合并 Linux 行，如果列少于 X。

huangapple
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英文:

linux merge rows if Columns less than X

问题

I need you support in creating the output.
我需要你的支持来创建输出。

I have output as
我的输出如下

1 ABC STATUS
7 PROHIBITED
8 PROHIBITED
2 CDE STATUS
7 PROHIBITED
8 PROHIBITED
3 FGH STATUS
142 PROHIBITED
2 HIJ STATUS
7 PROHIBITED

I am trying to achieve as
我试图实现以下结果

`1 ABC STATUS 7 PROHIBITED 8 PROHIBITED
2 CDE STATUS 7 PROHIBITED 8 PROHIBITED
3 FGH STATUS 142 PROHIBITED
2 HIJ STATUS 7 PROHIBITED

OR Either below where every first and second column is filled in below empty gaps.

或者下面的方式，其中每个第一列和第二列都填在下面的空白处。

1 ABC STATUS 1 ABC 7 PROHIBITED 1 ABC 8 PROHIBITED 2 CDE STATUS 2 CDE 7 PROHIBITED 2 CDE 8 PROHIBITED 3 FGH STATUS 3 FGH 142 PROHIBITED 2 HIJ STATUS 2 HIJ 7 PROHIBITED 2 HIJ 8 PROHIBITED

so far I have tried to achieve this in one single row with
到目前为止，我已经尝试在一行中实现这个结果

cat file | paste -sd'\t\n';

but it is not showing the result as expected. it would be appreciated if any one can provide me with guidance as to how to achieve
但它没有显示预期的结果。如果有人可以提供关于如何实现的指导，将不胜感激。

英文:

I need you support in creating the output.
I have output as

  1. 1  ABC  STATUS   	 
  2.    7   	PROHIBITED   
  3.    8   	PROHIBITED   
  4. 2  CDE  STATUS   	 
  5.    7    PROHIBITED   
  6.    8    PROHIBITED   
  7. 3  FGH  STATUS   	 
  8.    142  PROHIBITED   
  9. 2  HIJ  STATUS   	 
  10.    7   	PROHIBITED

I am trying to achieve as

  1. `1  ABC  STATUS   	 7   	PROHIBITED   8   	PROHIBITED   
  2. 2  CDE  STATUS   	 7    PROHIBITED   8    PROHIBITED   
  3. 3  FGH  STATUS   	 142  PROHIBITED   
  4. 2  HIJ  STATUS   	 7   	PROHIBITED

OR Either below where every first and second column is filled in below empty gaps.

  1. `1  ABC  STATUS   	 
  2. 1  ABC   7   	PROHIBITED   
  3. 1  ABC   8   	PROHIBITED   
  4. 2  CDE  STATUS   	 
  5. 2  CDE   7    PROHIBITED   
  6. 2  CDE  8    PROHIBITED   
  7. 3  FGH  STATUS   	 
  8. 3  FGH   142  PROHIBITED   
  9. 2  HIJ  STATUS   	 
  10. 2  HIJ    7   	PROHIBITED   
  11. 2  HIJ    8   	PROHIBITED   `

so far I have tried to achieve this in one single row with

  1. cat file | paste -sd'\t\n'

but it is not showing the result as expected. it would be appreciated if any one can provide me with guidance as to how to achieve

答案1

得分: 2

以下是代码的翻译部分：

  1.         { if (NF==3)
  2.              printf "%s%s",(NR>1 ? "\n" : ""),$0        # 如果不是输入的第一行，则在输出前加上"\n"（用于终止先前的输出行）
  3.           else      
  4.              printf "%s",$0
  5.         }
  6.     END {print ""}                                      # 终止最后一行的输出
  7.     ' file

这段代码的功能是生成以下输出：

  1. 1  ABC  STATUS   7    PROHIBITED   8    PROHIBITED
  2. 2  CDE  STATUS   7    PROHIBITED   8    PROHIBITED
  3. 3  FGH  STATUS   142  PROHIBITED
  4. 2  HIJ  STATUS   7    PROHIBITED   8    PROHIBITED
英文:

One awk idea for generating the first set of output:

  1. awk '
  2.     { if (NF==3)
  3.          printf "%s%s",(NR>1 ? "\n" : ""),$0        # if not 1st row of input then prefix the output with "\n" (to terminate previous line of output)
  4.       else      
  5.          printf "%s",$0
  6.     }
  7. END {print ""}                                      # terminate last line of output
  8. ' file

This generates:

  1. 1  ABC  STATUS   7    PROHIBITED   8    PROHIBITED
  2. 2  CDE  STATUS   7    PROHIBITED   8    PROHIBITED
  3. 3  FGH  STATUS   142  PROHIBITED
  4. 2  HIJ  STATUS   7    PROHIBITED   8    PROHIBITED

答案2

得分: 2

使用任何POSIX awk，您可以测试一行是否以空格开头：

  1. 1  ABC  STATUS          7    PROHIBITED      8    PROHIBITED
  2. 2  CDE  STATUS          7    PROHIBITED      8    PROHIBITED
  3. 3  FGH  STATUS          142  PROHIBITED
  4. 2  HIJ  STATUS          7    PROHIBITED
英文:

Using any POSIX awk you can just test if a line starts with a space or not:

  1. $ awk '{printf "%s%s", (/^[[:space:]]/ ? "" : sep), $0; sep=ORS} END{print ""}' file
  2. 1  ABC  STATUS          7    PROHIBITED      8    PROHIBITED
  3. 2  CDE  STATUS          7    PROHIBITED      8    PROHIBITED
  4. 3  FGH  STATUS          142  PROHIBITED
  5. 2  HIJ  STATUS          7    PROHIBITED

答案3

得分: 2

以下是已翻译的代码部分：

  1. awk -v RS='(^|\n)[0-9]+' -v OFS="\t" '
  2. RT && FNR>1{
  3.   gsub(/\n/,"")
  4.   $1=$1
  5.   print
  6. }
  7. END{
  8.   if($0){
  9.     gsub(/^ +|\n/,"")
  10.     $1=$1
  11.     print
  12.   }
  13. }
  14. '   Input_file
英文:

With your shown samples and attempts please try following GNU awk code.

  1. awk -v RS='(^|\n)[0-9]+' -v OFS="\t" '
  2. RT && FNR>1{
  3.   gsub(/\n/,"")
  4.   $1=$1
  5.   print
  6. }
  7. END{
  8.   if($0){
  9.     gsub(/^ +|\n/,"")
  10.     $1=$1
  11.     print
  12.   }
  13. }
  14. '   Input_file

答案4

得分: 1

如果要合并的行始终以空格开头，如示例输入中所示，那么可以使用 sed 解决方案：

  1. sed -e :a -e '$!N;s/\n\([[:blank:]]\)//;ta' -e 'P;D' file
英文:

If lines to be joined always begin with blanks as in the sample input, then a sed solution could be

  1. sed -e :a -e '$!N;s/\n\([[:blank:]]\)//;ta' -e 'P;D' file

答案5

得分: 1

这可能适用于你（GNU sed）：

  1. sed '/^\S/{:a;x;s/\n//gp;d};H;$!d;ba' 文件名

将记录存储在保持空间，并在新记录的开始或文件末尾时，切换回保持空间，移除任何换行符并打印。

注：新记录以行首的非空白字符为标志。

英文:

This might work for you (GNU sed):

  1. sed '/^\S/{:a;x;s/\n//gp;d};H;$!d;ba' file

Store records in the hold space and at the start of a new record or the end of the file, swap back to the hold space, remove any newlines and print.

N.B. A new record is designated by a non-white space character at the start of a line.

huangapple
  • 本文由 发表于 2023年4月4日 03:39:33
  • 转载请务必保留本文链接：https://go.coder-hub.com/75923200.html
  • awk
  • linux
  • merge
  • row
  • sed
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