将扁平的数据转为树形结构 层级部门人数统计 递归

 

def get_tree_ids(tree: dict, keyword='id', child_name="children") -> list:
    '''
    求树的全部节点的id
    方法：1、递归；2、数据库遍历parent_id查询
    '''
    ids = [tree[keyword]]
    children = tree.get(child_name)
    if children:
        for child in children:
            ids += get_tree_ids(child)
    return ids

def find_sub_tree(tree: dict, keyword_val, keyword='id', child_name="children") -> dict | None:
    '''
    求子树
    '''
    if tree[keyword] == keyword_val:
        return tree
    children = tree.get(child_name)
    ret = None
    if children:
        for child in children:
            ret = find_sub_tree(child, keyword_val)
            if ret:
                break
    return ret

case = {'id': 75, 'name': '测试用例', 'parent_id': None,
        'person_num': 22, 'children': [
        {'id': 76, 'name': '广告发行中心', 'parent_id': 75,
         'person_num': 14, 'children': [
            {'id': 77, 'name': '广告商务组', 'parent_id': 76, 'person_num': 0},
            {'id': 78, 'name': '广告投放组', 'parent_id': 76, 'person_num': 5},
            {'id': 79, 'name': '广告设计组', 'parent_id': 76, 'person_num': 7},
            {'id': 80, 'name': '发行运营组', 'parent_id': 76,
             'person_num': 1}]},
        {'id': 81, 'name': '技术部', 'parent_id': 75, 'person_num': 6},
        {'id': 82, 'name': '创意设计部', 'parent_id': 75, 'person_num': 0}]}

case1 = {'id': 13, 'name': 'A团队', 'parent_id': None, 'person_num': 12,
         'children': [{'id': 61, 'name': '项目二', 'parent_id': 13, 'person_num': 4,
                       'children': [{'id': 111, 'name': '测试部门', 'parent_id': 61,
                                     'person_num': 3, 'children': [
                               {'id': 10197, 'name': 'testDep0', 'parent_id': 111,
                                'person_num': 0, 'children': [
                                   {'id': 10199, 'name': 'testDep0-0',
                                    'parent_id': 10197, 'person_num': 0}]},
                               {'id': 10198, 'name': 'testDep1', 'parent_id': 111,
                                'person_num': 3, 'children': [
                                   {'id': 10200, 'name': 'testDep1-0',
                                    'parent_id': 10198, 'person_num': 3}]}]}]},
                      {'id': 62, 'name': '项目三', 'parent_id': 13, 'person_num': 4,
                       'children': [{'id': 64, 'name': '子项目一', 'parent_id': 62,
                                     'person_num': 4}]}]}

小结：

1、借助指针，2个循环搞定；

2、支持无限层级的树状结构。

 

 

	type T struct {
		domain.VOGoodsCat
		Children []*T
	}

	flat := func() []domain.VOGoodsCat {
		flat := []domain.VOGoodsCat{}
		for _, v := range out.List {
			u := domain.VOGoodsCat{}
			u.ParentId = v.ParentId
			u.CategoryId = v.CategoryId
			u.ParentId = v.ParentId
			u.Name = v.Name
			flat = append(flat, u)
		}
		return flat
	}()

	treeList := func() []T {
		// TODO 族谱树 多叉树
		treeList := []T{}
		flatPtr := []T{}
		for _, i := range flat {
			t := T{}
			t.CategoryId = i.CategoryId
			t.ParentId = i.ParentId
			t.Name = i.Name
			t.Children = []*T{}
			flatPtr = append(flatPtr, t)
		}
		for m := range flatPtr {
			for n := range flatPtr {
				if flatPtr[m].CategoryId == flatPtr[n].ParentId {
					flatPtr[m].Children = append(flatPtr[m].Children, &flatPtr[n])
				}
			}
		}
		for _, j := range flatPtr {
			if j.ParentId == -1 {
				treeList = append(treeList, j)
			}
		}
		return treeList
	}()

　　

如果Children无数据，从返回空数组调整为null

  []*T -->>  *[]*T  nil

 

	type T struct {
		domain.VOGoodsCat
		Children *[]*T
	}

	flat := func() []domain.VOGoodsCat {
		flat := []domain.VOGoodsCat{}
		for _, v := range out {
			u := domain.VOGoodsCat{}
			u.ParentId = v.ParentId
			u.CategoryId = v.CategoryId
			u.ParentId = v.ParentId
			u.Name = v.Name
			flat = append(flat, u)
		}
		return flat
	}()

	treeList := func() []T {
		// TODO 族谱树 多叉树
		treeList := []T{}
		flatPtr := []T{}
		for _, i := range flat {
			t := T{}
			t.CategoryId = i.CategoryId
			t.ParentId = i.ParentId
			t.Name = i.Name
			t.Children = nil
			flatPtr = append(flatPtr, t)
		}
		for m := range flatPtr {
			for n := range flatPtr {
				if flatPtr[m].CategoryId == flatPtr[n].ParentId {
					if flatPtr[m].Children == nil {
						flatPtr[m].Children = &[]*T{}
					}
					*(flatPtr[m].Children) = append(*(flatPtr[m].Children), &flatPtr[n])
				}
			}
		}
		for _, j := range flatPtr {
			if j.ParentId == -1 {
				treeList = append(treeList, j)
			}
		}
		return treeList
	}()

　　

 

def list_to_tree(lst: list, parent_id, keyword='id', parent_name="parent_id", child_name="children") -> list:
    hashmap = {}
    tree = []
    for data in lst:
        if data[parent_name] == parent_id:
            tree.append(data)
        hashmap[data[keyword]] = data
    for data in lst:
        parent = hashmap.get(data[parent_name])
        if parent is None:
            continue
        if parent.get(child_name) is None:
            parent[child_name] = []
        parent[child_name].append(data)
    return tree


def list_to_tree_with_count(lst: list, parent_id, keyword='id', parent_name="parent_id", child_name="children",
                            num_name="person_num") -> list:
    tree_list = list_to_tree(lst, parent_id, keyword=keyword, parent_name=parent_name, child_name=child_name)
    '''
    测试sql
    SELECT u.`name`,u.create_time,c.id,c.parent_id ,c.`name` 
FROM account_user u right JOIN center_company_department c ON u.dept_id=c.id WHERE 
c.id IN (70200,70198,70199,70197,61,111,13,62,64,10169) ORDER BY c.id DESC,c.parent_id DESC;

    '''
    def sum_tree(node):
        '''
        求树节点的和
        '''
        total = 0
        if num_name in node:
            total = node[num_name]

        if child_name in node:
            for i in node[child_name]:
                total += sum_tree(i)
        return total

    def r(node) -> None:
        '''
        递归
        求出各个节点的后代的和
        '''
        node[num_name] = sum_tree(node)
        if child_name in node:
            for i in node[child_name]:
                r(i)
        return

    for i in tree_list:
        r(i)
    return tree_list

 

 

应用场景：

层级部门人数统计

 

 

 

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