命名空间映射名称空间上的多方法分派

Question

Is it possible to dispatch based on a namespaced map namespace ie #:<this-thing>{} ? Without hacks like printing or inspecting key prefixes?

I consider the last one hacky because a key prefix can be overridden:

(:qux/bar #:qux{:bar :baz}); => :baz
(:foo/bar #:qux{:foo/bar :baz}); => :baz
(:qux/bar #:qux{:foo/bar :baz}); => nil

Answer 1

A map might contain all qualified keyword keys from a certain namespace, or it may contain a mix of unqualified keys or qualified keys from multiple namespaces. Here's a function to get the set of all namespaces (as keywords) from qualified keyword keys in a map:

(defn key-namespaces
  "Returns set of all namespaces of keys in m."
  [m]
  (->> (keys m)
       (keep (comp keyword namespace))
       (set)))

Now you can use that as a dispatch-fn on a multimethod:

(defmulti do-thing key-namespaces)
(defmethod do-thing #{:foo} [m] (prn m))
(do-thing #:foo{:bar 1})
;; #:foo{:bar 1}
(foo {:bar/bar 1})
;; no multimethod found exception

You could specify multiple namespace prefixes in that set, or you could use a different dispatch-fn based on your use case.

Answer 2

This is not possible as this is just the visual representation of the map produced by writer . You would have to do the checks by yourself if all the keys in your map share the same namespace. Also the last example won't be produced by the writer - it will emit namespaced map literal only when all the keys share the same namespace .