您如何使用 Clojure 读取文件然后将该文件拆分为多个文件？

Question

我有一个文本文件，需要按以下方式拆分：第一行是标识符，必须位于创建的每个文件的顶部，后面几行是由逗号分隔的一对数字：

例子

12345 12000,2 13000,5 12501,1 etc...

当脚本运行时，它以下列方式留下 3 个文件

p1.dat 12345 12000,2 etc...

p2.dat 12345 13000,5 etc...

p3.dat 12345 12501,1 etc...

我已经在 Ruby 中完成了一个脚本，但我在 中生锈了]Clojure and/or 可能永远学不好，请大神指教一下。这怎么能以 'functional' 的方式完成？

Ruby代码：

#!/usr/bin/env ruby
if ARGV.length < 1
  puts """To use this script give the path to the file and the number of parts to divide,
 if you dont give this number 2 is assumed"""
  exit
end

file_path = ARGV[0].to_s

if ARGV.length == 1
  n_parts = 2
else
  n_parts = ARGV[1].to_i
end

particiones = Array.new(n_parts)
j = Array(1..n_parts)

File.readlines(file_path).each_with_index do |linea, i|
  #first line is id, so it must be in every divided part
  if i == 0 #repeat first line in every file
    j.each do |particion|
      particiones[particion-1] = linea
    end
    next # After processing first line
  end
  particiones[(i-1)%n_partes] +=  linea # parts get stored
end

j.each do |particion| # files are written
  # Crear los archivos p1, p2,... etc
  File.open("p"+particion.to_s+".dat", 'w'){|f| f << particiones[particion-1]}
  puts "File p"+particion.to_s+".dat was created"
end

Answer 1

希望我正确理解了您的任务 - 试试这个：

代码：

(defn split-file
  "First argument is path to file,
  second argument is number of created files, default is 2"
  ([path] (split-file path 2))
  ([path parts]
   (let [numbers (-> (slurp path)
                     (clojure.string/replace #"\n|\r\n" " ")
                     (clojure.string/split #" "))
         identifier (first numbers)
         pairs (rest numbers)
         file-names (for [i (range 1 (inc parts))]
                      (str "p" i ".dat"))
         data-for-files (->> (iterate rest pairs)
                             (take parts)
                             (map #(take-nth parts %))
                             (map #(conj % identifier))
                             (map #(clojure.string/join "\n" %)))]
     (doall (map (fn [file-name text]
                   (spit file-name text))
                 file-names
                 data-for-files)))))

通话：

(split-file "data.txt" 3)

(returns(nil nil nil),不过没关系)

文件data.txt:

输出文件：

p1.dat:

p2.dat:

p3.dat:

您如何使用 Clojure 读取文件然后将该文件拆分为多个文件？

How do you do reading from a file and then split that file into multiple files with Clojure?

ruby

functional-programming

file

clojure