```
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):
"""Convert an entry a tuple of the indices of the entry's parts.
>>> sequence = testsequence
>>> assert partindices("c/2/x") == (sequence.index("c"), sequence.index("c/2"), sequence.index("c/2/x"))
"""
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[:])
```