从2列列表中创建一棵树 [英] Making a tree out of a 2 column list
问题描述
我有两个列列表如下:
2,131
6,335
7,6
8,9
10,131
131,99
5,10
我我希望将它存储在树状结构中。
因此,如果我要求131,它应该返回131的所有子项,如2,10,
和5(从那以后) 5是10)的孩子。
如果我要求335,它应该返回:6和7.
如果我要求9,它应该返回8.
我想我可以使用元组或词典来实现它,但我无法想象。
最好,
SB。< br $>
-
$ b $bSebastiánBassi
Diplomado CienciayTecnología。
Club delarazón( www.clubdelarazon.org )
< blockquote> 4月14日9:37 ??? am,Sebastian Bassi < sba ... @ clubdelarazon.org>
写道:
我有两列列表,如:
2,131
6,335
7,6
8,9
10,131
131,99
5,10
我想把它存放在树状结构中。
因此,如果我要求131,它应该返回131的所有孩子,例如2,10
和5(因为5是10岁的孩子)。
如果我要求335,它应该返回:6和7.
如果我要求9,它应该返回8.
我想我可以使用元组或字典来做到这一点,但我不知道怎么做。
可能有更好的方法。
def tree_path(密钥,树,缩进):
打印''\t''*缩进,键
如果tree.has_key(键):
for m in tree [key]:
tree_path(m,树,缩进+ 1)
返回
打印''原始数据''
打印
来源= [(2,131),(6,335),(7,6),(8,9),(10,131),(131,99),(5,1) 0)]
tree = {}
for s in source:
if tree.has_key(s [1]) :
tree [s [1]]。append(s [0])
else:
tree [s [1]] = [s [0]]
树中的
:
打印''%3d''%(t),树[t]
打印
打印
tree_path(99,树,0)
print
打印''扩展数据''
打印
source_extend = [(666,2),(777,2),( 888,2)]
对于source_extend中的s
:
if tree.has_key(s [1]):
tree [S [1]]。追加(s [0])
否则:
树[s [1]] = [s [0]]
for tree:
打印''%3d''%(t),树[t]
打印
打印
tree_path(99,树,0)
##原始数据
##
## 99 [131]
## 6 [7]
## 9 [8]
## 10 [5]
## 335 [6]
## 131 [2,10]
##
##
## 99
## 131
## 2
## 10
## 5
##
##扩展数据
##
## 2 [666 ,777,888]
## 99 [131]
## 6 [7]
## 9 [8] >
## 10 [5]
## 335 [6]
## 131 [2,10]
# #
##
## 99
## 131
## 2
## 666
## 777
## 888
## 10
## 5
>
最好,
SB。
-
$ b $bSebastiánBassi
Diplomado CienciayTecnología。
Club delarazón( www.clubdelarazon.org )
希望这有帮助
#对子列表[孩子,父母]
list = [[2,131],[6,335],[7,6],[8,9],[10,131],[131,99], [5,10]]
#循环列表
#list = [[2,131],[6,335],[7,6],[8 ,9],[10,131],[131,99],[5,10],[3,10],
[131,3]]
#将这些对放入字典中,以便于检索
d = {}
for c,p in list:
#必须是能够为每个键保持多个值
如果不是(d.has_key(p)):
d [p] = [c]
否则:
d [p] .append(c)
#函数用于使用递归检索子项。 max_depth确保
终止
def retrieve_recursive(key,result = [],max_depth = 10):
如果d.has_key(键)
和max_depth> 0:
我在d [key]中的
:
result.append(i)
retrieve_recursive(i,result,max_depth -1)
返回结果
print retrieve_recursive(131)
我发现的解决方案与 ma********@gmail.com
一个(它使用一个defaultdict,它会懒散地产生结果),但是它没有相当有用的max_depth(可以添加)$
>
来自收藏品导入defaultdict
def finder(el,stor):
如果el存储:
for v in stor [el]:
yield v
for v2 in finder(v,stor):
yield v2
data = [[131,2],[335,6],[6,7],[9,8],[131,10],[99,131],[10,
5]]
存储=默认dict(list)
for k,v in data:
storage [k] .append(v)
test_data = ([131,[2,10,5]],
[335,[6,7]],
[9,[8]],>
)
打印存储
for k,vals in test_data:
断言集(finder(k,存储))==设置(vals)
再见,
bearophile
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 return 8.
I guess I could use tuples or dictionaries to do it, but I can''t figure outhow.
Best,
SB.
--
Sebastián Bassi
Diplomado Ciencia y Tecnología.
Club de la razón (www.clubdelarazon.org)
On Apr 14, 9:37???am, "Sebastian Bassi" <sba...@clubdelarazon.org>
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 child of 10).
If I request 335, it should return: 6 and 7.
If I request 9, it should return 8.
I guess I could use tuples or dictionaries to do it, but I can''t figure out how.There are probably better ways.
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
print ''original data''
source = [(2,131),(6,335),(7,6),(8,9),(10,131),(131,99),(5,1 0)]
tree = {}
for s in source:
if tree.has_key(s[1]):
tree[s[1]].append(s[0])
else:
tree[s[1]] = [s[0]]
for t in tree:
print ''%3d '' % (t),tree[t]
tree_path(99,tree,0)
print ''extended data''
source_extend = [(666,2),(777,2),(888,2)]
for s in source_extend:
if tree.has_key(s[1]):
tree[s[1]].append(s[0])
else:
tree[s[1]] = [s[0]]
for t in tree:
print ''%3d '' % (t),tree[t]
tree_path(99,tree,0)
## original data
##
## 99 [131]
## 6 [7]
## 9 [8]
## 10 [5]
## 335 [6]
## 131 [2, 10]
##
##
## 99
## 131
## 2
## 10
## 5
##
## extended data
##
## 2 [666, 777, 888]
## 99 [131]
## 6 [7]
## 9 [8]
## 10 [5]
## 335 [6]
## 131 [2, 10]
##
##
## 99
## 131
## 2
## 666
## 777
## 888
## 10
## 5
>
Best,
SB.
--
Sebastián Bassi
Diplomado Ciencia y Tecnología.
Club de la razón (www.clubdelarazon.org)
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 able to hold multiple values for each key
if not(d.has_key(p)):
d[p]=[c]
else:
d[p].append(c)
# function to retrieve children using recursion. max_depth to ensure
termination
def retrieve_recursive(key,result=[],max_depth=10):
if d.has_key(key) and max_depth>0:
for i in d[key]:
result.append(i)
retrieve_recursive(i,result,max_depth-1)
return result
print retrieve_recursive(131)
The solution I have found is quite similar to the ma********@gmail.com
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]:
yield v
for v2 in finder(v, stor):
yield v2
data = [[131, 2], [335, 6], [6, 7], [9, 8], [131, 10], [99, 131], [10,
5]]
storage = defaultdict(list)
for k, v in data:
storage[k].append(v)
test_data = ([131, [2, 10, 5]],
[335, [6, 7]],
[9, [8]],
)
print storage
for k, vals in test_data:
assert set(finder(k, storage)) == set(vals)
Bye,
bearophile
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