B中A与函数规划的相对补集(relative complement of A in B with functional programming)

B中A与函数规划的相对补集(relative complement of A in B with functional programming)

我必须检索仅存在于阵列B上但在阵列A上不存在的值。

从我的研究来看，它被称为：

B中A的相对补充

数组中的值可能不是原始的。我需要一个有效且功能性的应用程序来解决这个问题。 我找到了lodash _.without函数，但它只支持原始数组。

数组A：

[{ id: 1 }, { id:2 }]

数组B：

[{ id:2 }, { id:3 }]

结果应该是：

[{ id:3 }]

此对象是唯一存在于阵列B上的对象，但不存在于阵列A上。

I have to retrieve the values that exist only on Array B, but do not exist on Array A.

From my research, It is called:

relative complement of A in B

Values in the arrays may not be primitives.I need an efficient and functional apporach to this problem. I have found lodash _.without function, but it supports only array of primitive numbers.

Array A:

[{ id: 1 }, { id:2 }]

Array B:

[{ id:2 }, { id:3 }]

result should be:

[{ id:3 }]

this object is the only one who exist on Array B, but not on Array A.

最满意答案

你可以使用一个比较函数，它接受两个对象并检查id是否不相等。

var aa = [{ id: 1 }, { id: 2 }],
    bb = [{ id: 2 }, { id: 3 }],
    comparison = (a, b) => a.id !== b.id,
    result = bb.filter(b => aa.every(a => comparison(a, b)));

console.log(result); 
  
 
 
 检查是否平等 
 
 
 
  
  
var aa = [{ id: 1 }, { id: 2 }],
    bb = [{ id: 2 }, { id: 3 }],
    comparison = (a, b) => a.id === b.id,
    result = bb.filter(b => aa.every(a => !comparison(a, b)));

console.log(result); 
  
 
You could use a comparison function which takes two objects and check the id for unequalness.
 
 
 
  
  
var aa = [{ id: 1 }, { id: 2 }],
    bb = [{ id: 2 }, { id: 3 }],
    comparison = (a, b) => a.id !== b.id,
    result = bb.filter(b => aa.every(a => comparison(a, b)));

console.log(result); 
  
 
 
With a check for equalness
 
 
 
  
  
var aa = [{ id: 1 }, { id: 2 }],
    bb = [{ id: 2 }, { id: 3 }],
    comparison = (a, b) => a.id === b.id,
    result = bb.filter(b => aa.every(a => !comparison(a, b)));

console.log(result);