On Nov 17, 2008, at 14:30, Rich Hickey wrote:
> It's best to do this as generally as possible, building on something
> like this:
>
> (defn map-same [f coll]
> (let [ret (into (empty coll) (map f coll))]
> (if (seq? coll)
> (reverse ret)
> ret)))
OK, so I have as a minimal set of cases (1) seq (2) other coll and
(3) everything else. That gives me
(defn replace-syms
[sym-map expr]
(cond (coll? expr) (let [replace #(replace-syms sym-map %)
r-expr (map replace expr)]
(if (seq? expr)
r-expr
(into (empty expr) r-expr)))
(contains? sym-map expr) (get sym-map expr)
:else expr))
This experience leaves me wondering if perhaps "seq" is the closest
equivalent to the list concept in other Lisps. Everything that test
positive with (seq?) seems to print like a list and behave like a
list. But then, what would be a valid reason to use the (list?) test?
Konrad.
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