Having exported from FontLab to Glyphs I’ve found that many outlines have a duplicate node one on top of another (start/end?). Is there a script to remove these across the whole font?
Are they marked in red? Simply tidy up the paths. There is a function of the same name in the Paths menu.
Tried ‘Tidy up paths’ but that also changes various curved lines to straight lines so breaking the interpolation.
Is there a version of ‘Tidy up paths’, perhaps as ‘Tidy up nodes’ or ‘Tidy up red nodes’?
- Select the red (duplicate) nodes, as many as you like.
- Right click and choose Reconnect Nodes from the context menu.
OK. So glyph by glyph – can’t do this globally in one hit via Font View.
I’ll put the coffee on . . .
A little Python always helps. Put this in Macro Window:
Glyphs.clearLog()
print "Processing %i glyphs..." % len(Font.selectedLayers)
for thisLayer in Font.selectedLayers:
duplicateCounter = 0
for thisPath in thisLayer.paths:
for i in range(len(thisPath.nodes),-1,-1):
thisNode = thisPath.nodes[i]
previousNode = thisNode.prevNode
if thisNode.type == LINE and previousNode.position == thisNode.position:
del thisPath.nodes[i]
duplicateCounter += 1
if duplicateCounter:
print "%s: deleted %s duplicate nodes." % (thisLayer.parent.name, duplicateCounter)
print "Done."
Select glyphs in Font view, press the Run button of Macro Window. Report in Macro window.
2 Likes
Great 
Thanks.
I’ll still brew coffee though 
1 Like
This goes through all layers of the selected glyphs.
Glyphs.clearLog()
print "Processing %i glyphs..." % len(Font.selectedLayers)
for layer in Font.selectedLayers:
glyph = layer.parent
for thisLayer in glyph.layers:
print thisLayer
duplicateCounter = 0
for thisPath in thisLayer.paths:
for i in range(len(thisPath.nodes),-1,-1):
thisNode = thisPath.nodes[i]
previousNode = thisNode.prevNode
if thisNode.type == LINE and previousNode.position == thisNode.position:
del thisPath.nodes[i]
duplicateCounter += 1
if duplicateCounter:
print "%s: deleted %s duplicate nodes." % (thisLayer.parent.name, duplicateCounter)
print "Done."
Great Thom!
Hold on. This script removes all double points. But sometimes a double point is needed in one master, because in an other master these points are not on top of each other. So lets rewrite this code a bit more…
This should keep it all interpolatable.
Only remove double nodes if these nodes are double in all layers 
#MenuTitle: Remove Double Nodes from Export FL
"""
https://forum.glyphsapp.com/t/removing-a-node-on-top-of-a-node/11323/7
"""
Glyphs.clearLog()
from AppKit import NSPoint
def getPointIndex(layer,thisnode):
for path in layer.paths:
for node in path.nodes:
if thisnode == node:
return (layer.paths.index(path), node.index)
def getPointPos(layer, pathIndex_nodeIndex):
pathIndex, nodeIndex = pathIndex_nodeIndex
p_i=-1
for path in layer.paths:
p_i += 1
n_i = -1
for node in path.nodes:
n_i += 1
if p_i == pathIndex and n_i == nodeIndex:
#print "posNode:",node
return [node.x,node.y]
print "Processing %i glyphs..." % len(Font.selectedLayers)
for thisLayer in Font.selectedLayers:
duplicateCounter = 0
pathIndex = -1
for thisPath in thisLayer.paths:
pathIndex+=1
nodeIndex = -1
for i in range(len(thisPath.nodes),-1,-1):
nodeIndex +=1
thisNode = thisPath.nodes[i]
previousNode = thisNode.prevNode
if thisNode.type == LINE and previousNode.position == thisNode.position:
# check double in all layers
checkPointIndex = getPointIndex(thisLayer,thisNode)
checkPrevPointIndex = getPointIndex(thisLayer,previousNode)
glyph = thisLayer.parent
allDouble = True
for layer in glyph.layers:
if getPointPos(layer, checkPointIndex) == getPointPos(layer, checkPrevPointIndex):
continue
else:
allDouble = False
if allDouble:
print glyph.name
for layer in glyph.layers:
del layer.paths[checkPointIndex[0]].nodes[checkPointIndex[1]]
print "Done."
Good point!
I dont know/understand how to do this!! please someone help