[英]Generic Tree in Recursion in Python
I am quite new in Python and stuck in a generic tree.我对 Python 很陌生,并被困在通用树中。
I have 2 columns.我有 2 列。 Assume names to be A and B. Now the thing is values of column A are parents of the corresponding value in column B. Now the value of column B is again searched in column A. Now it becomes the parent and its corresponding value a child.假设名称为 A 和 B。现在事情是 A 列的值是 B 列中相应值的父级。现在再次在 A 列中搜索 B 列的值。现在它成为父级,其相应值成为子级.
For example:例如:
A B
--------
1 2
5 3
2 5
3
Now here for value '1'.现在这里是值“1”。
1 is parent. 1 是父母。 2 is child of 1. Now 2 is again searched in column A. this now makes 2 a parent and 5 its child. 2 是 1 的孩子。现在再次在 A 列中搜索 2。这现在使 2 成为父母,5 成为孩子。 Now we search for again it gives us the Value 3 .现在我们再次搜索它给了我们值 3 。 Same goes for 3 but there is no value in front of 3 so tree ends here and 3 is the leaf node of tree. 3 也是一样,但是 3 前面没有值,所以树在这里结束,3 是树的叶节点。 Here in example I have displayed one child of each parent but in my dataset each parent can have many child and tree goes until all leaf nodes are reached.在示例中,我显示了每个父节点的一个子节点,但在我的数据集中,每个父节点可以有多个子节点,并且树会一直运行直到到达所有叶节点。 I know recursion is the solution here but how?我知道递归是这里的解决方案,但如何解决?
What I need in output is all possibles sets with the parent.我在输出中需要的是与父级的所有可能集。 In this case [1,2], [1,2,5], [1,2,5,3].在这种情况下 [1,2], [1,2,5], [1,2,5,3]。 Also if 1 had 2 children let's say 2 and 4.此外,如果 1 有 2 个孩子,我们假设 2 和 4。
Then [1,2,4] will also be counted as a set.那么 [1,2,4] 也会被算作一个集合。 Any help here?这里有什么帮助吗? I am struggling really hard!我真的很努力!
Venky = {}
def Venkat(uno):
x = df[df.ColA == uno]
data = list(zip(x['ColB'],x.ColA))
Venky[uno] = Node(uno)
for child,parent in data:
Venky[child] = Node(child, parent=Venky[uno])
print(Venky[child])
y = df[df['ColA'] == (Venky[child]).name]
if (y['ColB'].empty == False):
data = list(zip(y['ColB'],y.ColA,))
#y_unique = y['ColB'].unique()
for k in y['ColB']:
res = Venkat(k)
return res
udo = 'a'
Venkat(udo)
for pre, fill, node in RenderTree(Venky[udo]):
print("%s%s" % (pre, node.name))
Above is my sample code.以上是我的示例代码。 df is my colA and colB dataframe . df 是我的 colA 和 colB 数据框。 ColA is column A and same for colB. ColA 是 A 列,colB 是相同的。 I have also imported Anytree and pandas library for this code.我还为此代码导入了 Anytree 和 pandas 库。 The outbut I am getting is something like this:我得到的结果是这样的:
Node('/1/2')
Node('/4/nan')
1
└── 2
Node('/1/2')
Node('/4/nan')
None
Here 4 is sibling of 2 in the tree ie 1 has 2 child here 2 and 4 unlike dataset.这里 4 是树中 2 的兄弟,即 1 有 2 个孩子,这里 2 和 4 与数据集不同。 But currently I don't care about output.但目前我不关心输出。 I want to construct a tree now.我现在想建一棵树。 Then I will play around with output.然后我会玩弄输出。
I would suggest a solution where you first convert the table structure to a dictionary providing a list of child nodes (dictionary value) for each node (dictionary key).我会建议一个解决方案,您首先将表结构转换为字典,为每个节点(字典键)提供子节点(字典值)列表。
Then perform a depth first walk through that tree, using recursion, extending a path with the newly visited child.然后执行深度优先遍历该树,使用递归,使用新访问的子节点扩展路径。 Output the path at each stage.输出每个阶段的路径。 For this a generator is an ideal solution.为此,发电机是理想的解决方案。
Code:代码:
def all_paths(table, root):
# convert table structure to adjacency list
children = {}
for node, child in table:
if child:
children[node] = children.setdefault(node, []) + [child]
# generator for iteration over all paths from a certain node
def recurse(path):
yield path
if path[-1] in children:
for child in children[path[-1]]: # recursive calls for all child nodes
yield from recurse(path + [child])
return recurse([root])
# Sample data in simple list format
table = [
[1, 2],
[5, 3],
[2, 5],
[2, 6],
[2, 4],
[6, 7],
]
# Output all paths from node 1
for path in all_paths(table, 1):
print(path)
The above outputs:以上输出:
[1]
[1, 2]
[1, 2, 5]
[1, 2, 5, 3]
[1, 2, 6]
[1, 2, 6, 7]
[1, 2, 4]
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.