我刚开始学习Python,我知道有更好的方法来构建递归查询。希望更高级的人能够看看如何最优化下面的代码。
我在StackOverflow上查找了类似的例子,但没有与我尝试遍历的数据结构相同的例子。
样本数据:
[
{'categoryId': 100, 'parentId': 0, 'catName': 'Animals & Pet Supplies'},
{'categoryId': 103, 'parentId': 100, 'catName': 'Pet Supplies'},
{'categoryId': 106, 'parentId': 103, 'catName': 'Bird Supplies'},
{'categoryId': 500, 'parentId': 0, 'catName': 'Apparel & Accessories'},
{'categoryId': 533, 'parentId': 500, 'catName': 'Clothing'},
{'categoryId': 535, 'parentId': 533, 'catName': 'Activewear'}
]
Python 代码:
def returnChildren(categoryId):
cats = dict()
results = categories.find( { "parentId" : categoryId } )
for x in results:
cats[x['categoryId']] = x['catName']
return cats
children = returnChildren(cat_id)
#build list of children for this node
for x in children:
print (x, "-", children[x])
results = returnChildren(x)
if (len(results) > 0):
for y in sorted(results.keys()):
print(y, "--", results[y])
sub_results = returnChildren(y)
if (len(sub_results) > 0):
for z in sorted(sub_results.keys()):
print(z, "----", sub_results[z])
sub_sub_results = returnChildren(z)
if (len(sub_sub_results) > 0):
for a in sorted(sub_sub_results.keys()):
print(a, "------", sub_sub_results[a])
这段代码将生成一个类似下面的树:
100 Animals & Pet Supplies
103 - Pet Supplies
106 -- Bird Supplies
500 Apparel & Accessories
533 - Clothing
535 -- Activewear