Combine items based on their keys and corresponding values

Question

Combine items based on their keys and corresponding values

The original question has been updated for clarity. The desired output is to merge an array of objects into a single object with specific conditions.

I am attempting to consolidate three sets of keys per object into one instance, choosing the lowest number for the Rank value or '-' if all Ranks are '-'

Below is sample data:


let objs = [
  {
    Keyword: 'A keyword',
    'First Rank': '-',
    'Second Rank': '-',
    'Third Rank': 1,
  },
  {
    Keyword: 'A keyword',
    'First Rank': '-',
    'Second Rank': 2,
    'Third Rank': '-',
  },
  {
    Keyword: 'A keyword',
    'First Rank': '-',
    'Second Rank': '-',
    'Third Rank': 7,
  }
]

// Code to filter and merge the objects...

Currently struggling to include objects with no values in the final merged result. Desired output shown below:

{
  'Keyword': 'A keyword',
  'First Rank': '-',
  'Second Rank': 2,
  'Third Rank': 1,
}

Any suggestions on how to achieve this? Thank you!

javascript arrays ecmascript-6 javascript-objects

Answer 1

Answer №1

Why not give this a try?

let objects = [
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': '-',
    'Third Rank': '-',
    'Third Title': '-'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': 'https://example.com',
    'Third Rank': 7,
    'Third Title': 'Title'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': 'https://example.com',
    'Second Rank': 11,
    'Second Title': 'Title',
    'Third URL': '-',
    'Third Rank': '-',
    'Third Title': '-'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': 'https://example.com',
    'Third Rank': 1,
    'Third Title': 'Title overwritten'
  },
]


var output = objects.slice().reduce((accumulator, item, index) => {
  resultItem = accumulator.find(accItem => accItem['Keyword'] === item['Keyword']);
  if (!resultItem) {
    accumulator.push(item)
  } else {
    Object.keys(resultItem).map(k => {
      if (item[k] !== '-') {
        if (typeof item[k] === 'number') {
          resultItem[k] = resultItem[k] === '-' ? item[k] : Math.min(item[k], resultItem[k]);
        } else {
          resultItem[k] = item[k];
        }
      }
    })
  }
  return accumulator;  
}, []);

console.log(output);

Answer 2

Why not give this a try?

let objects = [
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': '-',
    'Third Rank': '-',
    'Third Title': '-'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': 'https://example.com',
    'Third Rank': 7,
    'Third Title': 'Title'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': 'https://example.com',
    'Second Rank': 11,
    'Second Title': 'Title',
    'Third URL': '-',
    'Third Rank': '-',
    'Third Title': '-'
  },
  {
    'Keyword': 'A keyword',
    'First URL': '-',
    'First Rank': '-',
    'First Title': '-',
    'Second URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Third URL': 'https://example.com',
    'Third Rank': 1,
    'Third Title': 'Title overwritten'
  },
]


var output = objects.slice().reduce((accumulator, item, index) => {
  resultItem = accumulator.find(accItem => accItem['Keyword'] === item['Keyword']);
  if (!resultItem) {
    accumulator.push(item)
  } else {
    Object.keys(resultItem).map(k => {
      if (item[k] !== '-') {
        if (typeof item[k] === 'number') {
          resultItem[k] = resultItem[k] === '-' ? item[k] : Math.min(item[k], resultItem[k]);
        } else {
          resultItem[k] = item[k];
        }
      }
    })
  }
  return accumulator;  
}, []);

console.log(output);

Answer 3

Answer №2

To streamline the process, you can utilize the reduce function and then within the handler, inspect the numerical value to determine which is lower among the two values.

Noteworthy in this code excerpt is the introduction of an extra object with Third Rank = 1.

let array = [{ Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': 'https://example.com', 'Third Rank': 7, 'Third Title': '2 third title' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': 'https://example.com', 'Second Rank': 11, 'Second Title': '3 Second title', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': 'https://example.com', 'Second Rank': 11, 'Second Title': '3 Second title', 'Third URL': '-', 'Third Rank': 1, 'Third Title': '-' }], merged = array.reduce(({ ...r }, o) => { Object.entries(o).forEach(([k, v]) => { if (v !== '-') { if (isNaN(+v) || isNaN(+r[k])) r[k] = v; else r[k] = Math.min(+r[k], +v); } }); return r; }); console.log(merged);

Answer 4

To streamline the process, you can utilize the reduce function and then within the handler, inspect the numerical value to determine which is lower among the two values.

Noteworthy in this code excerpt is the introduction of an extra object with Third Rank = 1.

let array = [{ Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': 'https://example.com', 'Third Rank': 7, 'Third Title': '2 third title' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': 'https://example.com', 'Second Rank': 11, 'Second Title': '3 Second title', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { Keyword: 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': 'https://example.com', 'Second Rank': 11, 'Second Title': '3 Second title', 'Third URL': '-', 'Third Rank': 1, 'Third Title': '-' }], merged = array.reduce(({ ...r }, o) => { Object.entries(o).forEach(([k, v]) => { if (v !== '-') { if (isNaN(+v) || isNaN(+r[k])) r[k] = v; else r[k] = Math.min(+r[k], +v); } }); return r; }); console.log(merged);

Answer 5

Answer №3

To simplify the array, you can analyze the keys of each entry.

let array = [{ 'Keyword': 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { 'Keyword': 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': 'https://for-seven.example.com', 'Third Rank': 7, 'Third Title': 'Title' }, { 'Keyword': 'A keyword...

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 6

To simplify the array, you can analyze the keys of each entry.

let array = [{ 'Keyword': 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': '-', 'Third Rank': '-', 'Third Title': '-' }, { 'Keyword': 'A keyword', 'First URL': '-', 'First Rank': '-', 'First Title': '-', 'Second URL': '-', 'Second Rank': '-', 'Second Title': '-', 'Third URL': 'https://for-seven.example.com', 'Third Rank': 7, 'Third Title': 'Title' }, { 'Keyword': 'A keyword...

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 7

Answer №4

Appreciation to all contributors who provided solutions to this query. However, I realized that I omitted certain requirements in my initial explanation and by the time I clarified them, it was too late to receive follow-up responses. Hence, I took matters into my own hands and devised a solution that, although functional, may not be the most optimal approach.

let objs = [
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Second URL': '-',
    'Third Rank': 1,
    'Third Title': 'Title for 1',
    'Third URL': 'https://for-one.example.com'
  },
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': 2,
    'Second Title': 'Title for 2',
    'Second URL': 'https://for-two.example.com',
    'Third Rank': '-',
    'Third Title': '-',
    'Third URL': '-'
  },
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Second URL': '-',
    'Third Rank': 7,
    'Third Title': 'Title for 7',
    'Third URL': 'https://for-seven.example.com'
  }
]

// Code logic for filtering and merging object data goes here

console.log(result)

Answer 8

Appreciation to all contributors who provided solutions to this query. However, I realized that I omitted certain requirements in my initial explanation and by the time I clarified them, it was too late to receive follow-up responses. Hence, I took matters into my own hands and devised a solution that, although functional, may not be the most optimal approach.

let objs = [
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Second URL': '-',
    'Third Rank': 1,
    'Third Title': 'Title for 1',
    'Third URL': 'https://for-one.example.com'
  },
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': 2,
    'Second Title': 'Title for 2',
    'Second URL': 'https://for-two.example.com',
    'Third Rank': '-',
    'Third Title': '-',
    'Third URL': '-'
  },
  {
    'Keyword': 'A keyword',
    'First Rank': '-',
    'First Title': '-',
    'First URL': '-',
    'Second Rank': '-',
    'Second Title': '-',
    'Second URL': '-',
    'Third Rank': 7,
    'Third Title': 'Title for 7',
    'Third URL': 'https://for-seven.example.com'
  }
]

// Code logic for filtering and merging object data goes here

console.log(result)

Combine items based on their keys and corresponding values

Answer №1

Answer №2

Answer №3

Answer №4

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