我有四个文件，每个文件 10 行，如何获得如下输出

Question

我有4个文件。我需要检查所有文件的行数是否相同。

如果行数不同，我需要检测它并输出，例如：

#file1 - 10 lines, file2 - 9 lines, file3 - 10 lines, file4 - 10 lines
Line are miss matched
Number of lines 10 = 9 = 10 = 10

如果它们相等，我想逐行合并文件，如下所示：

文件：

#file1
10 
12
11

#file2
Arun
kamal
babu

#file3
300
200
400

#file4
spot1
spot4
spot5

输出：

Set1
10
Arun
300
spot1

Set2
12
kamal
200
spot4

Set3
11
babu
400
spot5

我的代码：

#

id_name=`cat file2`
echo $id_name

id_list=`cat file1`
echo $id_list

#

id_count=`cat file3`
echo $id_count

id_spot=`cat spot_list`
echo $id_spot


SS=`cat id_list | wc -l`
DS=`cat id_name | wc -l`
SF=`cat id_count | wc -l`
DF=`cat id_spot | wc -l`

if [ $SS == $DS == $SF == $DF ] then

   echo " Line are matched"
   echo " Total line $SS"


   for i j in $id_list $id_name
   do
      for a b in $id_count $id_spot
      do
         k = 1
         echo " Set$k"
         $i
         $j
         $a
         $b
      done
   done

else

   echo " Line are Miss matched"
   echo " Total line $SS  = $DS = $SF = $DF"

fi

Answer 1

With a really straightforward approach:

#!/usr/bin/env bash

SS=$(wc -l < file1)
DS=$(wc -l < file2)
SF=$(wc -l < file3)
DF=$(wc -l < file4)


if [[ $SS -eq $DS && $DS -eq $SF && $SF -eq $DF ]]; then 
   echo "Lines are matched"
   echo "Total number of lines: $SS"

   num=1
   while (( num <= SS )); do
      echo "Set$num"
      tail -n +$num file1 | head -n 1
      tail -n +$num file2 | head -n 1
      tail -n +$num file3 | head -n 1
      tail -n +$num file4 | head -n 1

      ((num++))
      echo
   done

else
   echo "Line are miss matched"
   echo "Number of lines $SS = $DS = $SF = $DF"
fi

It is not very efficient as it calls tail 4*number_of_lines times but it is straightforward.

Another approach is to replace the while loop with awk:

awk '{
   printf("\nSet%s\n", NR)
   print; 
   if( getline < "file2" )
      print
   if( getline < "file3" )
      print
   if ( getline < "file4" )
      print
}' file1

To join files line by line, the paste command is very useful. You can use this instead of the while loop:

paste -d$'\n' file1 file2 file3 file4

Or maybe a little less obvious:

{ cat -n file1 ; cat -n file2 ; cat -n file3; cat -n file4; }  | sort -n  | cut -f2-

That will output the lines but with no formatting (no Set1, Set2, newlines, ...), so you have to format it afterwards with awk, for example:

awk '{ 
   if ((NR-1)%4 == 0) 
      printf("\nSet%s\n", (NR+3)/4) 
   print 
}' < <(paste -d$'\n' file1 file2 file3 file4)

Some final notes:

• Do not use uppercase variables as they could collide with environment and internal shell variables
• Do not use echo "$var" | cmd or cat file | cmd when you can redirect input: cmd <<< "$var" or cmd < file
• You can have only one variable name in for loop. for i in ... is valid, whereas for i j in ... is not
• It is better to use [[ ]] instead of [ ] for testing, see this answer
• There are a lot of ways to do this
• It's up to you which approach you choose to use but be aware of the efficiency differences:

Results of time, tested on files with 10000 lines:

#first approach
real    0m45.387s
user    0m5.904s
sys     0m3.836s

#second approach - significantly faster
real    0m0.086s
user    0m0.024s
sys     0m0.040s

#third approach - very close to second approach
real    0m0.074s
user    0m0.016s
sys     0m0.036s

Answer 2

你能弄清楚如何检查的行数为每个文件（提示：wc）

要获得集合的输出：

paste File{1,2,3,4} | awk -F'\t' -v OFS='\n' '{$1=$1; print "Set"NR, $0, ""}'

$1=$1 用于将输入字段分隔符转换为输出字段分隔符。