转换数组数组插入空，Javascript

Question

我以这种方式接收数据：

  [Sun Jan 01 2006 00:00:00 GMT-0600 (CST), 8.605613643557573]
  [Mon Jan 01 2007 00:00:00 GMT-0600 (CST), 4.639263458390814]
  [Tue Jan 01 2008 00:00:00 GMT-0600 (CST), 3.690424190442011]
  [Thu Jan 01 2009 00:00:00 GMT-0600 (CST), -6.068362055704255]
  [Fri Jan 01 2010 00:00:00 GMT-0600 (CST), 0.011317380895792274]
  [Sat Jan 01 2011 00:00:00 GMT-0600 (CST), 3.9984661908088825]
  [Sun Jan 01 2012 00:00:00 GMT-0600 (CST), 2.4211622308876173]
  [Tue Jan 01 2013 00:00:00 GMT-0600 (CST), -1.5740599423273804]
  [Wed Jan 01 2014 00:00:00 GMT-0600 (CST), 2.6624793033769967]
  [Thu Jan 01 2015 00:00:00 GMT-0600 (CST), 1744.9379869455643]
  ["err", NaN]
  [Wed Jan 01 2020 00:00:00 GMT-0600 (CST), 3417.1886283181875]
  [Wed Jan 01 2025 00:00:00 GMT-0600 (CST), 3331.7059850696196]
  [Tue Jan 01 2030 00:00:00 GMT-0600 (CST), 3237.940431993671]

我试着让数据集看起来像这样：

  [Sun Jan 01 2006 00:00:00 GMT-0600 (CST), 8.605613643557573, 0]
  [Mon Jan 01 2007 00:00:00 GMT-0600 (CST), 4.639263458390814, 0]
  [Tue Jan 01 2008 00:00:00 GMT-0600 (CST), 3.690424190442011, 0]
  [Thu Jan 01 2009 00:00:00 GMT-0600 (CST), -6.068362055704255, 0]
  [Fri Jan 01 2010 00:00:00 GMT-0600 (CST), 0.011317380895792274, 0]
  [Sat Jan 01 2011 00:00:00 GMT-0600 (CST), 3.9984661908088825, 0]
  [Sun Jan 01 2012 00:00:00 GMT-0600 (CST), 2.4211622308876173, 0]
  [Tue Jan 01 2013 00:00:00 GMT-0600 (CST), -1.5740599423273804, 0]
  [Wed Jan 01 2014 00:00:00 GMT-0600 (CST), 2.6624793033769967, 0]
  [Thu Jan 01 2015 00:00:00 GMT-0600 (CST), 1744.9379869455643, 0]
  ["err", NaN]
  [Wed Jan 01 2020 00:00:00 GMT-0600 (CST),0 , 3417.1886283181875]
  [Wed Jan 01 2025 00:00:00 GMT-0600 (CST),0 , 3331.7059850696196]
  [Tue Jan 01 2030 00:00:00 GMT-0600 (CST),0 , 3237.940431993671]

有办法通过查看“错误”的索引来做到这一点吗？

Answer 1

您可以使用一个变量来跟踪是否已经找到"err"，并使用该信息来决定.splice()新成员的位置。

var found = false
for (const a of data) {
  if (!found && a[0] === "err")
    found = true;
  else
    a.splice(found ? 1 : 2, 0, 0);
}

如果要使用找到err的索引，则在数组上使用.entries()并更改found变量来存储索引。

var foundIdx = -1;
for (const [idx, a] of data.entries()) {
  const found = foundIdx !== -1;

  if (!found && a[0] === "err")
    foundIdx = idx;
  else
    a.splice(found ? 1 : 2, 0, 0);
}

console.log("the err index was", foundIdx);

Answer 2

可以使用findIndex查看错误的索引。

var errIndex = data.findIndex(el => el[0] == "err");
assert(errIndex != -1);
return data.map(([date, val], i) => {
  if (i < errIndex)
    return [date, val, 0];
  if (i > errIndex)
    return [date, 0, val];
  return ["err", NaN] // or whatever
);

或者你可能会想做

var it = data.values();
for (const el of it) {
  if (el[0] == "err") break;
  el.push(0);
}
for (const el of it) {
  el.splice(1, 0, 0);
}