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authorchrysn <chrysn@web>2012-03-15 21:36:36 -0400
committeradmin <admin@branchable.com>2012-03-15 21:36:36 -0400
commit128234cde1ed0504d6beb0b2d10544feb2d77809 (patch)
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parent3ca2d6f45900edf36162c0a1269654095cae7914 (diff)
downloadikiwiki-128234cde1ed0504d6beb0b2d10544feb2d77809.tar
ikiwiki-128234cde1ed0504d6beb0b2d10544feb2d77809.tar.gz
the algorithms required for map plugin sorting
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+testdata = "c/3 a b d b/1 c/1 c/2/x c/2 c".split(" ")
+
+def strategy_byearlychild(sequence):
+ """Sort by earliest child
+
+ When this strategy is used, a parent is displayed with all its children as
+ soon as the first child is supposed to be shown.
+
+ >>> strategy_byearlychild(testdata)
+ ['c', 'c/3', 'c/1', 'c/2', 'c/2/x', 'a', 'b', 'b/1', 'd']
+ """
+
+ # first step: pull parents to top
+ def firstchildindex(item):
+ childindices = [i for (i,text) in enumerate(sequence) if text.startswith(item + "/")]
+ # distinction required as min(foo, *[]) tries to iterate over foo
+ if childindices:
+ return min(sequence.index(item), *childindices)
+ else:
+ return sequence.index(item)
+ sequence = sorted(sequence, key=firstchildindex)
+
+ # second step: pull other children to the start too
+ return strategy_byparents(sequence)
+
+def strategy_byparents(sequence):
+ """Sort by parents only
+
+ With this strategy, children are sorted *under* their parents regardless of
+ their own position, and the parents' positions are determined only by
+ comparing the parents themselves.
+
+ >>> strategy_byparents(testdata)
+ ['a', 'b', 'b/1', 'd', 'c', 'c/3', 'c/1', 'c/2', 'c/2/x']
+ """
+
+ def partindices(item):
+ return tuple(sequence.index(item.rsplit('/', i)[0]) for i in range(item.count('/'), -1, -1))
+
+ return sorted(sequence, key=partindices)
+
+def strategy_forcedsequence(sequence):
+ """Forced Sequence Mode
+
+ Using this strategy, all entries will be shown in the sequence; this can
+ cause parents to show up multiple times.
+
+ The only reason why this is not the identical function is that parents that
+ are sorted between their children are bubbled up to the top of their
+ contiguous children to avoid being repeated in the output.
+
+ >>> strategy_forcedsequence(testdata)
+ ['c/3', 'a', 'b', 'd', 'b/1', 'c', 'c/1', 'c/2', 'c/2/x']
+ """
+
+ # this is a classical bubblesort. other algorithms wouldn't work because
+ # they'd compare non-adjacent entries and move the parents before remote
+ # children. python's timsort seems to work too...
+
+ for i in range(len(sequence), 1, -1):
+ for j in range(1, i):
+ if sequence[j-1].startswith(sequence[j] + '/'):
+ sequence[j-1:j+1] = [sequence[j], sequence[j-1]]
+
+ return sequence
+
+def strategy_forcedsequence_timsort(sequence):
+ sequence.sort(lambda x,y: -1 if y.startswith(x) else 1)
+ return sequence
+
+if __name__ == "__main__":
+ import doctest
+ doctest.testmod()
+
+ import itertools
+
+ for perm in itertools.permutations(testdata):
+ if strategy_forcedsequence(testdata[:]) != strategy_forcedsequence_timsort(testdata[:]):
+ print "difference for testdata", testdata
+ print "normal", strategy_forcedsequence(testdata[:])
+ print "timsort", strategy_forcedsequence_timsort(testdata[:])