On Wed, Jan 25, 2012 at 6:12 AM, Edward K. Ream <[email protected]> wrote:
> We compute the result, merged_d, as follows::
Actually, d.update() updates d in place: it does not return a value.
Thus, the code must be::
inverted_old_d = self.invert(old_d)
inverted_new_d = self.invert(new_d)
inverted_old_d.update(inverted_new_d)
result = self.uninvert(inverted_old_d)
Here are the invert and uninvert methods, devoid of traces, asserts, etc.::
def invert (self,d):
'''Invert a shortcut dict whose keys are command names,
returning a dict whose keys are strokes.'''
result = {}
for commandName in d.keys():
if commandName == '_hash':
result['_hash'] = 'inverted %s' % d.get('_hash')
else:
for si in d.get(commandName,[]):
stroke = si.val # Already canonicalized.
si.commandName = commandName # Add for uninvert
aList = result.get(stroke,[])
if si not in aList:
aList.append(si)
result[stroke] = aList
return result
def uninvert (self,d):
'''Invert a shortcut dict whose keys are strokes,
returning a dict whose keys are command names.'''
result = {}
for stroke in d.keys():
if stroke == '_hash':
result['_hash'] = 'uninverted %s' % d.get('_hash')
else:
for si in d.get(stroke,[]):
commandName = si.commandName
aList = result.get(commandName,[])
if si not in aList:
aList.append(si)
result[commandName] = aList
return result
Before releasing the new code to the trunk a new set of unit tests
must pass. In the meantime, this post is my backup :-)
EKR
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