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I am new to typescript, and i've been having some problems in using my javascript skills. For example, can someone help me to write this exactly same javascript code below in typescript?

If not possible at all, any typescript function that will render the expected output (array without duplicate values).

This is just a simple way to remove duplicates from an array, but seems like typescript doesn't let me define an empty object... I'm not sure...

The output of the code below is: ['John', 'Paul', 'George', 'Ringo']

Thanks!

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];

function removeDups(names) {
  let unique = {};
  names.forEach(function(i) {
    if(!unique[i]) {
      unique[i] = true;
    }
  });
  return Object.keys(unique);
}

removeDups(names)

I am new to typescript, and i've been having some problems in using my javascript skills. For example, can someone help me to write this exactly same javascript code below in typescript?

If not possible at all, any typescript function that will render the expected output (array without duplicate values).

This is just a simple way to remove duplicates from an array, but seems like typescript doesn't let me define an empty object... I'm not sure...

The output of the code below is: ['John', 'Paul', 'George', 'Ringo']

Thanks!

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];

function removeDups(names) {
  let unique = {};
  names.forEach(function(i) {
    if(!unique[i]) {
      unique[i] = true;
    }
  });
  return Object.keys(unique);
}

removeDups(names)

• javascript
• typescript
• ecmascript-6

Share Improve this question Follow asked Aug 9, 2020 at 4:55 NandoNando 10311 silver badge55 bronze badges 2

• all javascript is proper typescript – frozen Commented Aug 9, 2020 at 4:56
• 1 @frozen not when you enable the strict pile flag, which is the default with tsc --init... – Patrick Roberts Commented Aug 9, 2020 at 5:18

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4 Answers 4

Sorted by: Reset to default 5

There's actually a very easy and performant way to remove duplicate elements from an array by leveraging the built-in behaviour of Set:

/**
 * Construct a copy of an array with duplicate items removed.
 * Where duplicate items exist, only the first instance will be kept.
 */
function removeDups<T>(array: T[]): T[] {
    return [...new Set(array)];
}

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];
console.log(removeDups(names)); // ["John", "Paul", "George", "Ringo"]

TypeScript Playground

That function converts your array to a Set, which removes duplicates faster than any native loop through the array will manage, and then uses spread syntax to convert that Set back into a new array.

By making the removeDups function use a generic type, it can be used with any type of array, not just string arrays like your example.

---

To convert the function you have written into TypeScript, here's what you can use:

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];

function removeDups(names: string[]): string[] {
  let unique: Record<string, boolean> = {};
  names.forEach(function(i) {
    if(!unique[i]) {
      unique[i] = true;
    }
  });
  return Object.keys(unique);
}

removeDups(names)

TypeScript Playground

This uses the utility type Record for your unique object's type. However, due to how Object.keys works, this function will only be able to work on arrays of strings. If you want to use this approach with all types of array, you could consider using a Map instead so you can index it by other types.

That would look something like this:

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];

function removeDups<T>(names: T[]): T[] {
  let unique: Map<T, boolean> = new Map();
  names.forEach(function(i) {
    if(!unique.has(i)) {
      unique.set(i, true);
    }
  });
  return Array.from(unique.keys());
}

console.log(removeDups(names)); // ["John", "Paul", "George", "Ringo"]

TypeScript Playground

names = ['John', 'Paul', 'George', 'Ringo', 'John']; 

removeDups(names) {
 let unique = {};
  this.names.forEach((i) => {
    if(!unique[i]) {
      unique[i] = true;
    }
  });
  return Object.keys(unique);
}

Then call removeDups from where you want

Here is another way:

names = ['John', 'Paul', 'George', 'Ringo', 'John']; 

removeDups(names) {

  return this.names.filter((elem, index, self)=> {
    return index === self.indexOf(elem);
})
}

In this case, you just have to add types

const names = ['John', 'Paul', 'George', 'Ringo', 'John'];

function removeDups(names: string[]) {
  let unique: any = {};
  names.forEach(function(i: string) {
    if(!unique[i]) {
      unique[i] = true;
    }
  });
  return Object.keys(unique);
}

removeDups(names)

typescript playground

Here is a simple and generic method to de-duplicate any array.

Js:
deDup = (arr) => arr.reduce((acc, current) => {
  if(!acc.includes(current)) acc.push(current)
  return acc
}, [])

Ts:
deDup = (arr: any[]) => arr.reduce((acc, current) => {
  if(!acc.includes(current)) acc.push(current)
  return acc
}, [] as any[])

Since this by definition is reducing an array by removing duplicate values, call reduce being sure to pass in an initial value of a new empty array (accumulator). On each index (current) if it's not in the accumulator array then add it, then return the accumulator array. This will quickly return a new array to the caller without any duplicates.

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