将两个循环转换为列表理解

如何解决将两个循环转换为列表理解

我正在尝试为该函数的第3-6行编写一个列表理解。它根据guests_diet中的选择返回相关的餐厅。结果应为字符串“对不起，没有餐厅符合您的限制”。该功能有效，但是我被要求做列表理解。我想出了下面的代码，但结果不正确。有人可以帮忙吗？

尝试的代码产生不正确的结果-

ans = [restaurant.append(key) for key,value in rest_names.items()\
                  for x in range(len(cuisine)) if cuisine[x] in value]

代码-

def no_you_pick(rest_names,cuisine):
    restaurant = []
    for key,value in rest_names.items():
        for x in range(len(cuisine)):
            if cuisine[x] in value:
                restaurant.append(key)
    restaurant.sort()
    if len(restaurant) == 0:
        return "Sorry,no restaurants meet your restrictions"
    elif len(restaurant) == 1:
        return ' '.join(restaurant)
    return ','.join(restaurant)

grading_scale = {"blossom": ["vegetarian","vegan","kosher","gluten-free","dairy-free"],\
             "jacob's pickles": ["vegetarian","gluten-free"],\
             "sweetgreen": ["vegetarian","kosher"]}
guests_diet = ["buttered-lobster"]
print(no_you_pick(grading_scale,guests_diet))

解决方法

以下代码将起作用-

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    var index = tag.edges[0].node.frontmatter.tags.indexOf(
      tag.fieldValue
    )

    return (
      <li key={idx}>
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          className="transition link"
        >
          {tag.fieldValue}
        </Link>
      </li>
    )
  })}
</ul>

输出：

def no_you_pick(rest_names,cuisine):
    restaurant = [key for key,value in rest_names.items() for x in range(len(cuisine)) if cuisine[x] in value]
    restaurant.sort()
    if len(restaurant) == 0:
        return "Sorry,no restaurants meet your restrictions"
    elif len(restaurant) == 1:
        return ' '.join(restaurant)
    return ','.join(restaurant)

grading_scale = {"blossom": ["vegetarian","vegan","kosher","gluten-free","dairy-free"],\
             "jacob's pickles": ["vegetarian","gluten-free"],\
             "sweetgreen": ["vegetarian","kosher"]}
guests_diet = ["buttered-lobster"]
print(no_you_pick(grading_scale,guests_diet))
print(no_you_pick(grading_scale,['vegan']))

---

您无需附加到Sorry,no restaurants meet your restrictions blossom,sweetgreen 。

restaurant 

由于您使用列表推导，因此可以直接构建列表-ans = [restaurant.append(key) for key,value in rest_names.items() for x in range(len(cuisine)) if cuisine[x] in value] ，并且只有符合给定条件的键才会添加到列表中。在这里，您无需编写restaurant，就可以像上面的代码中那样，在restaurant.append(key)上使用列表推导轻松地直接附加键

,

您可以将其替换为以下列表理解：

this.loadingController.create({
      message: 'Loading,please wait...'
    }).then((loading) => {

      loading.present();

      Promise.all([apiCallOne(),apiCallTwo(),apiCallThree()])
        .catch((err) => {
          console.log(err);
          loading.dismiss();
        })
        .then((results) => {
          // do something with results ...
          loading.dismiss();
        });

    });

无需声明restaurant = [key for key,value in rest_names.items()\ if any(item in cuisine for item in value)] 是其上方的空白列表。