在AWK / perl的分割文件，由线和模式的数 [英] Split file by number of lines and pattern in awk/perl

问题描述

我需要根据线的近似数（在本例中例如〜4，但数千在现实）的文件分割成块，而每个文件都有启动与也发生各块内的许多倍的图案。

I need to split a file into chunks based on approximate number of lines (e.g. ~4 in the example, but thousands in reality), whilst each file has to start with a pattern that also occurs many times within each chunk.

一个块需要先从开始而不是开始结束，是> 3系长

A block needs to start with START and not end with START and be >3 lines long

输入文件：

START
LINE
LINE
START
LINE 
LINE 
START
LINE 
LINE 
LINE 
START
LINE 
START
LINE

所需的输出文件：

Desired output files:

文件1

START
LINE
LINE
START
LINE
LINE

文件2

START
LINE 
LINE 
LINE 

文件3

START
LINE 
START
LINE

用下面的code的问题是，的第2次发生/ ^ START / 被包括在文件1的端部，当它应该在文件2.开始我不能工作的文件如何得到输出时，下一步的纪录是 / ^ START / 。还有，我可以使用没有尽头的模式。

The problem with the following code is that the 2nd occurrence of /^START/ is included at the end of file 1, when it should be at the start of file 2. I can't work out how get the file to output when the next record is /^START/. There is no end pattern that I can use.

awk '/^START/{f=1} f{ print $0 > "file_"n ; c++} c>3 && /^START/ { n++; c=1; close("file_"n) }' c=1 n=1 file

这是AWK或Perl的解决办法是多AP preciated！

An awk or perl solution would be much appreciated!

推荐答案

这会产生你想要的输出：

This produces the output that you want:

awk -v out=1 'NR>1 && ++i>3 && /^START/ {++out; i=0} {print > "file" out}' file

当所有的条件都满足，增量退出，这是输出文件名的一部分。

When all of the conditions are satisfied, increment out, which is part of the output filename.

输出：

$ cat file1
 0 START
 1 LINE
 2 LINE
 3 START
 4 LINE 
 5 LINE 
$ cat file2
 6 START
 7 LINE 
 8 LINE 
 9 LINE 
$ cat file3
10 START
11 LINE 
12 START
13 LINE

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