4

我试图过滤一些对象以更好地理解 JS，我正在使用 underscore.js

我来自 C# 背景并习惯于 LINQ，但下划线并不完全相同。

你能帮我根据定义的测试过滤掉这个数组吗，我遇到的问题是数组上的数组属性。Where运算符与我通常用来过滤项目的 C# 不同。

products = [
       { name: "Sonoma", ingredients: ["artichoke", "sundried tomatoes", "mushrooms"], containsNuts: false },
       { name: "Pizza Primavera", ingredients: ["roma", "sundried tomatoes", "goats cheese", "rosemary"], containsNuts: false },
       { name: "South Of The Border", ingredients: ["black beans", "jalapenos", "mushrooms"], containsNuts: false },
       { name: "Blue Moon", ingredients: ["blue cheese", "garlic", "walnuts"], containsNuts: true },
       { name: "Taste Of Athens", ingredients: ["spinach", "kalamata olives", "sesame seeds"], containsNuts: true }
    ];

it("given I'm allergic to nuts and hate mushrooms, it should find a pizza I can eat (functional)", function () {

      var productsICanEat = [];

      //This works but was hoping I could do the mushroom check as well in the same line
      var noNuts = _(products).filter(function (x) { return !x.containsNuts;});

      var noMushrooms = _(noNuts).reject(function(x){ return !_(x.ingredients).any(function(y){return y === "mushrooms";});});


      console.log(noMushrooms);

      var count = productsICanEat.length;
      expect(productsICanEat.length).toBe(count);
  });
4

4 回答 4

10

您只需!要从reject回调中删除 ，使其看起来像这样：

var noMushrooms = _(noNuts).reject(function(x){ 
    return _(x.ingredients).any(function(y){return y === "mushrooms";});
});

否则你会拒绝那些不含蘑菇的，而不是那些含有蘑菇的。

于 2013-01-15T14:14:19.007 回答
6

实现这一点的更简洁的方法是使用下划线的 chain() 函数：

var noMushrooms = _(products).chain()
    .filter(function (x) { 
        return !x.containsNuts;})
    .reject(function(x){ 
        return _(x.ingredients).any(function(y){
            return y === "mushrooms";
        });
    })
    .value();
于 2013-01-15T15:19:46.180 回答
4

我设法将我的解决方案全部包含在一个过滤器调用中，所以我想我会发布它：

products = [
       { name: "Sonoma", ingredients: ["artichoke", "sundried tomatoes", "mushrooms"], containsNuts: false },
       { name: "Pizza Primavera", ingredients: ["roma", "sundried tomatoes", "goats cheese", "rosemary"], containsNuts: false },
       { name: "South Of The Border", ingredients: ["black beans", "jalapenos", "mushrooms"], containsNuts: false },
       { name: "Blue Moon", ingredients: ["blue cheese", "garlic", "walnuts"], containsNuts: true },
       { name: "Taste Of Athens", ingredients: ["spinach", "kalamata olives", "sesame seeds"], containsNuts: true }
    ];

 it("given I'm allergic to nuts and hate mushrooms, it should find a pizza I can eat (functional)", function () {

      var productsICanEat = [];

      productsICanEat = _(products).filter(function (x) { return !x.containsNuts && !_(x.ingredients).any(function(y){return y === "mushrooms";});});


      expect(productsICanEat.length).toBe(1);
  });
于 2013-01-15T14:59:24.280 回答
4

这将给出所需的结果

var no_nuts = _.filter(products,function(item) {
         return !item.containsNuts;
       });

var no_mushroom = _.reject(no_nuts,function(item) {
        return _.any(item.ingredients,function(item1) {
            return item1 === "mushrooms"
        }); 
     });

console.log(no_mushroom);

reject()与 , 相反filter()，any()它等效于数组的某些方法，当数组中的任何元素通过回调时返回 true 时返回 true。

于 2014-02-20T17:46:27.500 回答