不在数组B中的数组A中的项目

问题描述

确定数组A中不在数组B中的项目的快速方法是什么?

What is a quick way to determine items in array A that are not in array B?

到目前为止，我有以下内容，但是可以一行完成吗?

So far I have the following, but can it be done in one line?

for $aname (@anames)
{
   if (not grep { $_ eq $aname } @hrefs)
   {
        push @anamesremove, $aname;
   }
}

推荐答案

如果您知道哪个数组包含另一个数组

If you know which array contains the other

my @small  = 10..12;
my @large  = 10..15;

my %ref = map { $_ => 1 } @small;

my @diff = grep { not exists $ref{$_} } @large;

行数不多，但效率很高.

Not quite one line but efficient.

如果不知道其中包含另一个，则必须双向进行.

If it's not known which contains the other then one has to do it both ways.

然后有各种用于阵列/列表操作的模块可以提供帮助.

And then there are various modules for array/list manipulation that can help.

其中之一，来自 List :: Compare 的get_complement正是需要的

For one, get_complement from List::Compare does precisely what is needed.

您可以使用 List :: Util 在一个语句中完成所有操作

What you have can be done in one statement, using List::Util

use List::Util qw(none);

my @diff = grep { my $e = $_; none { $e eq $_ } @small } @large;

但这具有 O(NM-M /2)复杂度，其中 N (大)和 M (小)是数组大小.

but this has O(NM-M/2) complexity, where N (large) and M (small) are array sizes.

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