如何使用ramda按数组对象中的最后一项排序

Question

以下是父母名单，我想按拉姆达分类按其第二个孩子的年龄进行排序：

[
  {
    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"
      },
    ]
  }
]

我该如何处理？ 我尝试按照以下方式做一些事情

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

Answer 1

使用R.sortBy并提取与函数与创造的价值R.pipe 。 该函数使用R.prop获取对象的children数组，获取最后一个孩子（ R.last ），使用R.propOr获取age （如果没有孩子则返回0），然后转换为Number 。 如果要R.negate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

在原始JavaScript中（使用一些相对格式较差的假设），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-如果父母少于两个孩子会怎样？

Answer 3

假设上面的JSON是手工生成的，包括语法错误，然后假设您的真实数据很好（一个父级数组，每个父级都有一个对象的children级数组），那么普通的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

我会用sortWith有ascend的功能。 使用sortWith允许您定义第一个排序顺序函数，第二个排序顺序函数等。

 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结合使用：

 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> 

其他人已经指出，这样做存在一些潜在的缺陷。 父母是否总是保证至少有两个孩子？ 我们是否真的要按字典顺序排序-即'11' < '2' 2'-还是要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']))) ，但这取决于您要执行的操作。 如果您只是用来学习sortBy ，那么sortBy和path都是有用的函数。 当您可以将要排序的项目转换为某种有序类型时， sortBy很有用-字符串，数字，日期或带有数值valueOf方法的任何内容。 您提供该转换函数和值列表，它将按此排序。 path只是对对象中的嵌套属性列表的null安全读取。