shell - 仅按包含“|”的行 sep 拆分文件

Question

我有一个巨大的文件结构：

>ABC_123|XX|YY|ID
CNHGYDGHA
>BBC_153|XX|YY|ID
ACGFDRER

我需要根据在线的第一个值拆分此文件

File1: ABC_123 -> should contain 
          >ABC_123|XX|YY|ID
          CNHGYDGHA

File2: BBC_153 -> should contain 
         >BBC_153|XX|YY|ID
          ACGFDRER

Answer 1

这会产生两个文件ABC_123，并BBC_153根据您的输入：

awk -F'|' 'NF > 1 { # when more than one field (i.e. line contains | )
    close(out)      # close the previous file (or do nothing, if none were open)
    out = $1        # assign first field to filename
    sub(/^>/, "", out) # remove the > from the start of the name
} 
{ print >> out }' file # print to the file, opening in append mode if needed

如果您确定文件名只会打开一次，那么您可以>使用>>.

Answer 2

awk方法：

awk -F'|' '/^>.+\|/{ fn = substr($1, 2) }{ print > fn }' file

查看 2 个创建的示例文件：

$ head [AB]BC_*
==> ABC_123 <==
>ABC_123|XX|YY|ID
CNHGYDGHA

==> BBC_153 <==
>BBC_153|XX|YY|ID
ACGFDRER