JavaScript - 删除两个数组之间的多个相同值

Question

我目前正在处理一个代码，我需要比较两个数组并删除多个同名元素。这是数组;

vacant = [
"FRAMIA420.2 - 0h 36 m",
"FRAMIA510.4 - 0h 36 m",
"FRAMIA320.7 - 0h 36 m",
"FRAMIA520.7 - 0h 36 m",
"FRAMIA450.3 - 1h 36 m",
"FRAMIA350.1 - 2h 21 m",
"FRAMIA210.2 - 2h 21 m",
"FRAMIA340.2 - 2h 36 m"]
booked = [
"FRAMIA440.5 - 13h 0 m",
"FRAMIA540.2 - 3h 45 m",
"FRAMIA340.2 - 5h 45 m",
"FRAMIA250.1 - 3h 45 m",
"FRAMIA420.2 - 3h 45 m",
"FRAMIA540.1 - 13h 0 m",
"FRAMIA520.5 - 3h 45 m",
"FRAMIA240.4 - 3h 45 m",
"FRAMIA510.2 - 7h 0 m",
"FRAMIA510.4 - 2h 45 m",
"FRAMIA520.7 - 2h 45 m",
"FRAMIA450.1 - 1h 45 m",
"FRAMIA450.3 - 2h 0 m"]

所以这两个数组之间的相似元素是：FRAMIA420.2、FRAMIA510.4、FRAMIA520.7、FRAMIA450.3 和 FRAMIA340.2

。

我已经过滤掉了元素的时间戳部分，所以我只需要比较名称部分;

var firstPart = [];
vacant.forEach(function (obj1) {
    firstPart.push(obj1.substring(0, obj1.indexOf('-')))
});
booked.forEach(function (obj2) {
    var c = firstPart.indexOf(obj2.substring(0, obj2.indexOf('-')));
});

最终结果应如下所示，仅将元素留在 vacant -array 内，与booked -array 没有相似之处：

FRAMIA320.7 - 0h 36 m
FRAMIA350.1 - 2h 21 m
FRAMIA210.2 - 2h 21 m

请注意，数组之间的相似之处每天都在变化，有时可能有 2 个相似的元素，有时可能有 8 个或更多。

有什么快速有效的方法可以做到这一点吗？

Answer 1

您可以构造 booked 数组中所有名称的列表，然后遍历空数组，检查该名称是否在预订名称列表中。

vacant = [
  "FRAMIA420.2 - 0h 36 m",
  "FRAMIA510.4 - 0h 36 m",
  "FRAMIA320.7 - 0h 36 m",
  "FRAMIA520.7 - 0h 36 m",
  "FRAMIA450.3 - 1h 36 m",
  "FRAMIA350.1 - 2h 21 m",
  "FRAMIA210.2 - 2h 21 m",
  "FRAMIA340.2 - 2h 36 m"
]
booked = [
  "FRAMIA440.5 - 13h 0 m",
  "FRAMIA540.2 - 3h 45 m",
  "FRAMIA340.2 - 5h 45 m",
  "FRAMIA250.1 - 3h 45 m",
  "FRAMIA420.2 - 3h 45 m",
  "FRAMIA540.1 - 13h 0 m",
  "FRAMIA520.5 - 3h 45 m",
  "FRAMIA240.4 - 3h 45 m",
  "FRAMIA510.2 - 7h 0 m",
  "FRAMIA510.4 - 2h 45 m",
  "FRAMIA520.7 - 2h 45 m",
  "FRAMIA450.1 - 1h 45 m",
  "FRAMIA450.3 - 2h 0 m"
]
function getName(str) {
  return str.substring(0, str.indexOf('-'));
}
var bookedNames = [];
booked.forEach(function (bookedStr) {
  bookedNames.push(getName(bookedStr))
});
var uniqueVacant = [];
vacant.forEach(function (vacantStr) {
  var vacantName = getName(vacantStr);
  if (!bookedNames.includes(vacantName))
    uniqueVacant.push(vacantStr)
});
console.log(uniqueVacant);

Answer 2

你需要 2 个循环，每个数组一个，并比较每个数组字符串的第一部分，如下所示：

vacant = [
"FRAMIA420.2 - 0h 36 m",
"FRAMIA510.4 - 0h 36 m",
"FRAMIA320.7 - 0h 36 m",
"FRAMIA520.7 - 0h 36 m",
"FRAMIA450.3 - 1h 36 m",
"FRAMIA350.1 - 2h 21 m",
"FRAMIA210.2 - 2h 21 m",
"FRAMIA340.2 - 2h 36 m"]
booked = [
"FRAMIA440.5 - 13h 0 m",
"FRAMIA540.2 - 3h 45 m",
"FRAMIA340.2 - 5h 45 m",
"FRAMIA250.1 - 3h 45 m",
"FRAMIA420.2 - 3h 45 m",
"FRAMIA540.1 - 13h 0 m",
"FRAMIA520.5 - 3h 45 m",
"FRAMIA240.4 - 3h 45 m",
"FRAMIA510.2 - 7h 0 m",
"FRAMIA510.4 - 2h 45 m",
"FRAMIA520.7 - 2h 45 m",
"FRAMIA450.1 - 1h 45 m",
"FRAMIA450.3 - 2h 0 m"]
for(i=0;i<vacant.length;i++) {
    item1 = vacant[i].split('-')[0];
    for(j=0;j<booked.length;j++) {
        item2 = booked[j].split('-')[0];
        if(item1===item2) {
                console.log('item number '+i+' in vacant is the same as item number '+j+' in booked');
        }
    }
}

https://jsfiddle.net/48hef0cz/

Answer 3

试试这个：

var vacant = [
"FRAMIA420.2 - 0h 36 m",
"FRAMIA510.4 - 0h 36 m",
"FRAMIA320.7 - 0h 36 m",
"FRAMIA520.7 - 0h 36 m",
"FRAMIA450.3 - 1h 36 m",
"FRAMIA350.1 - 2h 21 m",
"FRAMIA210.2 - 2h 21 m",
"FRAMIA340.2 - 2h 36 m"];
var booked = [
"FRAMIA440.5 - 13h 0 m",
"FRAMIA540.2 - 3h 45 m",
"FRAMIA340.2 - 5h 45 m",
"FRAMIA250.1 - 3h 45 m",
"FRAMIA420.2 - 3h 45 m",
"FRAMIA540.1 - 13h 0 m",
"FRAMIA520.5 - 3h 45 m",
"FRAMIA240.4 - 3h 45 m",
"FRAMIA510.2 - 7h 0 m",
"FRAMIA510.4 - 2h 45 m",
"FRAMIA520.7 - 2h 45 m",
"FRAMIA450.1 - 1h 45 m",
"FRAMIA450.3 - 2h 0 m"];
vacant = vacant.filter(function (element) {
  var roomName = element.split('-')[0];
  
  var index = booked.findIndex(function (booking) {
    return roomName === booking.split('-')[0];
  });
  
  return index == -1;
});
console.log(vacant);

Answer 4

您可以使用字典来跟踪唯一值。

vacant = [
"FRAMIA420.2 - 0h 36 m",
"FRAMIA510.4 - 0h 36 m",
"FRAMIA320.7 - 0h 36 m",
"FRAMIA520.7 - 0h 36 m",
"FRAMIA450.3 - 1h 36 m",
"FRAMIA350.1 - 2h 21 m",
"FRAMIA210.2 - 2h 21 m",
"FRAMIA340.2 - 2h 36 m"]
booked = [
"FRAMIA440.5 - 13h 0 m",
"FRAMIA540.2 - 3h 45 m",
"FRAMIA340.2 - 5h 45 m",
"FRAMIA250.1 - 3h 45 m",
"FRAMIA420.2 - 3h 45 m",
"FRAMIA540.1 - 13h 0 m",
"FRAMIA520.5 - 3h 45 m",
"FRAMIA240.4 - 3h 45 m",
"FRAMIA510.2 - 7h 0 m",
"FRAMIA510.4 - 2h 45 m",
"FRAMIA520.7 - 2h 45 m",
"FRAMIA450.1 - 1h 45 m",
"FRAMIA450.3 - 2h 0 m"]
vacantDict = {};
vacant.forEach(function(val) {
  var name = val.split(' - ')[0];
  vacantDict[name] = val;
});
booked.forEach(function(val) {
  var name = val.split(' - ')[0];
  if (vacantDict[name] !== undefined) {
    delete vacantDict[name];
  }
});
newVacantList = [];
for (var name in vacantDict) {
  newVacantList.push(vacantDict[name]);
}
console.log(newVacantList);