Clojure 确定是否存在空白键 编辑:
我的问题是如何过滤出zipmap中看似空白的键 虽然我有一个解决问题的方法,但知道如何过滤密钥将非常有帮助 结束编辑: 这个输出Clojure 确定是否存在空白键 编辑:,clojure,Clojure,我的问题是如何过滤出zipmap中看似空白的键 虽然我有一个解决问题的方法,但知道如何过滤密钥将非常有帮助 结束编辑: 这个输出 : [: [ ]] ([ ]) 3 ,, 是由 (println first-ent, " ", map-ent, " ", val-ent, " ", (count out-csv), " ", out-csv) 在这个函数中 (defn missing-accts "Prints accounts found in one repor
: [: [ ]] ([ ]) 3 ,,
是由
(println first-ent, " ", map-ent, " ", val-ent, " ", (count out-csv), " ", out-csv)
在这个函数中
(defn missing-accts
"Prints accounts found in one report but not the other."
[report-header mapped-data out-file]
(spit out-file (str "\n\n" report-header "\n\n") :append true)
(doseq [map-ent mapped-data]
(let [first-ent (first map-ent)
val-ent (rest map-ent)
out-csv (if first-ent
(str (name (key map-ent)) "," (first (val map-ent)) "," (last (val map-ent)) "\n")
nil)]
(println first-ent, " ", map-ent, " ", val-ent, " ", (count out-csv), " ", out-csv)
(if (> (count out-csv) 3)
(spit out-file out-csv :append true)
(println "Skipping: ", out-csv)))))
<空白键的输出计数为3的事实允许我过滤,因为它能检测出一个空白键,看起来并不干净。找到并过滤掉一个空白键是让我感到困惑的
谢谢。您可以使用以下方法创建空白关键字:
(关键字“”)
您可以使用此选项筛选列表并删除所有空白关键字:
(过滤器(fn[[key]](not=(关键字“”)key))映射ent)
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