通过数组元素的父子关系形成一个字典类型的树列表

Posted 2021-01-15 恍

# coding=utf-8
‘‘‘
通过父子key关系形成一个字典类型的列表
‘‘‘
deparetment = [{‘id‘: ‘03F50D0FE48A4C77896411D007C72A5B‘, ‘name‘: ‘2级部门‘, ‘p_id‘: ‘8F0095616A17484DAD2C17925C04B78E‘},
               {‘id‘: ‘07A592FE111E4B338AC967531AB99A7E‘, ‘name‘: ‘4级部门‘, ‘p_id‘: ‘A54DF3E5EBC14254ACF65BC20C2BF4B6‘},
               {‘id‘: ‘1F211E1C1D504EE0A1224F0757C67E8F‘, ‘name‘: ‘2级部门2‘, ‘p_id‘: ‘4D9190D6132441D19F9DF4DA19948AFD‘},
               {‘id‘: ‘3E449618073044819C1C729A7447DE13‘, ‘name‘: ‘5级部门‘, ‘p_id‘: ‘07A592FE111E4B338AC967531AB99A7E‘},
               {‘id‘: ‘4D9190D6132441D19F9DF4DA19948AFD‘, ‘name‘: ‘3333‘, ‘p_id‘: ‘-1‘},
               {‘id‘: ‘8F0095616A17484DAD2C17925C04B78E‘, ‘name‘: ‘111‘, ‘p_id‘: ‘-1‘},
               {‘id‘: ‘A54DF3E5EBC14254ACF65BC20C2BF4B6‘, ‘name‘: ‘3级部门‘, ‘p_id‘: ‘03F50D0FE48A4C77896411D007C72A5B‘},
               {‘id‘: ‘E11306E5EABB477C9098886CF7F63731‘, ‘name‘: ‘2222‘, ‘p_id‘: ‘-1‘}]


# 获取部门树
def get_all_department_tree(dp):
    top = []
     # 找出顶级部门，顶级部门p_id=-1
    for d in dp:
        info = {}
        if d[‘p_id‘] == ‘-1‘:
            info["value"] = d["id"]
            info["label"] = d["name"]
            top.append(info)
   
    #从每个顶级部门递归查找子部门
    for c in top:
        set_children(c, dp)
    return top

#递归查找部门
def set_children(c, dp):
    c["children"] = []
    n = 0
    for d in dp:
        info = {}
        if c["value"] == d["p_id"]:
            n = n + 1
            info["value"] = d["id"]
            info["label"] = d["name"]
            c["children"].append(info)
    if n:
        for cc in c["children"]:
            return set_children(cc, dp)  # 递归在下级继续查找
    else:
        c.pop(‘children‘)


print get_all_department_tree(deparetment)