如何将线性数组的元素组合起来？

Question

我的数据看起来是：

[ [ '0s', '0.200s' ],
  [ '0.200s', '0.600s' ],
  [ '1.600s', '2.500s' ],
  [ '3.500s', '3.900s' ],
  [ '3.900s', '4.400s' ],
  [ '4.400s', '4.600s' ],
  [ '4.600s', '4.700s' ],
  [ '4.700s', '5.200s' ],
  [ '5.200s', '5.400s' ],
  [ '5.400s', '5.800s' ],
  [ '5.800s', '6.100s' ],
  [ '6.100s', '6.800s' ],
  [ '6.800s', '7s' ],
  [ '7s', '7.300s' ],
  [ '7.300s', '7.500s' ]
]

第一个元素结束于0.200s，这是第二个元素开始的地方。所以我希望这两个组合成为['0s', '0.600s']。

下一个元素不是从这个元素结束的位置开始的，所以它应该继续。最终的结果应该是：

[ [ '0s', '0.600s' ],
  [ '1.600s', '2.500s' ],
  [ '3.500s', '7.500s' ]
]

我试图递归地做这件事，但是它会产生错误。这是我的功能：

function combineStartsEnds(timecodes) {
  if (timecodes[0][1] === timecodes[1][0]) {
    let combined = [
      [timecodes[0][0], timecodes[1][1]]
    ].concat(_.slice(timecodes, 2));

    return combineStartsEnds(combined);
  } else {
    return timecodes[0].concat(combineStartsEnds(_.slice(timecodes, 1)));
  }
};

这就产生了一个错误：

TypeError: Cannot read property '0' of undefined

对如何做到这一点有什么想法吗？

Answer 1

您在这里缺少括号，这是：

 return timecodes[0].concat(...)

必须：

 return [timecodes[0]].concat(...)

此外，还需要一个基本大小写来结束递归：

 function combineStartsEnds(timecodes) {
   if(!timecodes.length) return [];

我会怎么做：

 function combineStartsEnds(timecodes) {
   const result = []; let previous = [];
   for(const [start, end] of timecodes) {
     if(start === previous[/*end*/ 1]) {
       previous[/*end*/ 1] = end;
     } else {
       result.push(previous = [start, end]);
    }
  }
  return result;
 }

Answer 2

您也可以使用reduce来完成这一任务。

​

const times = [ [ '0s', '0.200s' ],
  [ '0.200s', '0.600s' ],
  [ '1.600s', '2.500s' ],
  [ '3.500s', '3.900s' ],
  [ '3.900s', '4.400s' ],
  [ '4.400s', '4.600s' ],
  [ '4.600s', '4.700s' ],
  [ '4.700s', '5.200s' ],
  [ '5.200s', '5.400s' ],
  [ '5.400s', '5.800s' ],
  [ '5.800s', '6.100s' ],
  [ '6.100s', '6.800s' ],
  [ '6.800s', '7s' ],
  [ '7s', '7.300s' ],
  [ '7.300s', '7.500s' ]
];

const merged = times.reduce((acc, [t3, t4]) => {
  const [t1, t2] = acc[acc.length - 1] || [null, null];
  if (t2 === t3) {
    acc.pop();
    acc.push([t1, t4]);
  } else {
    acc.push([t3, t4]);
  }
  return acc;
}, []);

console.log(merged);

​

Answer 3

您也可以尝试下面的方法来获得您想要的结果。

(1)使数组变平，您将得到

arr.flat()

["0s", "0.200s", "0.200s", "0.600s", "1.600s", "2.500s", "3.500s", "3.900s", "3.900s", "4.400s", "4.400s", "4.600s", "4.600s", "4.700s", "4.700s", "5.200s", "5.200s", "5.400s", "5.400s", "5.800s", "5.800s", "6.100s", "6.100s", "6.800s", "6.800s", "7s", "7s", "7.300s", "7.300s", "7.500s"]

(2)过滤和删除元素，如果在其位置之前和之后存在相同的元素，则将得到

arr.flat().filter((d,i,c) => d != c[i-1] && d != c[i+1])

["0s", "0.600s", "1.600s", "2.500s", "3.500s", "7.500s"]

(3)将上述结果降至所需的格式

arr.flat()
    .filter((d,i,c) => d != c[i-1] && d != c[i+1])
    .reduce((res, d, i, c) => (i%2 == 0 && res.push([d, c[i+1]]) , res) , [])

[["0s", "0.600s"]
["1.600s", "2.500s"]
["3.500s", "7.500s"]]