这会将输出转换为嵌套的字典形式.如果我也可以找到该路径，我会及时通知您.也许这是有帮助的.

This converts the output to the nested dictionary form. I will keep you updated if I can find the path as well. Maybe this, is helpful.

list_of_tuples = [('ROOT','ROOT', 'shot'),('nsubj','shot', 'I'),('det','elephant', 'an'),('dobj','shot', 'elephant'),('case','sleep', 'in'),('nmod:poss','sleep', 'my'),('nmod','shot', 'sleep')]

nodes={}

for i in list_of_tuples:
    rel,parent,child=i
    nodes[child]={'Name':child,'Relationship':rel}

forest=[]

for i in list_of_tuples:
    rel,parent,child=i
    node=nodes[child]

    if parent=='ROOT':# this should be the Root Node
            forest.append(node)
    else:
        parent=nodes[parent]
        if not 'children' in parent:
            parent['children']=[]
        children=parent['children']
        children.append(node)

print forest

输出为嵌套字典，

[{'Name': 'shot', 'Relationship': 'ROOT', 'children': [{'Name': 'I', 'Relationship': 'nsubj'}, {'Name': 'elephant', 'Relationship': 'dobj', 'children': [{'Name': 'an', 'Relationship': 'det'}]}, {'Name': 'sleep', 'Relationship': 'nmod', 'children': [{'Name': 'in', 'Relationship': 'case'}, {'Name': 'my', 'Relationship': 'nmod:poss'}]}]}]

[{'Name': 'shot', 'Relationship': 'ROOT', 'children': [{'Name': 'I', 'Relationship': 'nsubj'}, {'Name': 'elephant', 'Relationship': 'dobj', 'children': [{'Name': 'an', 'Relationship': 'det'}]}, {'Name': 'sleep', 'Relationship': 'nmod', 'children': [{'Name': 'in', 'Relationship': 'case'}, {'Name': 'my', 'Relationship': 'nmod:poss'}]}]}]

以下功能可以帮助您找到从根到叶的路径:

The following function can help you to find the root-to-leaf path:

def recurse_category(categories,to_find):
    for category in categories: 
        if category['Name'] == to_find:
            return True, [category['Relationship']]
        if 'children' in category:
            found, path = recurse_category(category['children'], to_find)
            if found:
                return True, [category['Relationship']] + path
    return False, []