Adds style to node serialization.

[jmilloy]

The Style serialization itself is updated so that the definition will be empty for the default style, which allows the node definition to omit the style when it is not necessary.

>>> style = Style()
>>> style.definition
OrderedDict()

>>> style=Style(name="test")
>>> style.definition
OrderedDict([('name', 'test')])

>>> node = Node()
>>> node.definition
OrderedDict([('Node', OrderedDict([('node', 'core.node.Node')]))])

>>> node = Node(style=Style(name="test"))
OrderedDict([('Node', OrderedDict([('node', 'core.node.Node'), ('style', OrderedDict([('name', 'test')]))]))])

In order to do that, I modified the Style cmap handling, so that colormap is string trait (that can be None) and cmap is a property that either uses colormap, enumeration_colors, or a default:

>>> style = Style()
>>> style.colormap # None
>>> style.cmap
<matplotlib.colors.ListedColormap object at 0x7ff855b7c9e8>
>>> style.cmap.name
'viridis'

>>> style = Style(colormap="hot")
>>> style.colormap
'hot'
>>> style.cmap
<matplotlib.colors.LinearSegmentedColormap object at 0x7ff855aa54a8>
>>> style.cmap.name
'hot`

>>> style = Style(enumeration_colors=('c', 'k'))
>>> style.colormap # None
>>> style.cmap
<matplotlib.colors.ListedColormap object at 0x7ff876eb8908>
>>> style.cmap.name
'from_list'

>>> style = Style(colormap="hot", enumeration_colors=('c', 'k'))
...
TypeError: Style can have a colormap or enumeration_colors, but not both

[coveralls]

Coverage decreased (-0.01%) to 90.619% when pulling 9cea677 on feature/node-serialization into 279eab9 on develop.

[coveralls]

Coverage decreased (-0.01%) to 90.619% when pulling 9cea677 on feature/node-serialization into 279eab9 on develop.