如何取消嵌套序列规范？

Question

使用 Clojure core.spec 我可以拥有以下内容：

(s/conform (s/cat :a even? :b (s/* odd?) :a2 even? :b2 (s/* odd?))  [2 3 5 12 13 15])
=> {:a 2, :b [3 5], :a2 12, :b2 [13 15]}

我想要的是通过外部化子规范来消除冗余：

(s/def ::even-followed-by-odds 
  (s/cat :a even? :b (s/* odd?)))

但是

(s/conform (s/tuple ::even-followed-by-odds ::even-followed-by-odds)  [2 3 5 12 13 15])
=> :clojure.spec/invalid

这个有效：

(s/conform (s/tuple ::even-followed-by-odds ::even-followed-by-odds)  [[2 3 5] [12 13 15]])
=> [{:a 2, :b [3 5]} {:a 12, :b [13 15]}]

所以我正在寻找的是一个函数或宏（比如 unnest），它会起作用：

(s/conform (s/tuple (unnest ::even-followed-by-odds) (unnest ::even-followed-by-odds))  [2 3 5 12 13 15])
=> [{:a 2, :b [3 5]} {:a 12, :b [13 15]}]

我怎样才能得到它？

Answer 1

您需要继续使用正则表达式：

(s/conform (s/cat :x ::even-followed-by-odds :y ::even-followed-by-odds) [2 3 5 12 13 15])

{:x {:a 2, :b [3 5]}, :y {:a 12, :b [13 15]}}

How to unnest sequence spec?