按 Javascript 中的索引删除组合二维数组的值

Question

一个js新手问题。我需要删除一个组合二维数组的值，该数组没有按索引配对或连接的值。抱歉，我不知道正确的术语。仅以我为例：

arr = [
    ["First Name", "Last Name", "Email", "Address", "Position", "Age", "Birthday"],
    ["John", "Doe", "john@doe.com", "", "", "34", ""]
];
res = arr.reduce((x, y) => x.map((v, i) => v + ': '+ y[i]));

console.log(res); //["First Name: John", "Last Name: Doe", "Email: john@doe.com", "Address: ", "Position: ", "Age: 34", "Birthday: "]

所以，我需要从数组中删除 "Address: " "Position: " "Birthday: "，剩下的是：

["First Name: John", "Last Name: Doe", "Email: john@doe.com", ""Age: 34"]

意思是，从另一个数组中删除那些不配对的。希望这是有道理的，感谢您的帮助！

Answer 1

像这样

arr = [
    ["First Name", "Last Name", "Email", "Address", "Position", "Age", "Birthday"],
    ["John", "Doe", "john@doe.com", "", "", "34", ""]
];
res = arr
   .reduce((x, y) => x
   .map((v, i) => { if (y[i]) return v + ': '+ y[i]}))
   .filter(Boolean);

console.log(res);

Answer 2

arr = [
    ["First Name", "Last Name", "Email", "Address", "Position", "Age", "Birthday"],
    ["John", "Doe", "john@doe.com", "", "", "34", ""]
];
res = arr.reduce(function(x, y){
  return x.map(function(v, i) {
    if (!['Birthday', 'Age'].includes(v)) return v + ': '+ y[i]; 
  }).filter(x => x !== undefined);
});

console.log(res);

Answer 3

您可以使用函数 Array.prototype.reduce 并对空的 space（这是虚假的）应用强制以跳过这些值。

const arr = [    ["First Name", "Last Name", "Email", "Address", "Position", "Age", "Birthday"],   ["John", "Doe", "john@doe.com", "", "", "34", ""]],
      [properties, values] = arr,
      result = values.reduce((a, v, i) => Boolean(v) ? a.concat([properties[i], ": ", v].join("")) : a, []);
      
console.log(result);

按 Javascript 中的索引删除组合二维数组的值

Remove value of a combined two dimensional array by index in Javascript

javascript

arrays

indexof