer_info = LayerInfo.new(lay,0)
    layer_index = layout.layer(layer_info)
    cv.cell.shapes(layer_index).insert(Box.new(crd[0],crd[1]))  
  
  circle = []
  for i in range(0, n):
     path = r * math.cos(i * da), r * math.sin(i * da)
     circle.append(path)
  
  lay1 = 9  
  lay2 = 5
  
  pth = coords(circle)
  create_polygon(pth,lay1)
  
  box = [(-6.818,-1.185), (-5.195,3.591)]
  crd = coords(box)
  create_box(crd,lay2)
  
  box = [(5.742,-2.333), (7.674,2.771)]
  crd = coords(box)
  create_box(crd,lay2)
  
  box = [(-2.827,0.849), (-1.326,2.469)]
  crd = coords(box)
  create_box(crd,lay2)
  
  box = [(0.730,-2.708), (2.231,-1.088)]
  crd = coords(box)
  create_box(crd,lay2)
  
  def cut_region(lay1,lay2):
    lv.transaction("Undo") 
  
    lif1 = LayerInfo.new(lay1,0)
    li1 = layout.layer(lif1)
  
    lif2 = LayerInfo.new(lay2,0)
    li2 = layout.layer(lif2)
    r1 = pya.Region(topcell.begin_shapes_rec(li1))
    r2 = pya.Region(topcell.begin_shapes_rec(li2))
  
    #Boolean Functions
    #outside  = r1 -  r2
    #outside  = r2 - r1
    #outside  = r2 + r1
    #outside  = r1 + r2
    #outside  = r1 & r2
    outside  = r2 & r1
    #outside  = r2 ^ r1
    #outside  = r1 ^ r2
  
  
    layout.clear_layer(li2)
    output = layout.layer(lif2)
    topcell.shapes(output).insert(outside)  
  
  
  lay1 = 9  
  lay2 = 5
  cut_region(lay1,lay2)
  
  
  lv.commit()
  
  
• sebastian

  April 2021

  If I understand @ahmedo correctly he said not overlapping, so instead of
  outside = r2 & r1
  it should be
  not_outside = r2.select_not_outside(r1)
  and then
  topcell.shapes(output).insert(not_outside)

  r1 & r2 will cut parts of boxes that are outside of the circle.
• tagger5896

  April 2021

  @sebastian

  I may have misread .. but good point and thanks for highlighting this, we all learn
• sebastian

  April 2021

  No worries ^^. I also think overlapping is not clearly defined in daily speech, so you may very well be correct. Regions were confusing when I started with KLayout, so I wanted to point out that there are fancy options in the regions included that might save a lot of time.
• ahmedo

  April 2021

  Thank you @tagger5896 and @sebastian for your valuable help.

  I will make a test with your code. I see that the code looks for the name of the layout at the beginning, so I guess I need to run the script using the command line:
  klayout -r <script.py> my.gds

  Is it right ?

  By the way I have implemented a solution that allows me to check the overlap:

  filter_poly = db.Polygon(circular_shape.polygon)
  for inst in layout_insts:       
          # Define the 4 points of the instance box
          llp = db.Point(inst.bbox().left,inst.bbox().bottom)
          lrp = db.Point(inst.bbox().right,inst.bbox().bottom)
          tlp = db.Point(inst.bbox().left,inst.bbox().top)
          trp = db.Point(inst.bbox().right,inst.bbox().top)
  
          let = 0
          if filter_poly.inside(llp) or filter_poly.inside(lrp) or filter_poly.inside(tlp) or filter_poly.inside(trp):
              let = 1
  
          # if let = 0 , then remove instance.
          if let == 0 :
              inst.delete()
  

  Regards
• ahmedo

  April 2021

  For my question I think that I have undestood that @tagger5896 's code should be executed from macro menu
  I'm used to creating scripts that I execute always in batch mode.

  Regards
• tagger5896

  April 2021

  @ahmedo
  You should have an open layout and run the code from the Macro IDE menu yes.
• ahmedo

  April 2021

  @tagger5896

  Thank you so much for your help.

  Regards
• tagger5896

  April 2021

  @sebastian
  thanks for pointing this out ... floodgates open and the sea's part,
  and yes this makes it easier

  not_outside = r2.select_not_outside(r1)
• tagger5896

  April 2021

  @ahmedo @sebastian

  New thread https://www.klayout.de/forum/discussion/1790/boolean-functions-based-on-regions-functions-to-add-box-circle-path-and-polygon#latest