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},
]

我想转换optimizedRoute成结果.我想用ES6做到这一点.reduce().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

您可以使用reduce获取路径的开始和结束部分并返回结束以进行下一次启动.

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了GrégoryNEUT 加入交代

// 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 2

另一种方法是map 结合使用方法slice.对于map函数,您必须传递一个callback函数作为参数,该函数将应用于给定数组中的每个项目.

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

Answer 3

我并不真正理解"with reduce"的要求,因为使用循环的相应代码可以立即读取,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,但是与此相比,有些答案非常复杂!

Answer 4

这是一个带有reduce. 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，在那种情况下（但这只是我的意见），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);