以下是输入内容：
list_child_parent= [
#first value is child, second is parent
(0, 1),
(1, 3),
(8, 7),
(3, 6),
(4, 3),
(5, 3)
]
output_dict = {
6: {3: {1: {0: {}}, 4: {}, 5: {}}}, 7: {8: {}}
}
这段代码可能不太美观，也不一定符合Pythonic风格，但它可以帮助你入门：
#!/usr/bin/env python3
def make_map(list_child_parent):
has_parent = set()
all_items = {}
for child, parent in list_child_parent:
if parent not in all_items:
all_items[parent] = {}
if child not in all_items:
all_items[child] = {}
all_items[parent][child] = all_items[child]
has_parent.add(child)
result = {}
for key, value in all_items.items():
if key not in has_parent:
result[key] = value
return result
if __name__ == '__main__':
list_child_parent = [
#first value is child, second is parent
(0, 1),
(1, 3),
(8, 7),
(3, 6),
(4, 3),
(5, 3)
]
actual = make_map(list_child_parent)
expected = {
6: {
3: {
1: {
0: {}
},
4: {},
5: {}
}
},
7: {
8: {}
}
}
print('OK' if expected == actual else 'FAIL')
这段代码将会把给定格式的树转换成一个树形字典。虽然看起来有点冗长，但它有助于跟踪发生了什么。从性能方面来说，它表现得相当不错。
LIST_CHILD_PARENTS = [
#first value is child, second is parent
(0, 1),
(1, 3),
(8, 7),
(3, 6),
(4, 3),
(5, 3)
]
class Node(object):
def __init__(self, value):
self.value = value
# List of references to Node()'s.
self.child = []
# Reference to parent Node()
self.parent = None
def set_parent(self, parent):
self.parent = parent
def set_child(self, child):
self.child.append(child)
def get_a_root(items):
"""Find a root node from items.
Grab some node and follow the parent pointers to a root.
"""
cur_key = list(items.keys())[0]
while items[cur_key].parent is not None:
cur_key = items[cur_key].parent.value
parent = items[cur_key]
return parent
def extract_tree(items, root):
"""Remove the tree from root in items.
"""
cur_key = root.value
this_node = items[cur_key]
if len(this_node.child) == 0:
items.pop(cur_key)
return
else:
for child in this_node.child:
extract_tree(items, child)
items.pop(cur_key)
def insert_from_root(tree, root):
"""Insert the dictionary items from a tree.
"""
current = root
if len(current.child) == 0:
tree[current.value] = {}
return
else:
table = {}
for child in current.child:
insert_from_root(table, child)
tree[current.value] = table
def build_graphs():
"""Map all input graphs into Node(object)'s.
Return: A hash table by value: Node(value, child, parent)
"""
items = {}
for child, parent in LIST_CHILD_PARENTS:
if not child in items:
c_n = Node(child)
items[child] = c_n
else:
c_n = items[child]
if not parent in items:
p_n = Node(parent)
items[parent] = p_n
else:
p_n = items[parent]
p_n.set_child(c_n)
c_n.set_parent(p_n)
return items
def run_dict_builder():
"""Map the graphs from input and map into a dict.
Sequence:
1- Map all graphs from input trees: list(tuple)
2- For each root node:
2A - Get a root node.
2B - Extract tree under this root from graphs list.
2C - Insert the tree from this root into dict.
3- Return the Dictionary Tree structure.
"""
graphs = build_graphs()
h_table = {}
while len(graphs) > 0:
root = get_a_root(graphs)
extract_tree(graphs, root)
insert_from_root(h_table, root)
return h_table
print(run_dict_builder())
AttributeError: 'dict' object has no attribute 'has_key' - Dan Safeeif parent in items: 来代替 if key in dict:。 - Mattparents = [None, 1, 2, 3, 3, 2, 6, 6, 1, 9, 10, 10, 9, 13, 13]
children = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15]
# {1: {2: {3: {4: {}, 5: {}}, 6: {7: {}, 8: {}}}, 9: {10: {11: {}, 12: {}}, 13: {14: {}, 15: {}}}}}
def create_tree(node_map, root=None):
""" Given a list of tuples (child, parent) return the nested dictionary representation.
"""
def traverse(parent, node_map, seen):
children = {}
for edge in node_map:
if edge[1] == parent and edge[0] not in seen:
seen.add(edge[0])
children[edge[0]] = traverse(edge[0], node_map, seen)
return children
return traverse(root, node_map, {root})
排除根节点的示例。
我们希望将根节点设置为“a”，但由于根节点应被指定为没有父节点（None），因此被排除在外。
parents = ["a", "b", "c", "a", "b", "e", "e"]
children = ["b", "c", "d", "e", "f", "g", "h"]
edges = list(zip(children, parents))
test_tree = create_tree(edges, "a")
print(test_tree)
# result {'b': {'c': {'d': {}}, 'f': {}}, 'e': {'g': {}, 'h': {}}}
将根节点添加到没有父节点的设计中，将该节点指定为根节点。
parents = [None, "a", "b", "c", "a", "b", "e", "e"]
children = ["a", "b", "c", "d", "e", "f", "g", "h"]
edges = list(zip(children, parents))
test_tree = create_tree(edges, None)
print(test_tree)
# result: {'a': {'b': {'c': {'d': {}}, 'f': {}}, 'e': {'g': {}, 'h': {}}}}
多个树可以通过有多个根节点来表示。
parents = [None, "a", "b", "c", None, "b", "e", "e"]
children = ["a", "b", "c", "d", "e", "f", "g", "h"]
edges = list(zip(children, parents))
test_tree = create_tree(edges, None)
print(test_tree)
# result: {'a': {'b': {'c': {'d': {}}, 'f': {}}}, 'e': {'g': {}, 'h': {}}}
edges = [
#first value is child, second is parent
(0, 1),
(1, 3),
(8, 7),
(3, 6),
(4, 3),
(5, 3)
]
roots = {edge[1] for edge in edges} - {edge[0] for edge in edges}
edges += [(root, None) for root in roots]
test_tree = create_tree(edges, None)
print(test_tree)
# result: {6: {3: {1: {0: {}}, 4: {}, 5: {}}}, 7: {8: {}}}
{6: {3: {1: {0: {}}, 4: {}, 5: {}}}, 7: {8: {}}}吗？你在元组中展示了3有子节点1、5、4，但输出结果似乎让人以为1有子节点5、4、0。 - Matt