Perl - 将数组元素句子与变量进行比较

Question

我使用grep在temporaryF文件和arrayWarning之间返回不匹配的数组

my @c =grep!${{map{$_,1}@temporaryF}{$_},@arrayWarning;

在@c里面有很多行例如:

Sun Sep 30 00:05:55 fibre channel DENY forever
Sun Sep 30 00:06:55 fibre channel ROOT cause
Sun Sep 30 00:08:55 fibre channel ROOT cause 
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:20:55 fibre channel DANN 
Sun Sep 30 00:30:55 fibre channel DANN  

你可以看到ROOT在@c中发生了3次.如何迭代@c仅输出最新出现的ROOT - > Sun Sep 30 00:10:55光纤通道ROOT而不是其他重复行.

所以它会变成:

Sun Sep 30 00:05:55 fibre channel DENY forever  
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:30:55 fibre channel DANN

Answer 1

注意：这是@RobEarl 答案的延伸 - 所以如果你喜欢它，请确保也给他学分！

这里的要点是也存储行数，以确保输出可以排序。

长版

#!/usr/bin/perl

use strict;
use warnings;

# store (with count)
my $count  = 0;
my %latest = map {
    my $source = (split /\s+/ => $_)[6];
    $source => {count => $count++, string => $_};
} <DATA>;

# output
print $_->{string} for sort {$a->{count} <=> $b->{count}} values %latest;

__DATA__
Sun Sep 30 00:05:55 fibre channel DENY forever
Sun Sep 30 00:06:55 fibre channel ROOT cause
Sun Sep 30 00:08:55 fibre channel ROOT cause 
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:20:55 fibre channel DANN 
Sun Sep 30 00:30:55 fibre channel DANN  

输出：

Sun Sep 30 00:05:55 fibre channel DENY forever
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:30:55 fibre channel DANN  

感觉有点像施瓦茨变换。

单线版本

这是一个很好的例子，可以通过简单的 oneliner 和 perl 强大的解释器开关来完成任务：

$ perl -nale '$l{$F[6]}={c=>$c++,s=>$_};END{print$_->{s}for sort{$a->{c}<=>$b->{c}}values%l}'
Sun Sep 30 00:05:55 fibre channel DENY forever
Sun Sep 30 00:06:55 fibre channel ROOT cause
Sun Sep 30 00:08:55 fibre channel ROOT cause 
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:20:55 fibre channel DANN 
Sun Sep 30 00:30:55 fibre channel DANN  

输出：

Sun Sep 30 00:05:55 fibre channel DENY forever
Sun Sep 30 00:10:55 fibre channel ROOT cause  
Sun Sep 30 00:30:55 fibre channel DANN