Map array items through several functions

Question

Is there a more elegant way then this to execute several functions in succession for each item in the array:

type Transform<T> = (o: T) => T;
type Item = { /* properties */ };

transform(input, transformers: Transform<Item>[]) {
  const items: Item[] = getItems(input);
  return items.map(item => {
    let transformed = item;
    tramsformers.forEach(t => transformed = t(transformed));
    return transformed;
  })
}

Answer 1

This is a great use case for reduce:

transform(input, transformers: Transform<Item>[]) {
  const items: Item[] = getItems(input);
  return items.map(item => transformers.reduce((val, transformer) => transformer(val), item));
}

Or perhaps more readably:

transform(input, transformers: Transform<Item>[]) {
  const items: Item[] = getItems(input);
  return items.map(
    item => transformers.reduce(
      (val, transformer) => transformer(val),
      item
    )
  );
}

Live Example:

function getItems(input) {
  return [
    "abcdefg",
    "1234567"
  ];
}
function transform(input, transformers) {
  const items = getItems(input);
  return items.map(item => transformers.reduce((val, transformer) => transformer(val), item));
}
const result = transform("x", [
  v => v.toUpperCase(),
  v => v.substring(1, v.length - 1)
]);
console.log(result);

As Nitzan Tomer points out, we could do away with the items constant:

transform(input, transformers: Transform<Item>[]) {
  return getItems(input).map(
    item => transformers.reduce(
      (val, transformer) => transformer(val),
      item
    )
  );
}

I frequently keep those sorts of things for debugging, but some good debuggers now make it easy to see the return value of functions before they return (Chrome's does), so if you removed it, you could step into getItems to see the items before the map.