使用reduce将数组转换为对象数组

Question

optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore']

result = [
  {start: bengaluru, end: salem},
  {start: salem, end: erode},
  {start: erode, end: tiruppur},
  {start: tiruppur, end: coimbatore},
]

我想将optimizedRoute 转换为结果。我想用 ES6 .reduce() 做到这一点。这是我尝试过的：

const r = optimizedRoute.reduce((places, place, i) => {
  const result: any = [];
  places = []
  places.push({
    startPlace: place,
    endPlace: place
  });
  // result.push ({ startplace, endplace, seats: 4 });
  // console.log(result);
  return places;
}, {});
console.log(r)

Answer 1

ReduceRight 的解决方案，如果有人正在寻找的话。

optimizedRoute.reduceRight((acc, d, i, arr) => 
             i == 0 
                ? acc 
                : [{ start: arr[i -1], end: d }, ...acc]  
            , [])

Answer 2

reduce 并不真正适合这里的账单，因为您并没有尝试将数组减少为单个值。

在一个完美的世界中，我们应该有multi-array map version，通常称为zip，我们可以像这样使用它

const result = zipWith(optimisedRoute.slice(0, -1),
                       optimisedRoute.slice(1),
                       (start, end) => ({start, end}));

但在 JavaScript 中没有。最好的替代方法是使用Array.from 将map 覆盖到路由的一系列索引中：

const result = Array.from({length: optimisedRoute.length - 1}, (_, index) => {
     const start = optimisedRoute[index];
     const end = optimisedRoute[index + 1];
     return {start, end};
});

Answer 3

我更喜欢可读性，而不仅仅是解决它的短代码

optimizedRoute.reduce((routes, city, index) => {
  const firstCity = index === 0;
  const lastCity = index === optimizedRoute.length - 1;
  if (!firstCity) {
    routes.last().end = city;
  }
  if (!lastCity) {
    routes.push({ start: city });
  }
  return routes;
}, []);

此外，该解决方案更短但可读性差（至少对我而言），可能是：

optimizedRoute.reduce((routes, city) => {
  routes.last().start = city;
  routes.push({ end: city });
  return routes;
}, [{}]).slice(1, -1);

关于last()，这是我通常用于提高可读性的功能：

Array.prototype.last = function() { 
  return this[this.length - 1] 
}

Answer 4

既然您要求reduce，这是一种方法：

let optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore']
   
let res = optimizedRoute.reduce((accum, item, i)=>{
   if(i == optimizedRoute.length - 1) 
      return accum;
   accum.push({start: item, end: optimizedRoute[i+1]})
   return accum;
}, [])

console.log(res);

Answer 5

我不太明白“with reduce”的要求，因为使用循环的相应代码是立即可读的，不需要解释：

const optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore'];
const result = new Array(optimizedRoute.length - 1);

for (let i = 0; i < result.length; ++i) {
  result[i] = {
    start: optimizedRoute[i],
    end: optimizedRoute[i + 1]
  };
}

console.log(result)

有时候做一些聪明的事情很有趣，但与此相比，有些答案非常复杂！

Answer 6

我简化了 nina scholz 的答案， 按照 nina 的想法，使用 reduce 获取路径的开始和结束部分，并返回 end 用于下一次开始。

getParts = a => {
  const result = [];
    a.reduce((start, end) => {
      result.push({ start, end });
      return end;
    });
    return result;
};
var optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore'];
console.log(this.getParts(optimizedRoute));

Answer 7

您可以使用reduce 获取路线的开始和结束部分，并返回结束以供下一次开始。

getParts = a => (                   // take a as array and return an IIFE
    r => (                          // with an initialized result array
        a.reduce((start, end) => (  // reduce array by taking two values
            r.push({ start, end }), // push short hand properties
            end                     // and take the last value as start value for next loop
        )),
        r                           // finally return result
    )
)([]);                              // call IIFE with empty array

const getParts = a => (r => (a.reduce((start, end) => (r.push({ start, end }), end)), r))([]);

var optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore']

console.log(getParts(optimizedRoute));

.as-console-wrapper { max-height: 100% !important; top: 0; }

---

@EDIT Grégory NEUT 添加解释

// Two thing to know first :

// When no initial value is provided,
// Array.reduce takes the index 0 as first value and start to loop at index 1

// Doing (x, y, z)
// Will execute the code x, y and z

// Equivalent to :

// x;
// y;
// z;

let ex = 0;

console.log((ex = 2, ex = 5, ex = 3));

// So about the code

const getParts = (a) => {
  // We are creating a new function here so we can have an array where to
  // push data to
  const func = (r) => {
    // Because there is no initial value
    //
    // Start will be the value at index 0 of the array
    // The loop is gonna start at index 1 of the array
    a.reduce((start, end) => {
      console.log(start, end);

      r.push({
        start,
        end,
      });

      return end;
    });

    return r;
  };

  return func([]);
};

// Equivalent
const getPartsEquivalent = (a) => {
  const r = [];

  // Because there is no initial value
  //
  // Start will be the value at index 0 of the array
  // The loop is gonna start at index 1 of the array
  a.reduce((start, end) => {
    console.log(start, end);

    r.push({
      start,
      end,
    });

    return end;
  });

  return r;
};

var optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore']

console.log(getPartsEquivalent(optimizedRoute));

.as-console-wrapper {
  max-height: 100% !important;
  top: 0;
}

Answer 8

这是reduce 的示例。不过，我不确定这是最自然的方法！

使用reduce 感觉有点矫枉过正，在这种情况下（但这只是我的看法），我自然会使用索引，好吧，我会选择一个简单的for 循环。

const optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore'];
let startCity;
const result = optimizedRoute.reduce((acc, city) => {
  if(startCity) {
    acc.push({start: startCity, end: city});
  }
  startCity = city;
  return acc;
}, []);

console.log(result);

Answer 9

以下代码使用Spread operator、Ternary operator和Array.reduce。

const optimizedRoute = [
  'Bengaluru',
  'Salem',
  'Erode',
  'Tiruppur',
  'Coimbatore',
];

// Look if we are at dealing with the last value or not
// If we do only return the constructed array
// If we don't, add a new value into the constructed array.

// tmp is the array we are constructing
// x the actual loop item
// xi the index of the item
const lastItemIndex = optimizedRoute.length - 1;

const ret = optimizedRoute.reduce((tmp, x, xi) => xi !== lastItemIndex ? [
  ...tmp,

  {
    start: x,
 
    // We access to the next item using the position of
    // the current item (xi)
    end: optimizedRoute[xi + 1],
  },
] : tmp, []);

console.log(ret);

Answer 10

另一种方法是将map 方法与slice 结合使用。对于 map 函数，您必须将 callback 函数作为 argument 传递，它将应用于给定 array 中的每个项目。

optimizedRoute = ['Bengaluru', 'Salem', 'Erode', 'Tiruppur', 'Coimbatore']
var result = optimizedRoute
                .slice(0, -1)
                .map((item, index) => ({start : item, end : optimizedRoute[index + 1]}));
console.log(result);