[英]Remove array from array of arrays using Ramda?
是否有任何速記選項可以使用 Ramda 庫從數組數組中刪除數組?
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]]
使用R.difference和R.flip :
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))
<script src="//cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.min.js"></script>
R.reject 、 R.either和R.equals 的組合可以實現這一點。
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>
這可能是最好的答案
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)
使用 vanilla JavaScript,您可以復制兩個數組中的每一個,使用.map()
返回數組元素的string
表示形式。
然后遍歷您的可移動項目數組並使用indexOf()
檢查您復制的數組中每個項目是否存在。
這是您的代碼應該如何:
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);
}
});
演示:
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);
只需使用以下代碼行:
const data = [ ['g', 'd'], [5, 11], [1, 2], [43, 4], ['a', 'b'] ] const itemsToRemove = [ [1, 2], ['a', 'b'] ]; console.log(R.difference(data, itemsToRemove));
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.