Hi all, just looking for a bit of clarification on sequential collections. I'm reading the Joy of Clojure, and it states they're ordered sets of values, and this link also confirms: http://insideclojure.org/2015/01/02/sequences/

but when I enter (sequential? [2 3 1) it returns true.

The term ordered in this context, does it just mean that the order matters, whereas with something like a hashmap, the order doesn't matter?

yes. that is basically it.

Ace, thanks - thought so too, but would rather get confirmation than assume I've got it 🙂

Is it possible to do something like this (I know syntax here is wrong, but you get the idea): (def m {:x 3, :y 2, :z (+ :x :y)})

@jdeisenberg so z = 5?

Hello, if I have two arrays of integers and I want to take the greater of the two, but also have a condition say the max < 4 then to double the max, but only the first x times that max < 4, what would be the idiomatic way to express that? Do I have to use a recur or is there a way to accomplish that w/ map/reduce?

@jdeisenberg you can use a let there

@tbaldridge The idea is I want to use a key that's already in the map to define something later on in the map. (Presume that instead of 2 and 3 I had some really long, complex expressions)

@greg_arcara you can do that in a reduce, although it gets a bit tricky. Basically have the accumulator of the reduce be a hashmap and update it as needed.

@tbaldridge Ah, of course! Why didn't I think of that?!:thinking_face:

@tbaldridge can I do the reduction with two arrays?

ah you have two arrays....

yea, that's my issue

I can do it with one 🙂

map can take two seqs, so you can do: (map vector a b)

then either operate on the first or second of those elements

(map vector a b) is zip in Python (or some other languages you may know)

ah

that's a good approach

@tbaldridge Still not sure how the code would look to get that value for :z since I'd have to access the map currently being defined.

@tbaldridge thanks, I'll give this a go

@jdeisenberg you'd have to do this:

(def m (let [x 5 y 4]
        {:x x :y y :z (+ x y)}))

@tbaldrige Got it; thanks.

(Sorry about the mistype on the name there)

@tbaldridge thanks so much for your help, I was able to make the solution work.