减去列中的数字

4 回答

• 1

  perly狗

  perl -ne '$a{$.}=$_;print $_-$a{$.-2}."\n" if $a{$.-2}' file
  

  创建数组如果存在两行之前的键，则打印该行减去数组中的值 .

  0.9
  -0.6786
  -0.6778
  0.7902
  4.2099
  1.0866
  1.769
  1.4565
  

  对于像肯特回答的要求一样

  perl -ne '$a{$.}=$_;print $_-$a{$.-2}.(eof()?"\n":",") if $a{$.-2}' file
  
  0.9,-0.6786,-0.6778,0.7902,4.2099,1.0866,1.769,1.4565
  

  回复于 2024-05-15T08:37:27+08:00
• 1

  又快又脏：

  $  awk '{a[NR]=0+$0}END{for(i=3;i<=NR;i++)print a[i]-a[i-2]}' file
  0.9
  -0.6786
  -0.6778
  0.7902
  4.2099
  1.0866
  1.769
  1.4565
  

  更新：想出了另一个有趣的方式：

  $ awk 'NF>1{print $1-$2}' <(paste  <(sed -n '3,$p' file) file)  
  0.9
  -0.6786
  -0.6778
  0.7902
  4.2099
  1.0866
  1.769
  1.4565
  

  update2，将结果设为CSV：

  kent$  awk '{a[NR]=0+$0}END{for(i=3;i<=NR;i++)
              printf "%s%s", a[i]-a[i-2],NR==i?RS:","}' file
  0.9,-0.6786,-0.6778,0.7902,4.2099,1.0866,1.769,1.4565
  

  回复于 2024-05-15T08:37:27+08:00
• 4

  #!/bin/bash
  #Create an array
  mapfile -t lines < inputFile
  
  
  output=()
  for index in "${!lines[@]}"; do
  # Check if the index + 2 exist
      if [[ ${lines[$(expr $index + 2)]} ]]; then
      #It does exist, do the math
          output+=("$(expr ${lines[$index]} + ${lines[$(expr $index + 2)]})")
      fi
  
  done
  
  printf "%s\n" "${output[@]}" > output
  

  回复于 2024-05-15T08:37:27+08:00
• 0

  有了awk，我会写

  awk -v ORS="" '
      {a=b; b=c; c=$0}                  # remember the last 3 lines
      NR >= 3 {print sep c-a; sep=","}  # print the difference
      END {print "\n"}                  # optional, add a trailing newline.
  ' file
  

  或者让粘贴做gruntwork

  awk '{a=b;b=c;c=$0} NR >= 3 {print c-a}' file | paste -sd,
  

  回复于 2024-05-15T08:37:27+08:00