如何从对象数组中删除重复项并组合该重复项的值？

Question

我有这样的数组，我的 id 和 name 对于多个对象将是相同的，但 organizations 的值可以不同

array= [
  {id: 1, name: "Test1", organizations: 1},
  {id: 1, name: "Test1", organizations: 2},
  {id: 1, name: "Test1", organizations: 3},
  {id: 2, name: "Test2", organizations: 4},
  {id: 2, name: "Test2", organizations: 5},
  {id: 2, name: "Test2", organizations: 6} 
];

我想把它变成这样：

expectedArray =  [
  {id: 1, name: "Test1", organizations: [1,2,3]},
  {id: 2, name: "Test2", organizations: [4,5,6]} 
];

有人可以帮忙

Answer 1

const array= [
  {id: 1, name: "Test1", organizations: 1},
  {id: 1, name: "Test1", organizations: 2},
  {id: 1, name: "Test1", organizations: 3},
  {id: 2, name: "Test2", organizations: 4},
  {id: 2, name: "Test2", organizations: 5},
  {id: 2, name: "Test2", organizations: 6} 
];

const mergeDuplicates= (field, uniqueField) => (source = [], value)=> {
    const target = source.find( item => item[uniqueField] == value[uniqueField] );
    if(target) target[field].push( value[field] );
    else source.push({...value, [field]: [ value[field] ] });
    return source;
}
const mergeOrganaizationsById = mergeDuplicates('organizations','id')
const result = array.reduce(mergeOrganaizationsById, [])
console.log(result)

Answer 2

你会想要减少。

array.reduce((acc, cur) => {
  const i = acc.findIndex(a => a.id === cur.id && a.name === cur.name);
  if (i === -1) return [...acc, {...cur, organizations: [cur.organizations]}];
  return [...acc.slice(0, i), {...cur, organizations: [...acc[i].organizations, cur.organizations]}, ...acc.slice(i +1)];
}, []);

Answer 3

您可以使用forEach 实现输出，方法是基于name 进行分组，然后将必要的字段推入输出数组。

const array = [
  { id: 1, name: "Test1", organizations: 1 },
  { id: 1, name: "Test1", organizations: 2 },
  { id: 1, name: "Test1", organizations: 3 },
  { id: 2, name: "Test2", organizations: 4 },
  { id: 2, name: "Test2", organizations: 5 },
  { id: 2, name: "Test2", organizations: 6 }
];

const current = Object.create(null);
const finalArr = [];
array.forEach(function (o) {
  if (!current[o.name]) {
   current[o.name] = [];
   finalArr.push({ id: o.id, name: o.name, organizations: current[o.name] });
  }
  current[o.name].push(o.organizations);
});
console.log(finalArr);

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Answer 4

ES6 风格的另一种解决方案

const array= [{id: 1, name: "Test1", organizations: 1},{id: 1, name: "Test1", organizations: 2},{id: 1, name: "Test1", organizations: 3},{id: 2, name: "Test2", organizations: 4},{id: 2, name: "Test2", organizations: 5},{id: 2, name: "Test2", organizations: 6}];

const result =  Object.values(array.reduce((acc, { id, name, organizations }) => {
  const hash = `${id}-${name}`;
  acc[hash] = acc[hash] 
    ? { ...acc[hash], organizations: [...acc[hash].organizations, organizations] } 
    : { id, name, organizations: [organizations] };
  return acc;
}, {}));

console.log(result);

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Answer 5

我认为这可能是解决这个问题的最简单方法，只使用 forEach 和基本数组的方法。

希望我回答了你的问题。

const array = [
  { id: 1, name: "Test1", organizations: 1 },
  { id: 1, name: "Test1", organizations: 2 },
  { id: 1, name: "Test1", organizations: 3 },
  { id: 2, name: "Test2", organizations: 4 },
  { id: 2, name: "Test2", organizations: 5 },
  { id: 2, name: "Test2", organizations: 6 }
];

const newArr = [];

// for keeping track for added item
const addedItems = [];

// for keeping track of added item idx
let addedItemIdx = 0;

array.forEach((item) => {
  if (!addedItems.includes(item.id)) {
    let tempOrg = item.organizations;
    newArr.push({ ...item, organizations: [tempOrg] });
    addedItems.push(item.id);
    addedItemIdx++;
  } else {
    newArr[addedItemIdx - 1].organizations.push(item.organizations);
  }
});
console.log(newArr);

Answer 6

您要查找的内容称为哈希图。您可以阅读它们，但基本思想是您制作一个键、值对并使用键访问数据非常有效（O(1) 摊销）。所以这是在python中解决这个问题的一种方法。我相信你可以用它来用你的语言解决它。

    array= [
  {"id": 1, "name": "Test1", "organizations": 1},
  {"id": 1, "name": "Test1", "organizations": 2},
  {"id": 1, "name": "Test1", "organizations": 3},
  {"id": 2, "name": "Test2", "organizations": 4},
  {"id": 2, "name": "Test2", "organizations": 5},
  {"id": 2, "name": "Test2", "organizations": 6}
]
# Initilize a hashmap
hash_map = {}
# Loop over all the items in array and create the hashmap
for item in array:
    # key will be id and name as both are needed to group the organizations
    # We have use comma to separate them as we assume that name or id cannot have comma
    key = str(item["id"])+","+item["name"]
    # IF key is already present then add the new organizations id to it
    if key in hash_map:
        hash_map[key].append(item["organizations"])
    # Else make a new list with the current organizations id
    else:
        hash_map[key] = [item["organizations"]]

# Create the expected array
expected_array = []
for key,value in hash_map.items():
    # Get the id and name by spliting the key that we created 
    idx,name = key.split(",")
    expected_array.append({"id":idx,"name":name,"organizations":value})
print(expected_array)

Answer 7

const array = [
  { id: 1, name: "Test1", organizations: 1 },
  { id: 1, name: "Test1", organizations: 2 },
  { id: 1, name: "Test1", organizations: 3 },
  { id: 2, name: "Test2", organizations: 4 },
  { id: 2, name: "Test2", organizations: 5 },
  { id: 2, name: "Test2", organizations: 6 }
];

const result=array.reduce((acc,curr)=>{
  const id=curr.id;
  const {organizations}=curr;
   const findIndex=acc.findIndex(item=> item.id===curr.id)
  if(findIndex===-1){
    acc.push({...curr,organizations:[organizations]});
  } else {
   
    acc[findIndex].organizations.push(curr.organizations)
  }
  return acc;
},[]);
console.log(result);

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Answer 8

这应该可以

const array = [{
    id: 1,
    name: "Test1",
    organizations: 1
  },
  {
    id: 1,
    name: "Test1",
    organizations: 2
  },
  {
    id: 1,
    name: "Test1",
    organizations: 3
  },
  {
    id: 2,
    name: "Test2",
    organizations: 4
  },
  {
    id: 2,
    name: "Test2",
    organizations: 5
  },
  {
    id: 2,
    name: "Test2",
    organizations: 6
  }
];
const reducedArray = array.reduce((resultArray, arrayElement) => {
  const elementIndex = resultArray.findIndex(element => element.id === arrayElement.id);

  if (elementIndex !== -1) {
    resultArray[elementIndex].organizations.push(arrayElement.organizations)
  } else {
    resultArray.push({
      ...arrayElement,
      organizations: [arrayElement.organizations],
    });
  }
  return resultArray;
}, []);

console.log(reducedArray)

Answer 9

const array = [{
    id: 1,
    name: "Test1",
    organizations: 1
  },
  {
    id: 1,
    name: "Test1",
    organizations: 2
  },
  {
    id: 1,
    name: "Test1",
    organizations: 3
  },
  {
    id: 2,
    name: "Test2",
    organizations: 4
  },
  {
    id: 2,
    name: "Test2",
    organizations: 5
  },
  {
    id: 2,
    name: "Test2",
    organizations: 6
  }
];


const formattedData = array.reduce((result, {
  id,
  name,
  organizations
}) => {
  let filteredRow = result.find(row => row.id === id && row.name === name);
  const org = filteredRow ? filteredRow.organizations : [];
  org.push(organizations);
  filteredRow = {
    id,
    name,
    organizations: org
  };
  if (org.length === 1) result.push(filteredRow);
  return result;
}, [])

console.log(formattedData)