Question

我很欣赏有关如何利用Clojure有效地解析和比较两个文件的建议/见解。有两个（日志）文件包含员工出勤率;从这些文件中，我需要确定两名员工在同一部门中工作同一时间的所有日子。以下是日志文件的示例。

注意：每个文件都有不同数量的条目。

第一档：

Employee Id     Name         Time In          Time Out          Dept.
mce0518         Jon     2011-01-01 06:00  2011-01-01 14:00       ER
mce0518         Jon     2011-01-02 06:00  2011-01-01 14:00       ER
mce0518         Jon     2011-01-04 06:00  2011-01-01 13:00       ICU
mce0518         Jon     2011-01-05 06:00  2011-01-01 13:00       ICU
mce0518         Jon     2011-01-05 17:00  2011-01-01 23:00       ER

第二档：

Employee Id     Name            Time In           Time Out          Dept.
pdm1705         Jane        2011-01-01 06:00  2011-01-01 14:00       ER
pdm1705         Jane        2011-01-02 06:00  2011-01-01 14:00       ER
pdm1705         Jane        2011-01-05 06:00  2011-01-01 13:00       ER
pdm1705         Jane        2011-01-05 17:00  2011-01-01 23:00       ER

Answer 1

如果你不打算定期这样做，


(defn data-seq [f]
  (with-open [rdr (java.io.BufferedReader. 
                   (java.io.FileReader. f))]
    (let [s (rest (line-seq rdr))]
      (doall (map seq (map #(.split % "\\s+") s))))))

(defn same-time? [a b]
  (let [a  (drop 2 a)
        b  (drop 2 b)]
    (= a b)))

(let [f1 (data-seq "f1.txt")
      f2 (data-seq "f2.txt")]

  (reduce (fn[h v]
            (let [f2 (filter #(same-time? v %) f2)]
              (if (empty? f2)
                h
                (conj h [(first v) (map first f2)]))))  [] f1) 
  )

会找到你，

 [["mce0518" ("pdm1705")] ["mce0518" ("pdm1705")] ["mce0518" ("pdm1705")]]

Answer 2

我来的时间更短，而且（恕我直言）更易读的版本

(use ; moar toolz - moar fun
  '[clojure.contrib.duck-streams :only (reader)]
  '[clojure.string :only (split)]
  '[clojure.contrib.str-utils :only (str-join)]
  '[clojure.set :only (intersection)])

(defn read-presence [filename]
  (with-open [rdr (reader filename)] ; file will be securely (always) closed after use
    (apply hash-set ; make employee's hash-set
      (map #(str-join "--" (drop 2 (split % #" [ ]+"))) ; right-to-left: split row by spaces then forget two first columns then join using "--"
        (drop 1 ; ommit first line
          (line-seq rdr)))))) ; read file content line-by-line

(intersection (read-presence "a.in") (read-presence "b.in")) ; now it's simple!
;result: #{"2011-01-01 06:00--2011-01-01 14:00--ER" "2011-01-02 06:00--2011-01-01 14:00--ER" "2011-01-05 17:00--2011-01-01 23:00--ER"}

假设a.in和b.in是您的文件。我还假设你每个员工都有一个哈希集 - （天真）对N名员工的概括需要接下来的六行：

(def employees ["greg.txt" "allison.txt" "robert.txt" "eric.txt" "james.txt" "lisa.txt"])
(for [a employees b employees :when (and
                                      (= a (first (sort [a b]))) ; thou shall compare greg with james ONCE
                                      (not (= a b)))] ; thou shall not compare greg with greg
  (str-join " -- " ; well, it's not pretty... nor pink at least
    [a b (intersection (read-presence a) (read-presence b))]))
;result: ("a.in -- b.in -- #{\"2011-01-01 06:00--2011-01-01 14:00--ER\" \"2011-01-02 06:00--2011-01-01 14:00--ER\" \"2011-01-05 17:00--2011-01-01 23:00--ER\"}")

实际上这个循环太丑了，它不会记住中间结果......需要改进。

- 编辑 -

我知道核心或贡献必须有优雅的东西！

(use '[clojure.contrib.combinatorics :only (combinations)])

(def employees ["greg.txt" "allison.txt" "robert.txt" "eric.txt" "james.txt" "lisa.txt"])
(def employee-map (apply conj (for [e employees] {e (read-presence e)})))
(map (fn [[a b]] [a b (intersection (employee-map a) (employee-map b))])
  (combinations employees 2))
;result: (["a.in" "b.in" #{"2011-01-01 06:00--2011-01-01 14:00--ER" "2011-01-02 06:00--2011-01-01 14:00--ER" "2011-01-05 17:00--2011-01-01 23:00--ER"}])

现在它被记忆了（在employee-map中解析了数据），一般而且......懒惰：D

如何解析和比较文件？

2 个答案: