John Hunter wrote:
 > Thanks for the suggestion -- I did this automagically with

# add british equivs
for k, v in cnames.items():
   if k.find('gray')>=0:
       k = k.replace('gray', 'grey')
       cnames[k] = v

Just noticed that 'lightgrey' is still in the cnames dict, which means that 'lightgray' is an invalid name. Here's the patch.

Cheers,

Martin
Index: lib/matplotlib/colors.py
===================================================================
--- lib/matplotlib/colors.py    (revision 3013)
+++ lib/matplotlib/colors.py    (working copy)
@@ -109,7 +109,7 @@
     'lightcyan'            : '#E0FFFF',
     'lightgoldenrodyellow' : '#FAFAD2',
     'lightgreen'           : '#90EE90',
-    'lightgrey'            : '#D3D3D3',
+    'lightgray'            : '#D3D3D3',
     'lightpink'            : '#FFB6C1',
     'lightsalmon'          : '#FFA07A',
     'lightseagreen'        : '#20B2AA',
@@ -188,7 +188,7 @@
     if k.find('gray')>=0:
         k = k.replace('gray', 'grey')
         cnames[k] = v
-        
+
 def looks_like_color(c):
     warnings.warn('Use is_color_like instead!', DeprecationWarning)
     if is_string_like(c):
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