是否有使用Ramda库从数组数组中删除数组的简写选项?
Items to remove: [[1, 2], [a, b]]
Remove from: [[g, d], [5, 11], [1, 2], [43, 4], [a, b]]
Result: [[g, d], [5, 11], [43, 4]]
答案 0 :(得分:2)
R.reject,R.either和R.equals的组合可以达到此目的。
const data = [['g', 'd'], [5, 11], [1, 2], [43, 4], ['a', 'b']]
const fn = R.reject(R.either(R.equals(['a', 'b']), R.equals([1, 2])))
console.log(fn(data))

<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>
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答案 1 :(得分:2)
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const data = [['g', 'd'], [5, 11], [1, 2], [43, 4], ['a', 'b']]
const itemsToRemove = [[1, 2], ['a', 'b']]
const fn = R.flip(R.difference)(itemsToRemove);
console.log(fn(data))
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答案 2 :(得分:0)
使用vanilla JavaScript,您可以复制两个数组中的每一个,.map()
返回数组元素的string
表示。
然后循环遍历可移动项目数组,并使用indexOf()
检查复制数组中每个项目是否存在。
这应该是你的代码:
var arr = [["g", "d"], [5, 11], [1, 2], [43, 4], ["a", "b"]];
var strArr = arr.map(function(a){
return a.join(",");
});
var toRemove = [[1, 2], ["a", "b"]];
var strToRemove = toRemove.map(function(el){
return el.join(",");
});
strToRemove.forEach(function(a){
if(strArr.indexOf(a)>-1){
arr.splice(strArr.indexOf(a), 1);
strArr.splice(strArr.indexOf(a), 1);
}
});
<强>演示:强>
var arr = [
["g", "d"],
[5, 11],
[1, 2],
[43, 4],
["a", "b"]
];
var strArr = arr.map(function(a) {
return a.join(",");
});
var toRemove = [
[1, 2],
["a", "b"]
];
var strToRemove = toRemove.map(function(el) {
return el.join(",");
});
strToRemove.forEach(function(a) {
if (strArr.indexOf(a) > -1) {
arr.splice(strArr.indexOf(a), 1);
strArr.splice(strArr.indexOf(a), 1);
}
});
console.log(arr);
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答案 3 :(得分:0)
这可能是最好的答案
var Itemstoremove = [[1, 2], [a, b]]
var Removefrom = [[g, d], [5, 11], [1, 2], [43, 4], [a, b]]
var Result = R.reject(R.contains(R.__, Itemstoremove), Removefrom)
console.log(Result)
答案 4 :(得分:0)
只需使用以下代码行:
const data = [
['g', 'd'],
[5, 11],
[1, 2],
[43, 4],
['a', 'b']
]
const itemsToRemove = [
[1, 2],
['a', 'b']
];
console.log(R.difference(data, itemsToRemove));