reduce

[value] reduce (acc, item) => expression |> value | nothing
[value] reduce default using (acc, item) => expression |> value

Reduces an iterator to a single value using an accumulator function.

The reduce function folds an iterator into a single value by applying an accumulator function. It comes in two forms:

Without default value — The first item becomes the initial accumulator, and the function is called from the second item onward. If the iterator is empty, nothing is returned. This is sometimes called foldl1 in other functional languages.

With default value — Use the using keyword to provide an initial accumulator. The function is then called for every item, including the first. If the iterator is empty, the default value is returned.

Examples

Basic reduce

import 'iterators'

from [1, 2, 3] reduce (acc, item) => acc + item

Result:

6

With default value

The using keyword separates the default value from the accumulator function:

import 'iterators'

from [1, 2, 3] reduce 0 using (acc, item) => acc + item

Result:

6

Empty iterator without default

import 'iterators'

let numbers: [number] = []
from numbers reduce (acc, item) => acc + item

Result:

nothing

Empty iterator with default

When a default is provided, it is returned for empty iterators:

import 'iterators'

let numbers: [number] = []
from numbers reduce 99 using (acc, item) => acc + item

Result:

99

Product

import 'iterators'

from [1, 2, 3, 4] reduce (acc, item) => acc * item

Result:

24

Building a record from an iterator

import 'iterators'

from
  [{ key = 'a', value = 1 }, { key = 'b', value = 2 }]
  reduce {} using (acc, item) => acc with { '{item.key}' = item.value }

Result:

{ a = 1, b = 2 }