如何从数组中删除哪些元素是另一个数组的元素？

Question

I have two arrays which values are:

const allValues = ["111", "111", "111", "222", "111", "222"]

const removingValues = ["111", "111", "222"]

And I want this array.

const remaningValues = ["111", "222", "111"]

I have to tried _.difference(allValues, removingValues) but it given empty array. How can I do that in simple way?

the first array includes "111" : 4 times and "222" : 2 times the removing values from this array are "111" : 2 times "222" : 1 times and the remaining values should be : "111" : 4-2 = 2 and "222" : 2-1 = 1

Answer 1

You could use this function:

 const allValues = ["111", "111", "111", "222", "111", "222"]; const removingValues = ["111", "111", "222"]; const result = allValues.filter(s => { let i = removingValues.indexOf(s); return i > -1 && removingValues.splice(i, 1); }); console.log(result); 

The above has a time complexity of O(n²) , so if you are working with large arrays it will be better to create a hash of some sort, for example using a Map , which will give a O(n) time complexity:

 const allValues = ["111", "111", "111", "222", "111", "222"]; const removingValues = ["111", "111", "222"]; const map = new Map(removingValues.map(s => [s, 0])); removingValues.forEach(s => map.set(s, map.get(s)+1)); const result = allValues.filter(s => { let i = map.get(s); return i && map.set(s, i-1); }); console.log(result); 

Answer 2

The order of your expected remaining values makes little sense to me. If you are ok with ['111', '111', '222'] (instead of ['111', '222', '111'] ), then this simple code should do it:

 const without = (i, xs) => i > -1 ? xs.slice(0, i).concat(xs.slice(i + 1)) : xs const removeAll = (all, rems, idx = all .indexOf (rems[0])) => rems .length ? removeAll (without (idx, all), rems .slice(1)) : all const allValues = ["111", "111", "111", "222", "111", "222"]; const removingValues = ["111", "333", "111", "222"]; console .log ( removeAll (allValues, removingValues) ) 

The helper function without just return a copy of the old array with the given index not included.

removeAll is a simple recursion on the things to remove. If it's empty, we return the remaining list. Otherwise, we remove from our list the first index that matches the first element to remove, and pass that and the remaining items to match back into removeAll .

Answer 3

You can do this in a very concise way with Array.reduce , Array.splice and Array.indexOf like this:

 const vals = ["111", "111", "111", "222", "111", "222"] const rms = ["111", "111", "222"] let r = rms.reduce((r,c) => (r.splice(r.indexOf(c), 1) && r), [...vals]) console.log(r) 

The idea is to start from the values to remove and in the Array.reduce put as a starting value for the accumulator an actual copy of the allValues . Then inside of the reduce just remove the values which match the indexOf via Array.splice

The copy of the vals is so that we do not mutate the original array. If you do not care about mutating that you can just have:

let r = rms.reduce((r,c) => (r.splice(r.indexOf(c), 1) && r), vals)