Question

我有一个带有这种信息的csv文件：

2013  Cat.1  10  Structure1  Code1  34.10
2014  Cat.1  25  Structure1  Code1  254.24
2013  Cat.2  250 Structure1  Code1  2456.4
2014  Cat.2  234 Structure1  Code1  2345.9
2013  Cat.1  5   Structure2  Code2  59
2013  Cat.1  1   Structure2  Code2  18
2014  Cat.1  8   Structure2  Code2  123
2014  Cat.1  1   Structure2  Code2  18
2013  Cat.2  64  Structure2  Code2  59
2013  Cat.2  8   Structure2  Code2  18
2014  Cat.2  70  Structure2  Code2  123
2014  Cat.2  11  Structure2  Code2  18

我想要的结果文件就是那种：

2013  Cat.1  10         Structure1  Code1  34.10
2014  Cat.1  25         Structure1  Code1  254.24
2013  Cat.2  250        Structure1  Code1  2456.4
2014  Cat.2  234        Structure1  Code1  2345.9
2013  Cat.1  6 (5+1)    Structure2  Code2  77 (59+18)
2014  Cat.1  9 (8+1)    Structure2  Code2  141 (123+18)
2013  Cat.2  72 (64+8)  Structure2  Code2  77 (59+18)
2014  Cat.2  81 (70+11) Structure2  Code2  141 (123+18)

使用awk可以吗？对于第二个结构，我在此示例中只有2个不同的字段，但可能更多......

我对编程非常陌生，特别是awk。

感谢您的回答！

Answer 1

awk救援！

不是完整的解决方案，但可能会给你一些想法

$awk  '{
    k = $1 FS $2 FS $4 FS $5
    a[k] += $3
    as[k] = as[k] ? as[k] "+" $3 : "(" $3
    b[k] += $6
    bs[k] = bs[k] ? bs[k] "+" $6 : "(" $6
  }

  END {
    for (k in a) {
      print k, a[k], as[k] ")", b[k], bs[k] ")"
    }
  }' file

会给你

2014 Cat.2 Structure2 Code2 81 (70+11) 141 (123+18)
2014 Cat.1 Structure2 Code2 9 (8+1) 141 (123+18)
2014 Cat.2 Structure1 Code1 234 (234) 2345.9 (2345.9)
2014 Cat.1 Structure1 Code1 25 (25) 254.24 (254.24)
2013 Cat.2 Structure2 Code2 72 (64+8) 77 (59+18)
2013 Cat.1 Structure2 Code2 6 (5+1) 77 (59+18)
2013 Cat.2 Structure1 Code1 250 (250) 2456.4 (2456.4)
2013 Cat.1 Structure1 Code1 10 (10) 34.1 (34.10)

请注意，列顺序已更改为重用k，单个条目值也包含在parans中。两者都可以轻松处理。

Answer 2

另一个awk答案，GNU awk具体。我假设您实际上并不想打印出添加公式。

gawk '
  { data[$1 OFS $2][$4 OFS $5][1] += $3
    data[$1 OFS $2][$4 OFS $5][2] += $6 }
  END {
    for (k1 in data) {
      for (k2 in data[k1]) {
        print k1, data[k1][k2][1], k2, data[k1][k2][2]
      }
    }
  }
' | sort -k4,5 -k2,2 -k1,1 | column -t

2013  Cat.1  10   Structure1  Code1  34.1
2014  Cat.1  25   Structure1  Code1  254.24
2013  Cat.2  250  Structure1  Code1  2456.4
2014  Cat.2  234  Structure1  Code1  2345.9
2013  Cat.1  6    Structure2  Code2  77
2014  Cat.1  9    Structure2  Code2  141
2013  Cat.2  72   Structure2  Code2  77
2014  Cat.2  81   Structure2  Code2  141

Answer 3

这是一个可能的答案：

import java.util.ArrayDeque;

import javax.swing.*;

public class QueueTest {

  public static MyQueue<String> displayMenu(MyQueue<String> queue) {
    // list of choices (array of Strings)
    String[] array = {"Offer Person", "Poll Person", "Peek Person", "Display Queue", "Exit Program"};
    int choice = 0;
    // display loop
    while (choice != array.length - 1) {
      choice = JOptionPane.showOptionDialog(null, // put in center of screen
                                            "Press a Button", // message to user
                                            "Queue (line) of People", // title of window
                                            JOptionPane.YES_NO_CANCEL_OPTION, // type of option
                                            JOptionPane.QUESTION_MESSAGE, // type of message
                                            null, // icon
                                            array, // array of strings
                                            array[array.length - 1]); // default choice (last one)

      switch(choice) {
        case 0:
          String name = JOptionPane.showInputDialog(null, "Enter person's name");
          queue.offer(name);
          break;
        case 1:
          JOptionPane.showMessageDialog(null, queue.poll() + " is next in line.");
          break;
        case 2:
          JOptionPane.showMessageDialog(null, queue.peek() + " is in front of the line.");
          break;
        case 3:
          String output = queue.toString();
          JOptionPane.showMessageDialog(null, output);
          break;
        case 4:
          System.out.println("Exiting program");
          break;
        default:
          System.out.println("Choice not valid");
          break;
      }
    }
    return queue;
  }//close displayMenu

  static class MyQueue<T>
    extends ArrayDeque<T> {
    public MyQueue() {
    }
  }

  public static void main(String[] args) {
    MyQueue<String> myQueue = new MyQueue<String>();
    displayMenu(myQueue);
  }
}

在此脚本中，一个或多个空格awk 'BEGIN{FS="[ ]+"; OFS="\t";} NR==FNR{ key = $1"-"$2"-"$4"-"$5 idx[key] = idx[key]+1 a[key][idx[key]] = $3 c[key][idx[key]] = $6 } NR!=FNR{ key = $1"-"$2"-"$4"-"$5 if(idx[key]==1){$1=$1; print ;next;} if(idx[key]<0){next;} line1 =" ("a[key][1] line2 =" ("c[key][1] sum1 = a[key][1] sum2 = c[key][1] for(i = 2; i< idx[key]; i++) { line1 = line1"+"a[key][i] line2 = line2"+"c[key][i] sum1 = sum1+a[key][i] sum2 = sum1+c[key][i] } sum1 = sum1 + a[key][idx[key]] sum2 = sum2 + c[key][idx[key]] line1 = sum1""line1"+"a[key][idx[key]]")" line2 = sum2""line2"+"c[key][idx[key]]")" print $1, $2, line1, $4, $5, line2 idx[key] = -1 }' inputFile inputFile被解释为字段分隔符（）。在输出中，字段由选项卡（FS="[ ]+"）分隔请注意，脚本以OFS="\t"两次作为参数调用如果您的输入确实是csv文件，请尝试使用inputFile作为字段分隔符将其导出并设置,。
问题中给出的输入的示例输出：

FS=OFS=","

Answer 4

这个单行将完成这项工作：

awk 'BEGIN{g=1;s="%4s %5s %-12s %10s %5s %-12s\n"} f{printf s ,$1,$2,$3+a" ("a"+"$3")",$4,$5,$6+b" ("b"+"$6")";f=0;g=0} /Structure2/{a=$3;b=$6;f=g;g=1} /Structure1/{printf s,$1,$2,$3,$4,$5,$6}' file

2013  Cat.1  10         Structure1  Code1  34.10
2014  Cat.1  25         Structure1  Code1  254.24
2013  Cat.2  250        Structure1  Code1  2456.4
2014  Cat.2  234        Structure1  Code1  2345.9
2013  Cat.1  6 (5+1)    Structure2  Code2  77 (59+18)
2014  Cat.1  9 (8+1)    Structure2  Code2  141 (123+18)
2013  Cat.2  72 (64+8)  Structure2  Code2  77 (59+18)
2014  Cat.2  81 (70+11) Structure2  Code2  141 (123+18)

我为对齐添加了一个格式，我在第三列和第六列中使用了12（%-12s） - 如果数字越高，你就可以增加它。

用awk计算特定字段

4 个答案: