我正在使用libpostal库在新闻文章中查找完整地址(街道,城市,州和邮政编码).给定输入文本时的libpostal:
位于CO 10566的Main Street Boulder发生了一起事故 - 位于Wilson的拐角处.
返回一个向量:
[{:label "house", :value "there was an accident at 5"}
{:label "road", :value "main street"}
{:label "city", :value "boulder"}
{:label "state", :value "co"}
{:label "postcode", :value "10566"}
{:label "road", :value "which is at the corner of wilson."}
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我想知道在Clojure中是否有一种聪明的方法来提取序列中出现:label值的序列:
[road unit? level? po_box? city state postcode? country?]
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where ?表示匹配中的可选值.
你可以用clojure.spec做到这一点.首先定义一些与地图的:label值匹配的规范:
(defn has-label? [m label] (= label (:label m)))
(s/def ::city #(has-label? % "city"))
(s/def ::postcode #(has-label? % "postcode"))
(s/def ::state #(has-label? % "state"))
(s/def ::house #(has-label? % "house"))
(s/def ::road #(has-label? % "road"))
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然后定义一个正则表达式规范,例如s/cat+ s/?:
(s/def ::valid-seq
(s/cat :road ::road
:city (s/? ::city) ;; ? = zero or once
:state ::state
:zip (s/? ::postcode)))
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现在你可以conform或者valid?你的序列:
(s/conform ::valid-seq [{:label "road" :value "Damen"}
{:label "city" :value "Chicago"}
{:label "state" :value "IL"}])
=>
{:road {:label "road", :value "Damen"},
:city {:label "city", :value "Chicago"},
:state {:label "state", :value "IL"}}
;; this is also valid, missing an optional value in the middle
(s/conform ::valid-seq [{:label "road" :value "Damen"}
{:label "state" :value "IL"}
{:label "postcode" :value "60622"}])
=>
{:road {:label "road", :value "Damen"},
:state {:label "state", :value "IL"},
:zip {:label "postcode", :value "60622"}}
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