Improve pin1 detection

If there are multiple pads that look like common first pad names only
the lexicographically smallest one will be considered first.
If no pads have common first pad name then all pads with
lexicographically smallest name will be considered first (previously
only one random pad with such name was chosen).

Fixes openscopeproject#172

InteractiveHtmlBom/ecad/kicad.py

@@ -236,7 +236,6 @@ def parse_pad(self, pad):
             layers.append("B")
         pos = self.normalize(pad.GetPosition())
         size = self.normalize(pad.GetSize())
-        is_pin1 = pad.GetPadName() in ['1', 'A', 'A1', 'P1', 'PAD1']
         angle = pad.GetOrientation() * -0.1
         shape_lookup = {
             pcbnew.PAD_SHAPE_RECT: "rect",
@@ -261,8 +260,6 @@ def parse_pad(self, pad):
             "angle": angle,
             "shape": shape
         }
-        if is_pin1:
-            pad_dict['pin1'] = 1
         if shape == "custom":
             polygon_set = pad.GetCustomShapeAsPolygon()
             if polygon_set.HasHoles():
@@ -332,14 +329,18 @@ def parse_modules(self, pcb_modules):
                 if pad_dict is not None:
                     pads.append((p.GetPadName(), pad_dict))
 
-            # If no pads have common 'first' pad name pick lexicographically.
-            pin1_pads = [p for p in pads if 'pin1' in p[1]]
-            if pads and not pin1_pads:
+            if pads:
+                # Try to guess first pin name.
                 pads = sorted(pads, key=lambda el: el[0])
+                pin1_pads = [p for p in pads if p[0] in ['1', 'A', 'A1', 'P1', 'PAD1']]
+                if pin1_pads:
+                    pin1_pad_name = pin1_pads[0][0]
+                else:
+                    # No pads have common first pin name, pick lexicographically smallest.
+                    pin1_pad_name = pads[0][0]
                 for pad_name, pad_dict in pads:
-                    if pad_name:
+                    if pad_name == pin1_pad_name:
                         pad_dict['pin1'] = 1
-                        break
 
             pads = [p[1] for p in pads]