Diez B. Roggisch wrote:
Adam Powell schrieb:
Hi!
I have seemingly simple problem, which no doubt someone out there has
already solved, but I'm stumped. Imagine I have a dictionary like
below, in which the keys are parent nodes and the values are a list of
children nodes.
dict = { 0: [1, 2], 1: [3], 2: [6], 3: [4,7], 4: [5,8], 8: [9] }
Is there an obvious way to produce a nested tree format for this
dictionary, like this:
[ 0 [ 1 [ 3 [ 4 [ 5 , 8 [ 9 ] ] , 7 ] ] , 2 [ 6 ] ]
Thanks for any help,
d = { 0: [1, 2], 1: [3], 2: [6], 3: [4,7], 4: [5,8], 8: [9] }
nodelists = dict((node ,[node, []]) for node in d.keys())
for node, children in d.iteritems():
for child in children:
nodelists[node][1].extend(nodelists.setdefault(child, [child]))
print nodelists[0]
Two remarks:
- don't use "dict" as name for a dictionary - look at the above code
*why* not...
- the code assumes that 0 is the root. if that wouldn't be the case,
you need to search for the root node. I've got an idea - you to?
Diez
Not quite. That gets you
[0, [1, [3, [4, [5, 8, [9]], 7]], 2, [6]]]
which probably isn't what you want. Note that the children of 0 are
1, [3, [4, [5, 8, [9]], 7]],
2,
[6]
which probably isn't what was desired.
You probably want
[0, [1, [3, [4, [5], [8, [9]]], [7]]], [2, [6]]]
so that each list is [node, [children]].
The original poster wanted
[ 0 [ 1 [ 3 [ 4 [ 5 , 8 [ 9 ] ] , 7 ] ] , 2 [ 6 ] ]
but that's not a meaningful Python list expression.
Try this recursive form:
d = { 0: [1, 2], 1: [3], 2: [6], 3: [4,7], 4: [5,8], 8: [9] }
def getsubtree(d, node) :
if d.has_key(node) :
return([node] + [getsubtree(d,child) for child in d[node]])
else :
return([node])
getsubtree(d,min(d.keys())) # smallest node is assumed to be the root
John Nagle
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