I have array like this, my id and name will be same for multiple objects but organizations values can be different

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} 
];

I want to convert this to be like this:

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

Can someone please help

CodePudding user response：

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)

CodePudding user response：

You'll want to reduce.

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)];
}, []);

CodePudding user response：

You can achieve the output using forEach by grouping based on name and then pushing the necessary fields into the output array.

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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CodePudding user response：

I think it may be the easiest way to tackle this problem, only using the forEach and basic arrays' method.

I hope I answered your question.

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);

CodePudding user response：

What you are looking for is called a hashmap. You can read about them but the basic idea is that you make a key,value pair and access to the data with keys is very efficient(O(1) amortized). So here is one way to solve this problem in python. I am sure you can use it to solve it in your language.

    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)

CodePudding user response：

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;
    }, [])

CodePudding user response：

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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