如何在带有嵌套字典的嵌套循环中正确使用字典“.update”函数[重复]

Question

因此，我从 Web 服务中提取字典格式的 JSON，以获取有关特定产品的各种指标。我有嵌套循环，可以拉出嵌套字典，我想将这些嵌套字典分配给唯一的产品（它们是键）。为了简单起见，我有一些示例代码显示了我遇到的问题（虽然荒谬，但香蕉很棒）。

import random
test = {}
fruit = ['apple','banana','watermelon']

color = ['red','blue','green']
shape = ['oval','round','long']
taste = ['good','bad','fresh']
farm = [{'US': 'GA'}, {"Peru": "Lima"},{'US': 'New York'}] # this data in my use is actually more varied

for i in fruit:
    this_color = random.choice(color) # these represent making URLs to pull from
    this_shape = random.choice(shape)
    this_taste = random.choice(taste)
    description = [this_color,this_shape,this_taste]
    for x in description:
        # this line would be pulling requests of dictionary formatted JSONs from the URLS
        test.update({i:{x: random.choice(farm)}})

这给出以下输出：

{   'apple': {'fresh': {'US': 'GA'}},
    'banana': {'bad': {'Peru': 'Lima'}},
    'watermelon': {'fresh': {'Peru': 'Lima'}}}

我的问题：我正在寻找一个字典，其中包含每个描述符 [this_color, this_shape, this_taste] 作为键，但我只将最后一项 this_taste 作为键。我敢打赌这是因为我错误地使用了.update，但我不确定实现这一点的“正确”方式。任何建议都会很棒！

Answer 1

import random
import collections.abc

#https://stackoverflow.com/questions/3232943/update-value-of-a-nested-dictionary-of-varying-depth
def update(d, u):
    for k, v in u.items():
        if isinstance(v, collections.abc.Mapping):
            d[k] = update(d.get(k, {}), v)
        else:
            d[k] = v
    return d

test = {}
fruit = ['apple','banana','watermelon']

color = ['red','blue','green']
shape = ['oval','round','long']
taste = ['good','bad','fresh']
farm = [{'US': 'GA'}, {"Peru": "Lima"},{'US': 'New York'}] # this data in my use is actually more varied

for i in fruit:
    this_color = random.choice(color) # these represent making URLs to pull from
    this_shape = random.choice(shape)
    this_taste = random.choice(taste)
    description = [this_color,this_shape,this_taste]
    for x in description:
        # this line would be pulling requests of dictionary formatted JSONs from the URLS
        update(test, {i: {x: random.choice(farm)}})