How to define border colour for rule-based style using PyQGIS?

Question

I use the following to create a Rule-based style for a given layer which includes defining a label, expression and colour fill:

from PyQt4.QtGui import QColor

layer = qgis.utils.iface.activeLayer()
style_rules = (
    ('First', 'expression_1', '#dbffdb'),
    ('Second', 'expression_2', '#f0ab64'),        
)
symbol = QgsSymbolV2.defaultSymbol(layer.geometryType())
renderer = QgsRuleBasedRendererV2(symbol)
root_rule = renderer.rootRule()
for label, expression, color_name in style_rules:
    rule = root_rule.children()[0].clone()
    rule.setLabel(label)
    rule.setFilterExpression(expression)
    rule.symbol().setColor(QColor(color_name))
    root_rule.appendChild(rule)

root_rule.removeChildAt(0)
layer.setRendererV2(renderer)
layer.triggerRepaint()

---

Can I define the colour border using the above method? If not, how could I iterate through the symbol layers and apply a border colour after a rule-based style has been set?

I tried incorporating it into style_rules but not sure how to set it (in this case, I just want the border colour to match the polygon fill):

style_rules = (
    ('First', 'expression_1', '#dbffdb', '#dbffdb'),
    ('Second', 'expression_2', '#f0ab64', '#f0ab64'),        
)
for label, expression, color_name, color_border in style_rules:
    rule = root_rule.children()[0].clone()
    rule.setLabel(label)
    rule.setFilterExpression(expression)
    rule.symbol().setColor(QColor(color_name))
    root_rule.appendChild(rule)

---

EDIT:

I have looked at the QgsSymbolV2 and QgsRuleBasedRendererV2 classes but neither seems to have anything related to the border. The QgsSimpleFillSymbolLayerV2 class does have a borderColor function so I'm now wondering if it is even possible to define a border colour using the QgsRuleBasedRendererV2 class.

---

I'm using QGIS 2.18.0 for Win7 64-bit.

Answer 1

Since you want the border color to match the corresponding polygon fill, you can use the data defined property @symbol_color, which returns the fill color for each polygon. Then, for the default polygon symbol (which is the base of your QgsRuleBasedRendererV2) you set its color_border property to @symbol_color:

ddp = QgsDataDefined( True, True, "@symbol_color" ) # active, useExpression, expression
symbol.symbolLayer( 0 ).setDataDefinedProperty( "color_border", ddp )

Your script would become this:

from PyQt4.QtGui import QColor

layer = iface.activeLayer()
style_rules = (
    ('First', 'expression_1', '#dbffdb'),
    ('Second', 'expression_2', '#f0ab64'),
)
symbol = QgsSymbolV2.defaultSymbol(layer.geometryType())
ddp = QgsDataDefined( True, True, "@symbol_color" )
symbol.symbolLayer( 0 ).setDataDefinedProperty( "color_border", ddp )
renderer = QgsRuleBasedRendererV2(symbol)
root_rule = renderer.rootRule()
for label, expression, color_name in style_rules:
    rule = root_rule.children()[0].clone()
    rule.setLabel(label)
    rule.setFilterExpression(expression)
    rule.symbol().setColor(QColor(color_name))
    root_rule.appendChild(rule)

root_rule.removeChildAt(0)
layer.setRendererV2(renderer)
layer.triggerRepaint()

On the other hand, to set the same border color for all polygons in a Rule based renderer, you would use setBorderColor() in this way (after initializing symbol variable):

symbol.symbolLayer( 0 ).setBorderColor( QColor(255,0,0) )

---

Just tested it on QGIS v2.14.8. Let me know if it works on your QGIS installation.

Answer 2

The general logic for setting border color is:

# black polygon with red border
symbol = QgsSymbolV2.defaultSymbol(layer.geometryType())
layer_style = {}
layer_style['color'] = '0, 0, 0'
layer_style['size'] = '2.5'
layer_style['color_border'] = '255, 0, 0'
symbol_layer = QgsSimpleFillSymbolLayerV2.create(layer_style)

Answer 3

If you just want the border to go away, you can define the symbol to have no border and that way it will effectively make it so that you don't have to do anything with it.

from:

symbol = QgsSymbolV2.defaultSymbol(layer.geometryType())

to:

symbol = QgsFillSymbolV2.createSimple({'style': 'solid', 'color': '255,0,0,255', 'style_border':'no'})  

This does not change the geometry type of your original layer (tried it).

Hope this helps!