python字典生成树状图的实例

目录

python字典生成树状图

python生成树结构

python字典生成树状图 from graphviz import Digraph # 获取所有节点中最多子节点的叶节点 def getMaxLeafs(myTree): numLeaf = len(myTree.keys()) for key, value in myTree.items(): if isinstance(value, dict): sum_numLeaf = getMaxLeafs(value) if sum_numLeaf > numLeaf: numLeaf = sum_numLeaf return numLeaf def plot_model(tree, name): g = Digraph("G", filename=name, format='png', strict=False) first_label = list(tree.keys())[0] g.node("0", first_label) _sub_plot(g, tree, "0") leafs = str(getMaxLeafs(tree) // 10) g.attr(rankdir='LR', ranksep=leafs) g.view() root = "0" def _sub_plot(g, tree, inc): global root first_label = list(tree.keys())[0] ts = tree[first_label] for i in ts.keys(): if isinstance(tree[first_label][i], dict): root = str(int(root) + 1) g.node(root, list(tree[first_label][i].keys())[0]) g.edge(inc, root, str(i)) _sub_plot(g, tree[first_label][i], root) else: root = str(int(root) + 1) g.node(root, tree[first_label][i]) g.edge(inc, root, str(i)) tree = { "tearRate": { "reduced": "no lenses", "normal": { "astigmatic": { "yes": { "prescript": { "myope": "hard", "hyper": { "age": { "young": "hard", "presbyopic": "no lenses", "pre": "no lenses" } } } }, "no": { "age": { "young": "soft", "presbyopic": { "prescript": { "myope": "no lenses", "hyper": "soft" } }, "pre": "soft" } } } } } } plot_model(tree, "tree.gv")

效果如下:

python生成树结构 # 生成树结构 def get_trees(data, key_column='elementId', parent_column='parentId', child_column='children'): """ :param data: 数据列表 :param key_column: 主键字段,默认id :param parent_column: 父ID字段名,父ID默认从0开始 :param child_column: 子列表字典名称 :return: 树结构 """ data_dic = {} for d in data: data_dic[d.get(key_column)] = d # 以自己的权限主键为键,以新构建的字典为值,构造新的字典 data_tree_list = [] # 整个数据大列表 for d_id, d_dic in data_dic.items(): pid = d_dic.get(parent_column) # 取每一个字典中的父id if not pid: # 父id=0,就直接加入数据大列表 data_tree_list.append(d_dic) else: # 父id>0 就加入父id队对应的那个的节点列表 try: # 判断异常代表有子节点,增加子节点列表=[] data_dic[pid][child_column].append(d_dic) except KeyError: data_dic[pid][child_column] = [] data_dic[pid][child_column].append(d_dic) return data_tree_list def recursion(data, l=None): if l is None: l = [] for i in data: if 'children' in i: children=i.pop('children') l.append(i) recursion(children,l) else: l.append(i) return l

以上为个人经验,希望能给大家一个参考,也希望大家多多支持易知道(ezd.cc)。

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