blocks|key|4758871|text|在Perl5中，您可以使用模块List::MoreUtils。或者成对使用，或者使用each_array返回的迭代器(可能需要两个以上的数组来并行迭代)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4758872|use+5.12.0;
use+List::MoreUtils+qw(pairwise+each_array);

my+@one+=+qw(a+b+c+d+e);
my+@two+=+qw(q+w+e+r+t);

my+@three+=+pairwise+{"$a:$b"}+@one,+@two;

say+join("+",+@three);

my+$it+=+each_array(@one,+@two);
while+(my+@elems+=+$it->())+{
++++say+"$elems[0]+and+$elems[1]";
}|code-block|syntax|javascript|4758873|entityMap|0|LINK|mutability|MUTABLE|url|http://search.cpan.org/dist/List-MoreUtils/lib/List/MoreUtils.pm^0|F|F|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

In Perl 5 you can use the module <a href="http://search.cpan.org/dist/List-MoreUtils/lib/List/MoreUtils.pm" rel="noreferrer">List::MoreUtils</a>. Either with pairwise or with the iterator returned by each_array (which can take more than two arrays to iterate through in parallel).

<pre><code>use 5.12.0;
use List::MoreUtils qw(pairwise each_array);

my @one = qw(a b c d e);
my @two = qw(q w e r t);

my @three = pairwise {"$a:$b"} @one, @two;

say join(" ", @three);

my $it = each_array(@one, @two);
while (my @elems = $it-&gt;()) {
 say "$elems[0] and $elems[1]";
}
</code></pre>

blocks|key|1069652|text|在Perl6中，Zip操作符是最好的选择。如果你想得到这两个值(而不是直接计算和)，你可以不带加号使用它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1069653|for+(10,+11,+12)+Z+(1,+2,+3)+->+$a,+$b+{
++++say+"$a+and+$b";
}|code-block|syntax|javascript|1069654|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In Perl 6, the Zip operator is the nicest choice. If you want to get both values (and not compute the sum directly), you can use it without the plus sign:

<pre><code>for (10, 11, 12) Z (1, 2, 3) -&gt; $a, $b {
 say "$a and $b";
}
</code></pre>

blocks|key|1069532|text|使用Zip操作符，您可以实现使用Scheme所做的操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1069533|>+.say+for+(10,+20,+30)+Z%2B+(1,+2,+3)
11
22
33|code-block|syntax|javascript|1069534|请参阅http://perlcabal.org/syn/S03.html#Zip_operators|offset|length|1069535|entityMap|0|LINK|mutability|MUTABLE|url|http://perlcabal.org/syn/S03.html#Zip_operators^0|0|0|3|1B|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

With the Zip operator you can achieve what you're doing with Scheme:

<pre><code>&gt; .say for (10, 20, 30) Z+ (1, 2, 3)
11
22
33
</code></pre>

See <a href="http://perlcabal.org/syn/S03.html#Zip_operators" rel="noreferrer">http://perlcabal.org/syn/S03.html#Zip_operators</a>

blocks|key|2971375|text|如果你确定数组的大小相同，你可以遍历数组的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2971376|foreach(+0+..+$#array1+)+{
++print+$array1[$_]+%2B+$array2[$_],+"\n";
}|code-block|syntax|javascript|2971377|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could just iterate over the indices of the arrays, if you're sure they're the same size:

<pre><code>foreach( 0 .. $#array1 ) {
 print $array1[$_] + $array2[$_], "\n";
}
</code></pre>

blocks|key|1069619|text|Algorithm::Loops提供了一个MapCar函数，用于迭代多个数组(具有不同的变种，以不同的方式处理大小不等的数组)。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1069620|entityMap|0|LINK|mutability|MUTABLE|url|http://p3rl.org/Algorithm%253a%253aLoops^0|0|G|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

<a href="http://p3rl.org/Algorithm%3a%3aLoops" rel="nofollow">Algorithm::Loops</a> offers a MapCar function for iterating over multiple arrays (with variants that deal differently with unequally sized arrays).

blocks|key|2685647|text|一种方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2685648|sub+for_each
{
++++my+$proc+=+shift+;

++++my+$len+=+@{$_[0]}+;

++++for+(+my+$i+=+0+;+$i+<+$len+;+$i%2B%2B+)
++++{
++++++++my+@args+=+map+$_->[$i]+,+@_+;

++++++++&$proc+(+@args+)+;
++++}
}

for_each+sub+{+say+$_[0]+%2B+$_[1]+}+,+([10,20,30],[1,2,3])|code-block|syntax|javascript|2685649|从List::MoreUtils使用each_arrayref|offset|length|style|CODE|2685650|sub+for_each
{
++++my+$proc+=+shift+;

++++my+$it+=+each_arrayref+(+@_+)+;

++++while+(+my+@elts+=+$it->()+)+{+&$proc+(+@elts+)+}+;
}

for_each+sub+{+say+$_[0]+%2B+$_[1]+}+,+([10,20,30],[1,2,3])|2685651|感谢Alex指出List::MoreUtils。|2685652|entityMap^0|0|0|1|F|I|D|0|0|8|F|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@$I|11|J|12|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

One way is:

<pre><code>sub for_each
{
 my $proc = shift ;

 my $len = @{$_[0]} ;

 for ( my $i = 0 ; $i &lt; $len ; $i++ )
 {
 my @args = map $_-&gt;[$i] , @_ ;

 &amp;$proc ( @args ) ;
 }
}

for_each sub { say $_[0] + $_[1] } , ([10,20,30],[1,2,3])
</code></pre>

Using <code>each_arrayref</code> from <code>List::MoreUtils</code>:

<pre><code>sub for_each
{
 my $proc = shift ;

 my $it = each_arrayref ( @_ ) ;

 while ( my @elts = $it-&gt;() ) { &amp;$proc ( @elts ) } ;
}

for_each sub { say $_[0] + $_[1] } , ([10,20,30],[1,2,3])
</code></pre>

Thanks to Alex for pointing out <code>List::MoreUtils</code>.

blocks|key|2685986|text|只要问题像加法一样简单(/multiplying，dividing，…)并且数组不是数组的数组，您也可以使用超运算符来执行任务：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2685987|<1+2+3>+«%2B»+<10+20+30>|code-block|syntax|javascript|2685988|(当然，这是一个Perl6的答案)|2685989|如果您没有访问法语报价«»的权限，可以将其重写为|offset|length|style|CODE|2685990|<1+2+3>+<<%2B>>+<10+20+30>|2685991|entityMap^0|0|0|0|B|2|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

As long as the question is as simple as just adding (/multiplying, dividing,…) and the 
Arrays are no Arrays of Arrays, you also could use the hyper-operators for your task:

<pre><code>&lt;1 2 3&gt; «+» &lt;10 20 30&gt;
</code></pre>

(of course this is a Perl6 answer)

if you don't have access to the French quotes <code>«»</code>, you could rewrite it as

<pre><code>&lt;1 2 3&gt; &lt;&lt;+&gt;&gt; &lt;10 20 30&gt;
</code></pre>

blocks|key|4759019|text|一种解决方案(有许多可供选择)可能如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4759020|my+@argle+=+(10,+20,+30);
my+@bargle+=+(1,+2,+3);
do+{
++++my+$yin+=+shift+@argle;
++++my+$yang+=+shift+@bargle;
++++say+$yin+%2B+$yang;
}+while+(@argle+&&+@bargle);|code-block|syntax|javascript|4759021|我感觉你是在问foreach，它可能看起来像这样：|offset|length|style|CODE|4759022|my+@argle+=+(10,+20,+30);
my+@bargle+=+(1,+2,+3);
for+my+$yin+(@argle)+{
++++my+$yang+=+shift+@bargle;
++++say+$yin+%2B+$yang;
}|4759023|但在这种情况下，它就不那么顺利了。如果任何一个数组更短，会发生什么？|4759024|entityMap^0|0|0|7|7|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

One solution (there are many to choose from) might look like this:

<pre><code>my @argle = (10, 20, 30);
my @bargle = (1, 2, 3);
do {
 my $yin = shift @argle;
 my $yang = shift @bargle;
 say $yin + $yang;
} while (@argle &amp;&amp; @bargle);
</code></pre>

I sense that you are asking about <code>foreach</code>, which might look like this:

<pre><code>my @argle = (10, 20, 30);
my @bargle = (1, 2, 3);
for my $yin (@argle) {
 my $yang = shift @bargle;
 say $yin + $yang;
}
</code></pre>

But it is not as smooth in this case. What happens if either array is shorter?

In Scheme you can iterate over multiple lists in with <code>for-each</code>:

<pre><code>&gt; (for-each (lambda (a b) (display (+ a b)) (newline)) '(10 20 30) '(1 2 3))
11
22
33
&gt; 
</code></pre>

I know that in Perl you can use <code>for</code> to iterate over a single list. What's a good way to iterate over multiple lists as in the Scheme example?

I'm interested in answers for Perl 5 or 6.

for with multiple arrays

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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在方案中，您可以使用for-each在中迭代多个列表> (for-each (lambda (a b) (display (+ a b)) (newline)) '(10 20 30) '(1 2 3))112233> 我知道在Perl语言中，您可以使用for遍历单个列表。像Scheme示例中那样迭代多个列表的好方法是什么？我对Perl5或6的答案感兴趣。

问对于具有多个数组的
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