On Sat, 14 Apr 2007 11:37:11 -0300, Sebastian Bassi <[EMAIL PROTECTED]> wrote:
> I have a two column list like:
>
> 2,131
> 6,335
> 7,6
> 8,9
> 10,131
> 131,99
> 5,10
>
> And I want to store it in a tree-like structure.
CS side note: your "columns" describe a graph;
http://en.wikipedia.org/wiki/Gr
On 15 Apr 2007 15:44:47 -0700, [EMAIL PROTECTED]
<[EMAIL PROTECTED]> wrote:
> But errors and bugs do happen, inside data too; so often it's better
> to be on safe side if the safe code is fast enough.
Yes, I agree since I've seen lot of errors in data. But this data
comes from a taxonomy tree made
[EMAIL PROTECTED] writes:
> > I guess this should make the program enter into a endless loop. But
> > the data won't have such a redundancy, because it was taken from a
> > philogenetic tree.
>
> But errors and bugs do happen, inside data too; so often it's better
> to be on safe side if the safe
Sebastian Bassi:
> I guess this should make the program enter into a endless loop. But
> the data won't have such a redundancy, because it was taken from a
> philogenetic tree.
But errors and bugs do happen, inside data too; so often it's better
to be on safe side if the safe code is fast enough.
On 4/15/07, Peter Otten <[EMAIL PROTECTED]> wrote:
> Depending on your input data you may need to add some cycle detection.
> For example, try it with
> tree_path(1, {1:[2], 2:[1]}, [])
I guess this should make the program enter into a endless loop. But
the data won't have such a redundancy, becau
Sebastian Bassi wrote:
> On 14 Apr 2007 09:32:07 -0700, [EMAIL PROTECTED] <[EMAIL PROTECTED]>
wrote:
>
> > def tree_path(key,tree,indent):
> > print '\t'*indent,key
> > if tree.has_key(key):
> > for m in tree[key]:
> > tree_path(m,tree,indent+1)
> > return
>
> Tha
On 14 Apr 2007 09:32:07 -0700, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> def tree_path(key,tree,indent):
> print '\t'*indent,key
> if tree.has_key(key):
> for m in tree[key]:
> tree_path(m,tree,indent+1)
> return
Thank you. It worked!.
I changed it a bit to re
The solution I have found is quite similar to the [EMAIL PROTECTED]
one (it uses a defaultdict, and it yields the result lazily), but it
lacks the quite useful max_depth (it can be added):
from collections import defaultdict
def finder(el, stor):
if el in stor:
for v in stor[el]:
Hope this helps
# list of pairs [child,parent]
list=[[2,131],[6,335],[7,6],[8,9],[10,131],[131,99],[5,10]]
# list with loop
#list=[[2,131],[6,335],[7,6],[8,9],[10,131],[131,99],[5,10],[3,10],
[131,3]]
# put the pairs in a dictionary, for easy retrieval
d={}
for c,p in list:
# must be abl
On Apr 14, 9:37�am, "Sebastian Bassi" <[EMAIL PROTECTED]>
wrote:
> I have a two column list like:
>
> 2,131
> 6,335
> 7,6
> 8,9
> 10,131
> 131,99
> 5,10
>
> And I want to store it in a tree-like structure.
> So if I request 131, it should return all the child of 131, like 2, 10
> and 5 (since 5 is
I have a two column list like:
2,131
6,335
7,6
8,9
10,131
131,99
5,10
And I want to store it in a tree-like structure.
So if I request 131, it should return all the child of 131, like 2, 10
and 5 (since 5 is child of 10).
If I request 335, it should return: 6 and 7.
If I request 9, it should retu
11 matches
Mail list logo
