Array.prototype.sort() 对象对

Question

我在一个对象数组中得到一个函数的结果：

let options = [
  {
    "aCol": 0,
    "aRow": 10,
    "bCol": 12,
    "bRow": 4,
    "cCol": 5,
    "cRow": 1,
  },
  {
    "aCol": 4,
    "aRow": 10,
    "bCol": 3,
    "bRow": 4,
    "cCol": 1,
    "cRow": 1,
  },
  {
    "aCol": 4,
    "aRow": 10,
    "bCol": 3,
    "bRow": 0,
    "cCol": 0,
    "cRow": 1,
  }
];

我想检查值对，即：

aCol
aRow

并按三对中零的数量按升序对它们进行排序。 结果应该是：

let options = [
  {
    "aCol": 4,
    "aRow": 10,
    "bCol": 3,
    "bRow": 0, // b contains a zero
    "cCol": 0, // c contains a zero
    "cRow": 1,
  },
  {
    "aCol": 0, // a contains a zero
    "aRow": 0,
    "bCol": 12,
    "bRow": 4,
    "cCol": 5,
    "cRow": 1,
  },
  {
    "aCol": 4,  // no zeroes in this object in any of the pairs
    "aRow": 10,
    "bCol": 3,
    "bRow": 4,
    "cCol": 1,
    "cRow": 1,
  },
];

我一直在利用零和这样的快捷方式搞砸：

options.sort((x,y) => 
  (x.aCol * x.aRow < y.aCol * y.aRow) ||
  (x.bCol * x.bRow < y.bCol * y.bRow) ||
  (x.cCol * x.cRow < y.cCol * y.cRow) 
) 

console.log(...options); 

Answer 1

let options = [
  {
    "aCol": 1,
    "aRow": 10,
    "bCol": 12,
    "bRow": 4,
    "cCol": 0,
    "cRow": 1,
  },
  {
    "aCol": 4,
    "aRow": 10,
    "bCol": 3,
    "bRow": 4,
    "cCol": 1,
    "cRow": 1,
  },
  {
    "aCol": 4,
    "aRow": 10,
    "bCol": 3,
    "bRow": 0,
    "cCol": 0,
    "cRow": 1,
  }
];


const sorted = options.sort((x, y) => {
   const xZeros = +!(x.aCol * x.aRow) + +!(x.bCol * x.bRow) + +!(x.cCol * x.cRow);
   const yZeros = +!(y.aCol * y.aRow) + +!(y.bCol * y.bRow) + +!(y.cCol * y.cRow);
   return yZeros - xZeros
})

console.log(sorted)

对我有用测试

+!(x.aCol * x.aRow) 

因此，如果乘法为零，则将其反转为“真”，然后“+”将其转换为 1；

如果乘法大于 0，则将其反转为 false，然后为 0

Answer 2

 function compareColAndRowZeroCounts(a, b) { const productZeroCountA = [ (a.aCol * a.aRow), (a.bCol * a.bRow), (a.cCol * a.cRow) ].filter(num => !num).length; const productZeroCountB = [ (b.aCol * b.aRow), (b.bCol * b.bRow), (b.cCol * b.cRow) ].filter(num => !num).length; const totalZeroCountA = Object.values(a).filter(num => !num).length; const totalZeroCountB = Object.values(b).filter(num => !num).length; // ascending order return ((productZeroCountA > productZeroCountB) && -1) || ((productZeroCountA < productZeroCountB) && 1) || ((totalZeroCountA > totalZeroCountB) && -1) || ((totalZeroCountA < totalZeroCountB) && 1) || 0; } const options = [{ "aCol": 0, "aRow": 0, "bCol": 12, "bRow": 4, "cCol": 5, "cRow": 1, }, { "aCol": 4, "aRow": 10, "bCol": 3, "bRow": 4, "cCol": 1, "cRow": 1, }, { "aCol": 4, "aRow": 10, "bCol": 3, "bRow": 0, "cCol": 0, "cRow": 1, }]; console.log( 'options.sort(compareColAndRowZeroCounts)', options.sort(compareColAndRowZeroCounts) );

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