将多个 .gz 文件中的特定列粘贴在一起

发布于 2024-12-12 02:15:58 字数 221 浏览 0 评论 0原文

未压缩的具有相同的行数和相同的列数。

所有文件都位于同一目录中。

是否可以从每个文件中提取每 5-6 列并将它们粘贴在一起，而无需转储临时文件？

一样的东西

for i in *.gz
do
   gunzip -c $i |cut -f5-6 >$i.tmp;
done

paste *.tmp

像谢谢

原文

that uncompressed have the same number of lines, and the same number of columns.

All files are located in the same dir.

Is it possible to pullout every 5-6 column from every file and paste them together without having to dump temp files?

Something like

for i in *.gz
do
   gunzip -c $i |cut -f5-6 >$i.tmp;
done

paste *.tmp

Thanks

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

网白 2024-12-19 02:15:58

您可以创建一个命令行来评估和使用 bash 进程替换以避免临时文件：

command=paste
for i in *.gz; do
    command="$command <(gzip -cd $i | cut -f5-6)"
done
eval $command

You can create a command line to evaluate and use bash process substitution to avoid tempfiles:

command=paste
for i in *.gz; do
    command="$command <(gzip -cd $i | cut -f5-6)"
done
eval $command

回复收藏 0 原文

左秋 2024-12-19 02:15:58

for f in *.gz ; do
  printf '_fn %s\n' "$f" 
  gunzip -c "$f"  
done | 
  awk 'END {
    for (i = 0; ++i <= m;)
      for (j = 0; ++j <= fc;)
        printf "%s", (d[j, i] (j < fc ? OFS : RS))    
    }
  $1 == "_fn" {
    fn = $2; fnr = x; ++fc
    next
    }
  { 
    d[fc, ++fnr] = $5 OFS $6
    fnr > m && m = fnr
    }' OFS='\t'

for f in *.gz ; do
  printf '_fn %s\n' "$f" 
  gunzip -c "$f"  
done | 
  awk 'END {
    for (i = 0; ++i <= m;)
      for (j = 0; ++j <= fc;)
        printf "%s", (d[j, i] (j < fc ? OFS : RS))    
    }
  $1 == "_fn" {
    fn = $2; fnr = x; ++fc
    next
    }
  { 
    d[fc, ++fnr] = $5 OFS $6
    fnr > m && m = fnr
    }' OFS='\t'

回复收藏 0 原文

~没有更多了~