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