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从列表中删除列表

list javascript

-1

Preston · 技术社区 · 7 年前

我有如下清单:

myList = [
["ooooh", "that"],
["dog ", "of"],
["mine", "word...."]
]

我现在要删除列表项: ["ooooh", "that"]

假设我知道元素0是 "ooooh" 元素1是 "that" ,但是 我不知道元素在所附列表中的位置 ,这是否可以在没有循环的情况下实现?

我的研究得出的一般回答似乎是“delete mylist['1']”,或者知道数组中元素的数量——在我的情况下,需要元素0&1都匹配才能删除。

在伪代码中,我希望:

myList.destroy(["ooooh", "that"])

如何才能做到这一点?

7 回复 | 直到 7 年前

Gerardo BLANCO 7 年前

你可以用 splice 从列表中删除项目。

myList.splice(
   myList.findIndex( item => 
     item[0] == "ooooh" && item[1] == "that"
   ), 
1);

希望这有帮助:>

myList = [
["ooooh", "that"],
["dog ", "of"],
["mine", "word...."]];

myList.splice(myList.findIndex(item => item[0] == "ooooh" && item[1] == "that"), 1);

console.log(myList)

Nelson Teixeira 7 年前

正如其他答案所指出的,有几种方法可以做到这一点,但我认为最简单的方法是使用过滤器。这样地::

myList = [
    ["ooooh", "that"],
    ["dog ", "of"],
    ["mine", "word"]
];
    
myList = myList.filter((element)=>
    //filter condition 
    (!
       ((element[0] == 'ooooh') && (element[1] == 'that'))
    )
);

console.log(myList)

Luis felipe De jesus Munoz 7 年前

只需过滤数组

var myList = [
["ooooh", "that"],
["dog ", "of"],
["mine", "word...."]
]

deleteFromList = (val, list) => list.filter(a=>JSON.stringify(a)!=JSON.stringify(val))

console.log(
  deleteFromList(["ooooh", "that"], myList)
)

/* Same as */

myList = myList.filter(a=>JSON.stringify(a)!=JSON.stringify(["ooooh", "that"]))

console.log(myList)

Vanojx1 7 年前

你可以使用 filter 函数将数组作为字符串进行比较

myList = [
  ["ooooh", "that"],
  ["dog ", "of"],
  ["mine", "word...."]
];

let findMe = ["ooooh", "that"]

myList = myList.filter((curr) => curr.join() !== findMe.join())

console.log(myList)

baao 7 年前

你可以使用过滤器和每一个一起有它的灵活性。

const myList = [
  ["ooooh", "that"],
  ["dog ", "of"],
  ["mine", "word...."]
];

const toDelete = ["ooooh", "that"];

const res = myList.filter(e => ! toDelete.every(x => e.includes(x)));
console.log(res);

Alex 7 年前

我来晚了,但我的意见是:

let myList = [
  ["ooooh", "that"],
  ["dog ", "of"],
  ["mine", "word...."]
];

Array.prototype.destroy = function(array) {
  this.forEach((el, i) => {
    if (JSON.stringify(el) === JSON.stringify(array)) {
      this.splice(i, 1);
    }
  });
}

console.log(myList);

myList.destroy(["ooooh", "that"]);

console.log(myList);

Pitto 7 年前

你也可以使用 pop 从数组中移除项。
Here 它的文档。

myArray = [
["ooooh", "that"],
["dog ", "of"],
["mine", "word..."]
]

console.log("Original Array:");
console.log(myArray);

/**
 * Removes one array from an array of arrays 
 *
 * @param  {String} myArray       - The original array of arrays
 * @param  {String} arrayToRemove - The array that will be removed from myArray
 */
function destroy(myArray, arrayToRemove) {
        arrayIndexToBeRemoved = myArray.findIndex(item => item[0] == arrayToRemove[0] && item[1] == arrayToRemove[1]);
        if (arrayIndexToBeRemoved != -1){
            removedArray = myArray.pop(arrayIndexToBeRemoved);
            console.log("Removed: " + removedArray);
        } else {
            console.log("Couldn't find: " + arrayToRemove + " in: " + myArray);
        }
}

destroy(myArray, ["mine", "word..."]);

console.log("Final Array:");
console.log(myArray);