基于关键字和值数组解析数组答案

【问题标题】：Parsing an array based on keyword AND array of values基于关键字和值数组解析数组
【发布时间】：2021-05-20 14:06:24
【问题描述】：

假设我有一个数组和 2 个变量；一个是简单的字符串，一个是字符串值数组：

var street = "Fernwood Avenue";
var fips = ["34011", "34007"];

var test_array =[["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]];

我在这里要做的是遍历我的 test_array 并在 sub_array[6] 中找到与 street 匹配的子数组，并且在 fips 列表中匹配 sub_array[3] (["34011", "34007" ]）。有没有办法使用过滤器（或任何其他方法）来实现这一点并返回以下结果？

var new_array = [["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"], 
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"]];

*奖励：假设有一个函数，其中街道和 fips 作为列表参数进入，其中街道名称将始终存在，但 fips 值可能会或可能不会。所以，

var keyword = ["Fernwood Avenue", "34011", "34007"];

或

var keyword = ["Fernwood Avenue"];

或

var keyword = ["Fernwood Avenue", "34011"];

使用类似这样的函数（*这不适用于场景 1）

function matcher(array1, array2) {
    return array1.every(value => array2.includes(value));
}

function array_parser(array, keywords) {
    return array.filter(values => matcher(keywords, values));
}

var new_array = array_parser(test_array, keywords);

【问题讨论】：

标签： javascript arrays parsing filter

【解决方案1】：

您的第一个问题是一个简单的过滤器

var street = "Fernwood Avenue";
var fips = ["34011", "34007"];

var test_array =[["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]];

const result = test_array.filter( values => values[6] == street && fips.includes(values[3]))
console.log(result);

关于你的额外问题：

var test_array =[["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]];


function matcher(array1, array2) {
    const street = array1[0];
    const fips = array1.slice(1);
    return array2[6] == street && (fips.length === 0 || fips.includes(array2[3]));
}

function array_parser(array, keywords) {
    return array.filter(values => matcher(keywords, values));
}

console.log(array_parser(test_array, ["Fernwood Avenue", "34011", "34007"]));

console.log(array_parser(test_array, ["Fernwood Avenue", "34011"]));

console.log(array_parser(test_array, ["Fernwood Avenue"]));

【讨论】：

所有这些答案都很棒。我去了这个，因为在我的大型应用程序的上下文中，匹配器功能发挥了最大的作用。谢谢大家！

【解决方案2】：

这回答了问题的第一部分

var street = "Fernwood Avenue";
var fips = ["34011", "34007"];

var test_array =[["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]];

console.log(test_array.filter(e => e[6]===street && (fips.indexOf(e[3]) > -1)))

这回答了问题的第二部分，但我不确定它的可读性如何，简单地说，Array.indexOf() 的第二个参数指定了在列表中搜索的起始索引。但是如果在keyword 中只指定街道，它总是会返回-1，因为没有第二个元素可以开始搜索。

添加对keyword.length 的检查通过在长度为 1 时始终返回 true 来解决此问题（又名，忽略 fip 的过滤）

var keyword = ["Fernwood Avenue","34011"];

var test_array =[["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]];

console.log(test_array.filter(e => (e[6]===keyword[0]) && (keyword.length > 1 ? keyword.indexOf(e[3],1) > -1 : true)))

【讨论】：

【解决方案3】：

使用两个变量：您可以使用该函数并检查Array#filter() 中的每个变量的已知索引。如果您想在传递一个空列表时匹配所有内容，那么您可以进行一个允许所有内容的模拟查找：

function findStuff(array, street, fips) {
  const streetIndex = 6;
  const fipsIndex = 3;
  
  let fipsLookup;
  if (fips.length > 0)
    fipsLookup = new Set(fips);
  else
    fipsLookup = { has() { return true; } };

  return array.filter(item => 
    item[streetIndex] === street && fipsLookup.has(item[fipsIndex])
  )
}

var test_array =[
  ["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
  ["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
  ["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
  ["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
  ["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
  ["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
  ["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]
];

var street = "Fernwood Avenue";
var fips = ["34011", "34007"]

let result = findStuff(test_array, street, fips);
for (const item of result)
  console.log(JSON.stringify(item)); //more concise display
  
console.log("-----");

fips = []

result = findStuff(test_array, street, fips);
for (const item of result)
  console.log(JSON.stringify(item)); //more concise display

*奖励：假设有一个函数，其中街道和 fips 作为列表参数进入，其中街道名称将始终存在，但 fips 值可能会或可能不会。

从此更改函数声明：

function findStuff(array, street, fips)

到这里：

function findStuff(array, [street, ...fips])

为了利用destructuring 将第一项作为街道，将collect the rest into an array 作为fips。函数的主体保持不变。

function findStuff(array, [street, ...fips]) {
  const streetIndex = 6;
  const fipsIndex = 3;
  
  let fipsLookup;
  if (fips.length > 0)
    fipsLookup = new Set(fips);
  else
    fipsLookup = { has() { return true; } };

  return array.filter(item => 
    item[streetIndex] === street && fipsLookup.has(item[fipsIndex])
  )
}

var test_array =[
  ["06101123", "0", "0.2", "34011", "Cumberland", "Local", "Fernwood Avenue"],
  ["02271163", "0.1", "0.22", "34003", "Bergen", "Local", "Fernwood Avenue"],
  ["04351186", "0.3", "0.59", "34007", "Camden", "Local", "Fernwood Avenue"],
  ["07131150", "0", "0.3", "34013", "Essex", "Local", "Beacon Street"],
  ["03041026", "0", "0.13", "34005", "Burlington", "Local", "Beacon Street"],
  ["20121109", "0.18", "0.43", "34039", "Union", "Local", "Jones Lane"],
  ["15141139", "0", "0.27", "34029", "Ocean", "Local", "Jones Lane"]
];

var keywords = ["Fernwood Avenue", "34011", "34007"];

let result = findStuff(test_array, keywords);
for (const item of result)
  console.log(JSON.stringify(item)); //more concise display
  
console.log("-----");

keywords = ["Fernwood Avenue"];

result = findStuff(test_array, keywords);
for (const item of result)
  console.log(JSON.stringify(item)); //more concise display

【讨论】：