如何在JavaScript中检查一个数组是否包含另一个数组的所有元素，包括count？

Question

I have two arrays:

arr1 = [1, 2, 3, 1, 2, 3, 4]
arr2 = [1, 3, 1, 1]
arr3 = [1, 1, 2, 2, 3]

Using arr1 as the baseline, arr2 does not match because it contains three 1 's whereas arr1 only has two. arr3 , however should return true because it has elements from arr1 .

I tried

if(_.difference(arr2, arr1).length === 0)

But this does not take into account the number of occurrences

Answer 1

You could count all value from the first array and iterate the second with every and return early if a value is not given or zero, otherwise decrement the count and go on.

 function check(a, b) { var count = a.reduce((o, v) => (o[v] = (o[v] || 0) + 1, o), {}); return b.every(v => count[v] && count[v]--); } var arr1 = [1, 2, 3, 1, 2, 3, 4], arr2 = [1, 3, 1, 1], arr3 = [1, 1, 2, 2, 3]; console.log(check(arr1, arr2)); // false console.log(check(arr1, arr3)); // true 

Answer 2

You can try to loop through second array and compare t against main array if value found make main array cell to false and set flag as true

 arr1 = [1, 2, 3, 1, 2, 3, 4] arr2 = [1, 3, 1, 1] arr3 = [1, 1, 2, 2, 3] function checkArray(compareMe, toThis){ var flag = false; for(var i = 0; i < toThis.length; i++){ flag = false; for(var j = 0; j < compareMe.length; j++){ if(compareMe[j] == toThis[i]){ compareMe[j] = false; flag = true; break; } } } return flag; } console.log(checkArray(arr1, arr2)); console.log(checkArray(arr1, arr3)); 

Answer 3

Try this solution:

 const arr1 = [1, 2, 3, 1, 2, 3, 4], arr2 = [1, 3, 1, 1], arr3 = [1, 1, 2, 2, 3], arr4 = [1, 2, 3, 1, 2, 3, 4, 1]; function containsFn(src, trg) { const srcCp = src.slice(); for(let i = 0; i < trg.length; i++) { const index = srcCp.indexOf(trg[i]); if(index > - 1) { srcCp.splice(index, 1); } else { return false; } } return true; } console.log(containsFn(arr1, arr2)); console.log(containsFn(arr1, arr3)); console.log(containsFn(arr1, arr4)); 

Answer 4

Looks like I'm already late to the party but this would be a recursive solution:

 arr1 = [1, 2, 3, 1, 2, 3, 4] arr2 = [1, 3, 1, 1] arr3 = [1, 1, 2, 2, 3] function findAllIn(find, search) { if (find.length == 0) { return true; } i = search.indexOf(find[0]); if (i == -1) { return false; } return findAllIn(find.slice(1,find.length), search.slice(0,i).concat(search.slice(i+1, search.length))); } console.log(findAllIn(arr2, arr1)); // false console.log(findAllIn(arr3, arr1)); // true 

Answer 5

This should do the trick

Not efficient but I think it is easy to understand

 const count = (x, xs) => xs.reduce((y, xi) => y + (xi === x ? 1 : 0), 0) const isInclusive = (xs, ys) => xs.every((xi) => count(xi, xs) >= count(xi, ys)) const arr1 = [1, 2, 3, 1, 2, 3, 4] const arr2 = [1, 3, 1, 1] const arr3 = [1, 1, 2, 2, 3] console.log(isInclusive(arr1, arr2)) console.log(isInclusive(arr1, arr3)) 

Answer 6

Based on this answer: https://stackoverflow.com/a/4026828/3838031

You can do this:

 Array.prototype.diff = function(a) { return this.filter(function(i) {return a.indexOf(i) < 0;}); }; arr1 = [1, 2, 3, 1, 2, 3, 4] arr2 = [1, 3, 1, 1] arr3 = [1, 1, 2, 2, 3] console.log((arr1.diff(arr2).length === 0) ? "True" : "False"); console.log((arr1.diff(arr3).length === 0) ? "True" : "False"); console.log((arr2.diff(arr1).length === 0) ? "True" : "False"); console.log((arr2.diff(arr3).length === 0) ? "True" : "False"); console.log((arr3.diff(arr1).length === 0) ? "True" : "False"); console.log((arr3.diff(arr2).length === 0) ? "True" : "False");