Javascript reduce（），传入初始值

Question

我正在使用下划线库完成一项练习，以对项目出现次数进行计数。

具体来说，我有这个数组：

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 }
    ];

并且正在尝试使用map（），flatten（）和reduce（）来确定每种成分出现的次数。

当我使用下面的代码时，IngredientCount返回未定义的值：

var ingredientCount = { "{ingredient name}": 0 };

    _(products).chain()
    .map(function(obj) {return obj.ingredients;})
    .flatten().reduce(function(memo, x) {
      return memo[x] = (memo[x] || 0) + 1;
    }, ingredientCount)
    .value();

但是，如果我删除第二个参数reduce（）并将初始对象包含在函数主体中，则一切都会按预期工作，即：

var IngredientCount = {“ {成分名称}”：0};

_(products).chain()
.map(function(obj) {return obj.ingredients;})
.flatten().reduce(function(memo, x) {
  return ingredientCount[x] = (ingredientCount[x] || 0) + 1;
})
.value();

有人可以帮忙解释为什么会这样吗？

谢谢！

Answer 1

return memo[x] = (memo[x] || 0) + 1; 返回一个数字，供循环的下一个迭代用作备忘录。

代替做

memo[x] = (memo[x] || 0) + 1;
return memo;

如果您真的想在单个语句中执行此操作

return memo[x] = (memo[x] || 0) + 1, memo;

Answer 2

看到一个普通的JS替代品总是很有趣：

var counts = p.reduce(function(acc, prod) {
  prod.ingredients.forEach(function(i) {
    acc[i] = (acc[i] || 0) + 1;
  });
  return acc;
},{});

这似乎更清晰，尽管在一般情况下最好使用：

  if (!acc.hasOwnProperty(i)) acc[i] = 0;
  acc[i]++

对于柜台部分。

 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 } ]; function countIngredients(p) { var counts = p.reduce(function(acc, prod) { prod.ingredients.forEach(function(i) { acc[i] = (acc[i] || 0) +1; }) return acc; },{}) return counts; } document.write(JSON.stringify(countIngredients(products))); 

Answer 3

我会这样强调

  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 } ]; ingredientCount = {}; _(products).chain() .map(function(obj) {return obj.ingredients;}).flatten().each(function(memo, x) { ingredientCount[memo] = (ingredientCount[memo] || 0) + 1; }); document.write(JSON.stringify(ingredientCount)); //this will produce this "{"artichoke":1,"sundried tomatoes":2,"mushrooms":2,"roma":1,"goats cheese":1,"rosemary":1,"black beans":1,"jalapenos":1,"blue cheese":1,"garlic":1,"walnuts":1,"spinach":1,"kalamata olives":1,"sesame seeds":1}" 

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