add withEncode function

[EricCrosson]

withEncode is similar to withValidate:

withEncode clones an existing codec and overrides the existing
encode function with the given encode function. This can be useful
when you want to use t.type to combinatorially create codecs but
want to leverage the encode function to encode into a different type.

[gcanti]

Thanks @EricCrosson, I would add an optional name argument though:

export function withEncode<A, O, I, P>(
  codec: t.Type<A, O, I>,
  encode: (a: A) => P,
  name: string = codec.name
): t.Type<A, P, I> {
  return new t.Type(name, codec.is, codec.validate, encode)
}

[EricCrosson]

Excellent! That's a much better implementation too. Thanks for the suggestion @gcanti , I've pushed these updates to the PR