试图拿出一个递归函数来展开树在Python

Question

我有表，看起来像这样：

id | parentid | name 
--------------------- 
1 | 0  | parent1 
--------------------- 
2 | 0  | parent2 
--------------------- 
3 | 1  | child1 
--------------------- 
4 | 3  | subchild1 

现在我试图拿出一个有效的方式来采取数据库中的数据和创建一个Python字典。

基本上，我希望能够做到：

tree = expand(Session.query(mytable).all()) 
print tree['parent2']['child2'] 

# result would be 'subchild1' 

我在与如何做到这一点完全丧失......我一直在用下面的函数乱搞，但我可以”让它工作。任何帮助，将不胜感激。

def expand(tree): 

    parents = [i for i in tree if i.parentid == 0] 

    for parent in parents: 
     children = expand(parent) 

Answer 1

如果我理解正确，那么其父ID为0的项目是根号还是第一级？

如果是这样，你的方法应该是这样的：

def expand(tree, id): 
    expanded_tree = {} 

    parents = [i for i in tree if i.parentid == id] 

    for parent in parents: 
     expanded_tree[parent.name] = expand(tree, parent.id) 

    return expanded_tree 

，你会开始它像：

tree = expand(Session.query(mytable).all(), 0) 

Answer 2

您的例子不匹配的数据给出，但这个应该是你想。

它不是递归的，因为递归在这里没有意义：输入数据没有递归结构（这就是我们正在创建的），所以你可以写成递归的循环...这是一个漂亮的毫无意义的事情要做的Python。

data = [ (1,0,"parent1"), (2,0,"parent2"), (3,1,"child1"), (4,3,"child2")] 

# first, you have to sort this by the parentid 
# this way the parents are added before their children 
data.sort(key=lambda x: x[1]) 

def make_tree(data): 
    treemap = {} # for each id, give the branch to append to 
    trees = {} 
    for id,parent,name in data: 
     # {} is always a new branch 
     if parent == 0: # roots 
      # root elements are added directly 
      trees[name] = treemap[id] = {} 
     else: 
      # we add to parents by looking them up in `treemap` 
      treemap[parent][name] = treemap[id] = {} 

    return trees 

tree = make_tree(data) 
print tree['parent1']['child1'].keys() ## prints all children: ["child2"]