Arrange the names in ascending order within arrays in an array

Question

Arrange the names in ascending order within arrays in an array

If we consider a scenario where an array of names is provided,

let names = [
  [
    ['firstName', 'Rachel'],
    ['age', 10],
    ['gender', 'female'],
  ],
  [
    ['firstName', 'Sam'],
    ['lastName', 'Smith'],
    ['age', 20],
    ['gender', 'male'],
  ],
];

The desired output would resemble the code snippet below.

let output = nameInOrder(names);
console.log(output); // --> ['Rachel', 'Sam Smith'];

In order to successfully implement the nameInOrder function,

function nameInOrder(arr) {
// insert code here
}

Would it be appropriate to start with the .map method for transforming the data into objects within an array, then utilize sort(a,b) and .join(' ')? How should they be sorted in ascending order?

javascript arrays object higher-order-functions

Answer 1

Answer №1

.map is a step in the right direction for data transformation and sorting.

let members = [
  [
    ['name', 'Alice'],
    ['age', 30],
    ['gender', 'female'],
  ],
  [
    ['name', 'Bob'],
    ['lastName', 'Johnson'],
    ['age', 25],
    ['gender', 'male'],
  ],
];

function sortNames(arr) {
  return arr.map(member => {
    const attributes = member.map(info => info[0])
    if (attributes.includes('name') && attributes.includes('lastName')) {
      return [
        member.find(info => info[0] == 'name')[1],
        member.find(info => info[0] == 'lastName')[1],
      ].join(' ')
    } else if (attributes.includes('name')) {
      return member.find(info => info[0] == 'name')[1]
    } else if (attributes.includes('lastName')) {
      return member.find(info => info[1] == 'lastName')[1]
    } else {
      return ''
    }
  }).sort()
}

console.log(sortNames(members))

Answer 2

.map is a step in the right direction for data transformation and sorting.

let members = [
  [
    ['name', 'Alice'],
    ['age', 30],
    ['gender', 'female'],
  ],
  [
    ['name', 'Bob'],
    ['lastName', 'Johnson'],
    ['age', 25],
    ['gender', 'male'],
  ],
];

function sortNames(arr) {
  return arr.map(member => {
    const attributes = member.map(info => info[0])
    if (attributes.includes('name') && attributes.includes('lastName')) {
      return [
        member.find(info => info[0] == 'name')[1],
        member.find(info => info[0] == 'lastName')[1],
      ].join(' ')
    } else if (attributes.includes('name')) {
      return member.find(info => info[0] == 'name')[1]
    } else if (attributes.includes('lastName')) {
      return member.find(info => info[1] == 'lastName')[1]
    } else {
      return ''
    }
  }).sort()
}

console.log(sortNames(members))

Answer 3

Answer №2

If you want to sort names in JavaScript arrays, you can utilize the .sort() method by referring to the documentation here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort.

Create a function that compares the names when using .sort().

names.sort((a,b) => {
  var nameA = a[0][1].toUpperCase(); // disregard case sensitivity
  var nameB = b[0][1].toUpperCase(); // disregard case sensitivity
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names are equal
  return 0;
});

It is recommended not to structure your data like this and instead consider the following format:

let names = [
  [
    { 'firstName': 'Rachel' },
    { 'age': 10 },
    { 'gender': 'female' }
  ]
];

In this format, sorting can still be done similarly without depending on the attribute order in the array of names.

names.sort((a,b) => {
  var nameA = a.firstName.toUpperCase(); // ignore case sensitivity
  var nameB = b.firstName.toUpperCase(); // ignore case sensitivity
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names are equal
  return 0;
});

Answer 4

If you want to sort names in JavaScript arrays, you can utilize the .sort() method by referring to the documentation here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort.

Create a function that compares the names when using .sort().

names.sort((a,b) => {
  var nameA = a[0][1].toUpperCase(); // disregard case sensitivity
  var nameB = b[0][1].toUpperCase(); // disregard case sensitivity
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names are equal
  return 0;
});

It is recommended not to structure your data like this and instead consider the following format:

let names = [
  [
    { 'firstName': 'Rachel' },
    { 'age': 10 },
    { 'gender': 'female' }
  ]
];

In this format, sorting can still be done similarly without depending on the attribute order in the array of names.

names.sort((a,b) => {
  var nameA = a.firstName.toUpperCase(); // ignore case sensitivity
  var nameB = b.firstName.toUpperCase(); // ignore case sensitivity
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names are equal
  return 0;
});

Answer 5

Answer №3

If the data source is limited to arrays only.

const items = [
  [
    ['itemName', 'Apple'],
    ['price', 1.50],
    ['category', 'fruit'],
  ],
  [
    ['itemName', 'Banana'],
    ['price', 0.75],
    ['category', 'fruit'],
  ],
];

const result = items.map(item => {
    const details = item.reduce((obj, [key, value]) => {
        obj[key] = value;
        return obj;
      }, {});
    return `${details.itemName || ''} - $${details.price || ''}`.trim();
});

console.log(result);

Answer 6

If the data source is limited to arrays only.

const items = [
  [
    ['itemName', 'Apple'],
    ['price', 1.50],
    ['category', 'fruit'],
  ],
  [
    ['itemName', 'Banana'],
    ['price', 0.75],
    ['category', 'fruit'],
  ],
];

const result = items.map(item => {
    const details = item.reduce((obj, [key, value]) => {
        obj[key] = value;
        return obj;
      }, {});
    return `${details.itemName || ''} - $${details.price || ''}`.trim();
});

console.log(result);

Arrange the names in ascending order within arrays in an array

Answer №1

Answer №2

Answer №3

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