let operator for mathjs?

[gwhitney]

Sometimes it is useful to define a sub-expression and then use the result of that, perhaps multiple times, in another expression. For example, you might calculate the index of the largest triangular number less than n via floor((sqrt(1+8*n)-1)/2) and then want to call that t and compute how much larger n is than that triangular number with n - (t+1)t/2.

A while back someone posted on one of the issues or discussions here a trick to do just this (but sadly I can't find the exact reference at the moment). It was to write

[t = floor((sqrt(1+8*n)-1)/2), n - (t+1)t/2][2]

It works, but this syntax is hard to read, and not suggestive of what is really going on, and if you add another preparatory assignment between the one for t and the value you want, you then need to change the 2 to a 3, etc. In other words, this is a workaround/hack.

A more elegant method is just to add a function called let that takes any number of arguments and returns its last one unchanged:

import {all, create} from 'mathjs'
const math = create(all)
math.import({
    'let': math.typed('let', {'...any': args => args[args.length - 1]}),
})

and now you can write let(t = floor((sqrt(1+8*n)-1)/2), n - (t+1)t/2) and it looks and works the way you might expect. So I am opening this discussion to consider whether it is worth adding something like let from other languages into the mathjs expression language: either exactly this function, or a different syntax, say one modeled on Ocaml like let t = Math.floor((Math.sqrt(1+8*n)-1)/2 in n - (t+1)t/2.

If there is some other existing way of doing this sort of thing that I have overlooked, I'd be happy to hear. I am aware you can write t = floor((sqrt(1+8*n)-1)/2); n - (t+1)t/2 and that works, but it produces the answer wrapped in a ResultSet, and there are no facilities for manipulating that ResultSet or converting it into a value, within the mathjs expression language.

[gwhitney]

The source of the "array trick" for let-like behavior was #3128, it turns out.