javascript - 比较对象值并返回新数组

Question

我有这两个数组:

const data = [ 
  { type: 'type1', location: 23 },
  { type: 'type2', location: 37 },
  { type: 'type3', location: 61 },
  { type: 'type4', location: 67 } 
]

const com = [ 
  { name: "com1", location: 36 },
  { name: "com2", location: 60 } 
]

我想测试 com +1 数组中的 locationComment 是否等于 data 数组中的 locationMethod ，如果是，那么我想要这样的东西:

const array = [
 {type: 'type2', name: "com1"},
 {type: 'type3', name: "com2"}
]

这是我的代码:

const result = data.map((x)=>{
  const arr = [];
  com.map((y)=>{
    if(x.location == y.location+1){
      arr.push({
        type: x.type,
        name: y.name,
        location: x.location
      })
    }
  })
  return arr;
});

这是我得到的输出:

[ [],
  [ { type: 'type2', name: 'com1', location: 37 } ],
  [ { type: 'type3', name: 'com2', location: 61 } ],
  [] ]

Answer 1

因为您不确定 com 数组中的每个元素是否都会匹配，所以您应该使用 reduce:

const data = [ 
  { type: 'type1', location: 23 },
  { type: 'type2', location: 37 },
  { type: 'type3', location: 61 },
  { type: 'type4', location: 67 } 
];

const com = [ 
  { name: "com1", location: 36 },
  { name: "com2", location: 60 } 
];

const output = com.reduce((a, { name, location }) => {
  const found = data.find(item => item.location === location + 1);
  if (found) {
    a.push({ type: found.type, name });
  }
  return a;
}, []);
console.log(output);

如果 location 是唯一的，您可以通过将 data 转换为由 索引的对象来将时间复杂度降低到 O(N) code>location 第一:

const data = [ 
  { type: 'type1', location: 23 },
  { type: 'type2', location: 37 },
  { type: 'type3', location: 61 },
  { type: 'type4', location: 67 } 
];

const com = [ 
  { name: "com1", location: 36 },
  { name: "com2", location: 60 } 
];

const dataByLoc = data.reduce((a, item) => {
  a[item.location] = item;
  return a;
}, {});

const output = com.reduce((a, { name, location }) => {
  const found = dataByLoc[location + 1];
  if (found) {
    a.push({ type: found.type, name });
  }
  return a;
}, []);
console.log(output);