如何使用 ramda 按数组对象中的最后一项进行排序

Question

这是一个家长列表，我想用ramda按第二个孩子的年龄对父母进行排序：

[
  {
    name: "Alicia",
    age: "43",
    children: [{
        name: "Billy",
        age: "3"
      },
      {
        name: "Mary",
        age: "8"
      },
    ]
  },
  {
    name: "Felicia",
    age: "60",
    children: [{
        name: "Adrian",
        age: "4"
      },
      {
        name: "Joseph",
        age: "5"
      },
    ]
  }
]

我该怎么做？ 我尝试做一些类似的事情

parents.sort(
                sortBy("-children.age"))
            );

Answer 1

使用 R.sortBy 并使用函数 create with R.pipe 提取值。该函数获取带有 R.prop 的对象 的子数组，获取最后一个子数组 (R.last(，获取带有 R.propOr 的age(如果没有子项则返回 0(，并转换为 Number 。如果要颠倒顺序，可以使用R.negate。

const { sortBy, pipe, prop, last, propOr } = R
const fn = sortBy(pipe(
  prop('children'),
  last,
  propOr(0, 'age'),
  Number,
  // negate - if you want to reverse the order
))
const parents = [{"name":"Alicia","age":"43","children":[{"name":"Billy","age":"3"},{"name":"Mary","age":"8"}]},{"name":"Felicia","age":"60","children":[{"name":"Adrian","age":"4"},{"name":"Joseph","age":"5"}]}]
const result = fn(parents)
console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.js"></script>

Answer 2

在原版 JavaScript 中(对格式相对较差的输入做出一些假设(，使用 Array.prototype.sort 方法：

let parents = [ .... ]; // What you have above
parents = parents.sort((a, b) => {
  return a.children[1].age - b.children[1].age; // Change - to + for ascending / descending
});

不过要小心 - 如果父母少于 2 个孩子会发生什么？

Answer 3

假设上面的JSON是手动生成的，包括语法错误，然后假设你的真实数据很好(一个父数组，每个父级都有一个children的对象数组(，那么正常的JS排序就可以了：

const compareC2(parent1, parent2) {
  let c1 = parent1.children;
  let c2 = parent2.children;
  if (!c1 || !c2) {
    // what happens if someone has no children?
  }
  let l1 = c1.length;
  let l2 = c2.length;
  if (l1 === 0 || l2 === 0) {
    // different symptom, but same question as above
  }
  if (l1 !== l2) {
    // what happens when the child counts differ?
  }
  if (l1 !== 2) {
    // what happens when there are fewer, or more than, 2 children?
  }
  // after a WHOLE LOT of assumptions, sort based on
  // the ages of the 2nd child for each parent.
  return c1[1].age - c2[1].age;
}  
let sorted = parents.sort(compareC2);

Answer 4

我会将sortWith与ascend函数一起使用。使用 sortWith 可以定义第一排序顺序函数、第二排序顺序函数等。

const people = [
  {
    name: "Alicia",
    age: "43",
    children: [{
        name: "Billy",
        age: "3"
      },
      {
        name: "Mary",
        age: "8"
      },
    ]
  },
  {
    name: "Felicia",
    age: "60",
    children: [{
        name: "Adrian",
        age: "4"
      },
      {
        name: "Joseph",
        age: "5"
      },
    ]
  }
];
const by2ndChildAge = ascend(pathOr(0, ['children', 1, 'age']));
const by1stChildAge = ascend(pathOr(0, ['children', 0, 'age']));
console.log(sortWith([by2ndChildAge, by1stChildAge], people));

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.26.1/ramda.min.js"></script>
<script>const {sortWith, ascend, pathOr} = R;</script>

Answer 5

我认为

最简单的解决方案是将sortBy与path结合起来：

const sortBy2ndChildAge = sortBy(path(['children', 1, 'age']))
const people = [{name: "Alicia", age: "43", children: [{name: "Billy", age: "3"}, {name: "Mary", age: "8"}]}, {name: "Felicia", age: "60", children: [{name: "Adrian", age: "4"}, {name: "Joseph", age: "5"}]}]
console.log(sortBy2ndChildAge(people))

<script src="https://bundle.run/ramda@0.26.1"></script><script>
const {sortBy, path} = ramda    </script>

其他人已经注意到了这一点，这有几个潜在的缺陷。 父母总是保证至少有两个孩子吗？ 我们真的想要一个词典排序吗 - 即 '11' < '2' -- 或者您想将这些值转换为数字？

解决这两个问题很容易：sortBy(compose(Number, pathOr(0, ['children', 1, 'age'])))，但这取决于您要做什么。 如果您只是使用它来了解Ramda，那么sortBy和path都是有用的功能。 当您可以将要排序的项目转换为某种有序类型(字符串、数字、日期或使用数字valueOf方法的任何内容(时，sortBy很有用。 您提供该转换函数和值列表，它将按此排序。 path只是对对象中嵌套属性列表的 null 安全读取。