如何在[文本文件]中为每一行创建一个以第一列命名的新文件，并包含每行的所有剩余列？

I have a file like the following

我有一个像下面这样的文件

Columns are delimited by tabs

列由制表符分隔

    Group1  name1:ENSG0365  name4:ENSMU0000196  name2:ENSMU00001826
    Group2  name2:ENS000132622  name4:ENSUS00074793 
    Group3  name3:ENSFP000844   name1:ENSMU000025

I want a list of files for every line, named after the 1st Column (Group1, Group2, etc) and looking like this

我想要每行的文件列表，以第1列（Group1，Group2等）命名并且看起来像这样

name1:ENSG0365
name4:ENSMU0000196
mouse2:ENSMU00001826

name1：ENSG0365 name4：ENSMU0000196 mouse2：ENSMU00001826

I don't know which tools is better for this task. Thanks for your help.

我不知道哪个工具更适合这项任务。谢谢你的帮助。

2 个解决方案

#1

With bash 4:

用bash 4：

while read -r name rest; do echo "$rest" >"$name"; done </path/to/your/file

Update:

更新：

while read -r name rest; do printf "%s\n" $rest >"$name"; done < /path/to/your/file

#2

perl -n -e '@x=split;open(F,">".shift(@x));$,="\n";print F @x' < groups.txt

Taking a wild guess at what you mean in the comment, you could try:

对评论中的含义进行疯狂猜测，您可以尝试：

perl -n -e '@x=split /\t/;open(F,">".shift(@x));$,="\n";print F @x' < groups.txt

#1