Javascript 获取关于组的所有uniq值组合

我有一组属性值：

例如：

[
  {
    memberAttribute: { attributeName: 'a' },
    value: '1'  
  },
  {
    memberAttribute: { attributeName: 'a' },
    value: '2'  
  },
  {
    memberAttribute: { attributeName: 'b' },
    value: '1'  
  },
  {
    memberAttribute: { attributeName: 'b' },
    value: '2'  
  }
]

现在我想获得给定成员属性的所有唯一组合

因此，如果我想要成员属性“a”和“b”的唯一组合，结果将是：

[
   {
     memberAttribute: { attributeName: 'a' },
     value: '1'  
   },
   {
     memberAttribute: { attributeName: 'b' },
     value: '1'  
   }
],
[
   {
     memberAttribute: { attributeName: 'a' },
     value: '1'  
   },
   {
     memberAttribute: { attributeName: 'b' },
     value: '2'  
   }
],
[
   {
     memberAttribute: { attributeName: 'a' },
     value: '2'  
   },
   {
     memberAttribute: { attributeName: 'b' },
     value: '1'  
   }
],
[
   {
     memberAttribute: { attributeName: 'a' },
     value: '2'  
   },
   {
     memberAttribute: { attributeName: 'b' },
     value: '2'  
   }
]

我需要能够给出n个输入成员属性，但只能获得2个输入属性的期望结果

当前的解决方案：

export const getAttributeCombinations = (
  attributes: MemberAttributeValue[]
) => {
  // TODO - This algorithm only supports 2 attribute types
  // It should support any number of attribute types

  const combinations = new Array<Array<MemberAttributeValue>>();

  for (const attribute of attributes) {
    let unusedAttributes = allExcept(attribute, attributes);

    const permutate = () => {
      const combination = [attribute];
      const toRemove = new Array<Number>();

      for (let i = 0; i < unusedAttributes.length; i++) {
        const unusedAttribute = unusedAttributes[i];

        if (!attributeTypeAlreadyExists(unusedAttribute, combination)) {
          toRemove.push(i);
          combination.push(unusedAttribute);
        }
      }

      for (const index of toRemove) {
        unusedAttributes = remove(index, 1, unusedAttributes);
      }

      combinations.push(combination);
    };

    permutate();

    while (unusedAttributes.length > 0) {
      permutate();
    }
  }

  const sortedCombinations = map(sortByAttributeName, combinations);
  return uniqByCombination(sortedCombinations);
};

---

这在Ramda中相对简单，除了Ramda的

xprod

函数只能在两个列表上工作。如果它在列表上工作，我们只需几个步骤就可以完成。但是写我们自己的很容易：

const xproduct = reduce(pipe(xprod, map(unnest)), [[]])

const transform = pipe(
  groupBy(path(['memberAttribute', 'attributeName'])),
  values,
  xproduct
)

const inclusionAttributes = [
  {"memberAttribute": {"attributeName": "gender"}, "value": "Male"}, 
  {"memberAttribute": {"attributeName": "gender"}, "value": "Female"}, 
  {"memberAttribute": {"attributeName": "age band"}, "value": "0-50"}, 
  {"memberAttribute": {"attributeName": "age band"}, "value": "51+"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "cats"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "dogs"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "goats"}
]

console.log(transform(inclusionAttributes))
//=> Male/0-50/cats, Male/0-50/dogs, Male/0-50/goats, Male/51+/cats,...

---

您可以在

上看到这一点。当前解决方案有效，但仅适用于2个属性组。我需要能够支持n组b的输入属性。因此，可能存在成员属性c和d，每个属性都有x个唯一值。您应该能够按a、b、c和d进行分组。@lukejkw您能为3或4属性组提供一个示例对象（输入对象）吗？它看起来与给出的示例非常相似，但属性名称不同。你想让我更新这个问题吗？还需要注意的是，一个属性类型可以有多个值。所以“a”可以有很多不同的值@KoushikChatterjeehm，不要改变这一点，如果需要，可以为3个属性添加另一个示例输入来更新问题possible@KoushikChatterjee我已经为3个属性添加了失败的jest测试，增加了答案。我知道有一种简单优雅的方法可以做到这一点，拉姆达。

const xproduct = reduce(pipe(xprod, map(unnest)), [[]])

const transform = pipe(
  groupBy(path(['memberAttribute', 'attributeName'])),
  values,
  xproduct
)

const inclusionAttributes = [
  {"memberAttribute": {"attributeName": "gender"}, "value": "Male"}, 
  {"memberAttribute": {"attributeName": "gender"}, "value": "Female"}, 
  {"memberAttribute": {"attributeName": "age band"}, "value": "0-50"}, 
  {"memberAttribute": {"attributeName": "age band"}, "value": "51+"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "cats"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "dogs"}, 
  {"memberAttribute": {"attributeName": "likes"}, "value": "goats"}
]

console.log(transform(inclusionAttributes))
//=> Male/0-50/cats, Male/0-50/dogs, Male/0-50/goats, Male/51+/cats,...