array reduce not correctly inferring types

Question

i am having some troubles with array reduce in Typescript. To simplify, let's say I have a simple array of numbers where I want to remove duplicates and return a new array without them I was used to do something like this using reduce:

const new = nums.reduce((acc, item) => acc.includes(item) ? acc : [...acc, item], [])

where: nums = [0, 0, 1, 1, 1, 2, 2, 3, 3, 4] and new should be: new = [0, 1, 2, 3, 4]

I have tried to type the function like this:

const new: number[] = nums.reduce((acc: number[], item:number) => acc.includes(item) ? acc : [...acc, item], [])

I receive the following error for "new":

TS2322: Type 'number' is not assignable to type 'number[]'.

and an error on the accumulator:

TS2769: No overload matches this call.

Seems like there is no way of telling typescript that the accumulator should be an array of numbers, any solution?

Try with the Array constructor const new: number[] = nums.reduce((acc: number[], item:number) => acc.includes(item) ? acc : [...acc, item], new Array<number>()); — Kunal Mukherjee
– Kunal Mukherjee, Commented Oct 29, 2020 at 10:03

KiraLT · Accepted Answer · 2020-10-29 10:14:31Z

3

According to reduce function types:

reduce<U>(callbackfn: (previousValue: U, currentValue: T, currentIndex: number, array: T[]) => U, initialValue: U): U;

The return value is inferred from the initialValue. So you can either cast initialValue:

nums.reduce((acc, item) => acc.includes(item) ? acc : [...acc, item], [] as number[])

Or rewrite template argument:

nums.reduce<number[]>((acc, item) => acc.includes(item) ? acc : [...acc, item], [])

answered Oct 29, 2020 at 10:14

KiraLT

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1 Comment

Filippo Rivolta Over a year ago

Thanks that was something I missed

Orinayo Oyelade · Accepted Answer · 2020-10-29 10:06:56Z

2

Do this nums.reduce<number[]>(...) to tell TypeScript what reduce will return

answered Oct 29, 2020 at 10:06

Orinayo Oyelade

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