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grep: give different labels to + and - symbols...
grep: give different labels to + and - symbols I find it more useful to give different colours to plus and minus, but it's difficult to do so if the default output uses the same label for both. The following augments the names of columns with some extra labels, akin to the diff.inserted and diff.deleted labels for the diff command. This is done by adding an extra label field to the columns tuples.

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ro.py
67 lines | 2.0 KiB | text/x-python | PythonLexer
##############################################################################
#
# Copyright (c) 2003 Zope Foundation and Contributors.
# All Rights Reserved.
#
# This software is subject to the provisions of the Zope Public License,
# Version 2.1 (ZPL). A copy of the ZPL should accompany this distribution.
# THIS SOFTWARE IS PROVIDED "AS IS" AND ANY AND ALL EXPRESS OR IMPLIED
# WARRANTIES ARE DISCLAIMED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
# WARRANTIES OF TITLE, MERCHANTABILITY, AGAINST INFRINGEMENT, AND FITNESS
# FOR A PARTICULAR PURPOSE.
#
##############################################################################
"""Compute a resolution order for an object and its bases
"""
from __future__ import absolute_import
__docformat__ = 'restructuredtext'
def _mergeOrderings(orderings):
"""Merge multiple orderings so that within-ordering order is preserved
Orderings are constrained in such a way that if an object appears
in two or more orderings, then the suffix that begins with the
object must be in both orderings.
For example:
>>> _mergeOrderings([
... ['x', 'y', 'z'],
... ['q', 'z'],
... [1, 3, 5],
... ['z']
... ])
['x', 'y', 'q', 1, 3, 5, 'z']
"""
seen = {}
result = []
for ordering in reversed(orderings):
for o in reversed(ordering):
if o not in seen:
seen[o] = 1
result.insert(0, o)
return result
def _flatten(ob):
result = [ob]
i = 0
for ob in iter(result):
i += 1
# The recursive calls can be avoided by inserting the base classes
# into the dynamically growing list directly after the currently
# considered object; the iterator makes sure this will keep working
# in the future, since it cannot rely on the length of the list
# by definition.
result[i:i] = ob.__bases__
return result
def ro(object):
"""Compute a "resolution order" for an object
"""
return _mergeOrderings([_flatten(object)])