我在JavaScript中创建了一个多维数组,我想找到具体值的确切索引。该值将是用户输入。
var array=[];
for(var i=0;i<10;i++)
{
array[i] = [];
for(var j=0;j<100;j++)
{
k = k +1
array[i].push(k);
}
}
var index=array.indexOf(**"What to insert here???"**);
答案 0 :(得分:15)
/**
* Index of Multidimensional Array
* @param arr {!Array} - the input array
* @param k {object} - the value to search
* @return {Array}
*/
function getIndexOfK(arr, k) {
for (var i = 0; i < arr.length; i++) {
var index = arr[i].indexOf(k);
if (index > -1) {
return [i, index];
}
}
}
// Generate Sample Data
var k = 0;
var array = [];
for (var i = 0; i < 10; i++) {
array[i] = [];
for (var j = 0; j < 100; j++) {
k = k + 1;
array[i].push(k);
}
}
var needle = 130;
var result = getIndexOfK(array, needle);
console.log('The value #' + needle + ' is located at array[' + result[0] + '][' + result[1] + '].');
&#13;
答案 1 :(得分:2)
这个例子似乎也适用于IRREGULAR多维数组:
function findIndex(valueToSearch, theArray, currentIndex) {
if (currentIndex == undefined) currentIndex = '';
if(Array.isArray(theArray)) {
for (var i = 0; i < theArray.length; i++) {
if(Array.isArray(theArray[i])) {
newIndex = findIndex(valueToSearch, theArray[i], currentIndex + i + ',');
if (newIndex) return newIndex;
} else if (theArray[i] == valueToSearch) {
return currentIndex + i;
}
}
} else if (theArray == valueToSearch) {
return currentIndex + i;
}
return false;
}
var a = new Array();
a[0] = new Array(1, 2, 3, 4, 5);
a[1] = 'ciao';
a[2] = new Array(new Array(6,7),new Array(8,9),10);
var specificIndex = findIndex('10', a);
我写得很快,所以每个人都被邀请改进这个功能!
P.S。现在该函数返回一个STRING值,所有索引用逗号分隔,您可以简单地编辑它以返回一个对象
答案 2 :(得分:1)
function indexOf2d(arr, val) {
var index = [-1, -1];
if (!Array.isArray(arr)) {
return index;
}
arr.some(function (sub, posX) {
if (!Array.isArray(sub)) {
return false;
}
var posY = sub.indexOf(val);
if (posY !== -1) {
index[0] = posX;
index[1] = posY;
return true;
}
return false;
});
return index;
}
console.log(indexOf2d(array, 50));
答案 3 :(得分:0)
我的代码就像MS Excel中的PROCV ...,并且仅在第一列中标识索引搜索。也许可以帮助您(或其他人)。
var convertToRoman = function (valueLimitTen) {
var convertTable = [ [1, "I"],
[2, "II"],
[3, "III"],
[4, "IV"],
[5, "V"],
[6, "VI"],
[7, "VII"],
[8, "VIII"],
[9, "IV"],
[10, "X"],
];
var myIndex;
for(var i in convertTable){
if(convertTable[i][0] == valueLimitTen){
myIndex = i;
return convertTable[i][1];
}
}
}
console.log(convertToRoman(2)); //Result II
console.log(convertToRoman(10)); //Result X
答案 4 :(得分:0)
ES6 让这件事变得相当容易。
function findIndexOfNestedArray(nestedArray, searchArray) {
return searchArray.findIndex(item => {
return item.length === nestedArray.length
&& item.every((a, i) => a === nestedArray[i])
})
}
答案 5 :(得分:0)
复制并粘贴此代码
替换 <masterArray> 与您的主阵列(嵌套所有其他阵列)
代码:
//Remove Duplicates
let stringArray = masterArray.map(item=>item.join('')); // transform all nested arrays into strings (for comparation)
let nonDuplicated = [] // temporary array to compare data
stringArray.forEach(nestedArrayString=>{
let reapeatedIndex = nonDuplicated.indexOf(nestedArrayString); // this is the comparator
if(reapeatedIndex>0){
indexToRemove = stringArray.indexOf(nestedArrayString);
console.log('---indexToRemove',indexToRemove)
stringArray.splice(indexToRemove,1);
masterArray.splice(indexToRemove,1);
}else{nonDuplicated.push(nestedArrayString)}
});
//
let lastString = stringArray.filter(foundString=>foundString===stringArray[stringArray.length-1])
if(lastString.length>1){
stringArray.splice(stringArray.length-1,1)
masterArray.splice(masterArray.length-1,1)
};
console.log('>> masterArray',masterArray)
我一直在寻找删除数组内的数组的解决方案,并提供了这个丑陋但高效的解决方案: 处理数组数组(两次),但不能进一步嵌套。
这段代码会把嵌套的数组转成字符串 => 一一记录成一个临时数组 => 每次记录比较当前记录的字符串是否已经在数组中 .then 如果重复,则从 中删除masterArray(你的)和 stringArray 元素