如何根据javascript中的条件订购安排？答案

【问题标题】：how to order arrangements based on a condition in javascript?如何根据javascript中的条件订购安排？
【发布时间】：2020-10-10 06:02:46
【问题描述】：

我有一个结构，其中排列的数量可以变化：

array1 = [
{local: {id: 1, name: 'local1'}},
{local: {id: 2, name: 'local2'}},
{local: {id: 3, name: 'local3'}},
{local: {id: 4, name: 'local4'}},
{local: {id: 5, name: 'local5'}}
];

array2 = [
{local: {id: 1, name: 'local1'}},
{local: {id: 3, name: 'local3'}},
{local: {id: 3, name: 'local4'}},
{local: {id: 3, name: 'local5'}},
];

array3 = [
{local: {id: 1, name: 'local1'}},
{local: {id: 3, name: 'local2'}},
{local: {id: 3, name: 'local3'}},
{local: {id: 3, name: 'local5'}},
];

我需要从这些创建一个新数组，其中这个新数组首先按在所有数组中重复的 id 排序，然后是不重复的，应该是这样的：

newArray = [
    {local: {id: 1, name: 'local1'}},
    {local: {id: 3, name: 'local3'}},
    {local: {id: 5, name: 'local5'}},
    {local: {id: 2, name: 'local2'}},
    {local: {id: 4, name: 'local4'}}
  ]

谁能帮帮我！！

【问题讨论】：

对于重复的 id 情况，您希望将所有数组中的哪个元素推送到结果数组中。
对于所有相似的 id，name 的值是否相同？
到目前为止您尝试过什么？你研究过你的问题吗？我在您的问题中没有看到任何代码尝试。虽然我们都渴望提供帮助，但同样重要的是要注意 Stack Overflow 不是编码服务。请向我们展示您迄今为止所做的尝试，以便我们可以为您指明正确的方向。

标签： javascript arrays sorting dictionary filter

【解决方案1】：

将所有数组转换为对象以便快速搜索。

const array1 = [{
    local: {
      id: 1,
      name: 'local1'
    }
  },
  {
    local: {
      id: 2,
      name: 'local2'
    }
  },
  {
    local: {
      id: 3,
      name: 'local3'
    }
  },
  {
    local: {
      id: 4,
      name: 'local4'
    }
  },
  {
    local: {
      id: 5,
      name: 'local5'
    }
  }
];

const array2 = [{
    local: {
      id: 1,
      name: 'local1'
    }
  },
  {
    local: {
      id: 3,
      name: 'local3'
    }
  },
  {
    local: {
      id: 3,
      name: 'local4'
    }
  },
  {
    local: {
      id: 3,
      name: 'local5'
    }
  },
];

const array3 = [{
    local: {
      id: 1,
      name: 'local1'
    }
  },
  {
    local: {
      id: 3,
      name: 'local2'
    }
  },
  {
    local: {
      id: 3,
      name: 'local3'
    }
  },
  {
    local: {
      id: 3,
      name: 'local5'
    }
  },
];

const obj1 = array1.reduce((acc, item) => {
  acc[item.local.id] = item;
  return acc;
}, {});

const obj2 = array2.reduce((acc, item) => {
  acc[item.local.id] = item;
  return acc;
}, {});


const obj3 = array3.reduce((acc, item) => {
  acc[item.local.id] = item;
  return acc;
}, {});

const result = {
  ...obj3,
  ...obj2,
  ...obj1
};

const output = [];
const temp = [];

for (let key in result) {
  if (obj1[key] && obj2[key] && obj3[key]) {
    output.push(result[key]);
  } else temp.push(result[key]);
}

console.log([...output, ...temp]);

【讨论】：

【解决方案2】：

我会这样做（可能不是最佳解决方案）：

/* Same Arrays as yours */ const array1=[{local:{id:1,name:"local1"}},{local:{id:2,name:"local2"}},{local:{id:3,name:"local3"}},{local:{id:4,name:"local4"}},{local:{id:5,name:"local5"}}],array2=[{local:{id:1,name:"local1"}},{local:{id:3,name:"local3"}},{local:{id:3,name:"local4"}},{local:{id:3,name:"local5"}}],array3=[{local:{id:1,name:"local1"}},{local:{id:3,name:"local2"}},{local:{id:3,name:"local3"}},{local:{id:3,name:"local5"}}];

function myFunc(arrays) {
  // All items, with duplicates
  const allItems = [].concat.apply([], arrays);
  // All IDs, without duplicates thanks to `Set`
  const allIDs = Array.from(
    allItems.reduce((set, item) => set.add(item.local.id), new Set())
  );
  
  // Helper function used for sorting
  const isInAllArrays = id => arrays.every(
    arr => arr.some(item => item.local.id === id)
  );
  // Sort the IDs based on whether they are in all arrays or not
  allIDs.sort((a, b) => {
    const _a = isInAllArrays(a), _b = isInAllArrays(b);
    if (_a !== _b) return _a ? -1 : 1;
    return 0;
  });
  // Map all IDs to the first element with this ID
  return allIDs.map(id => allItems.find(item => item.local.id === id));
}

const newArray = myFunc([array1, array2, array3]);

// Just for readability in the demo below
console.log(JSON.stringify(newArray).split('},{').join('},\n{'));

【讨论】：

【解决方案3】：

1) 遍历所有数组并构建一个对象，其键为id，值包括对象，并保持出现频率（计数）。 2) 现在，上述对象的Object.values 并根据“计数”对它们进行排序。您将在顶部获得最常见的项目。

const sort = (...arrs) => {
  const all = {};
  arrs
    .flat()
    .forEach(
      (obj) =>
        (all[obj.local.id] =
          obj.local.id in all
            ? { ...all[obj.local.id], count: all[obj.local.id].count + 1 }
            : { ...obj, count: 1 })
    );
  return Object.values(all)
    .sort((a, b) => b.count - a.count)
    .map(({ count, ...rest }) => rest);
};

array1 = [
  { local: { id: 1, name: "local1" } },
  { local: { id: 2, name: "local2" } },
  { local: { id: 3, name: "local3" } },
  { local: { id: 4, name: "local4" } },
  { local: { id: 5, name: "local5" } },
];

array2 = [
  { local: { id: 1, name: "local1" } },
  { local: { id: 3, name: "local3" } },
  { local: { id: 3, name: "local4" } },
  { local: { id: 3, name: "local5" } },
];

array3 = [
  { local: { id: 1, name: "local1" } },
  { local: { id: 3, name: "local2" } },
  { local: { id: 3, name: "local3" } },
  { local: { id: 3, name: "local5" } },
];


console.log(sort(array1, array2, array3))

【讨论】：