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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