我有一个CSV文件,我使用split
解析为N
个项目数组,其中N
是3
的倍数。
有没有办法可以做到这一点
foreach my ( $a, $b, $c ) ( @d ) {}
类似于Python?
答案 0 :(得分:14)
我在CPAN上的模块List::Gen中解决了这个问题。
use List::Gen qw/by/;
for my $items (by 3 => @list) {
# do something with @$items which will contain 3 element slices of @list
# unlike natatime or other common solutions, the elements in @$items are
# aliased to @list, just like in a normal foreach loop
}
您还可以导入mapn
函数,List::Gen
用于实现by
:
use List::Gen qw/mapn/;
mapn {
# do something with the slices in @_
} 3 => @list;
答案 1 :(得分:12)
您可以使用List::MoreUtils::natatime。来自文档:
my @x = ('a' .. 'g');
my $it = natatime 3, @x;
while (my @vals = $it->()) {
print "@vals\n";
}
natatime
在XS中实现,因此您应该更喜欢它以提高效率。仅用于说明目的,以下是如何在Perl中实现三元素迭代器生成器的方法:
#!/usr/bin/perl
use strict; use warnings;
my @v = ('a' .. 'z' );
my $it = make_3it(\@v);
while ( my @tuple = $it->() ) {
print "@tuple\n";
}
sub make_3it {
my ($arr) = @_;
{
my $lower = 0;
return sub {
return unless $lower < @$arr;
my $upper = $lower + 2;
@$arr > $upper or $upper = $#$arr;
my @ret = @$arr[$lower .. $upper];
$lower = $upper + 1;
return @ret;
}
}
}
答案 2 :(得分:4)
@z=(1,2,3,4,5,6,7,8,9,0);
for( @tuple=splice(@z,0,3); @tuple; @tuple=splice(@z,0,3) )
{
print "$tuple[0] $tuple[1] $tuple[2]\n";
}
产生
1 2 3
4 5 6
7 8 9
0
答案 3 :(得分:4)
my @list = (qw(one two three four five six seven eight nine));
while (my ($m, $n, $o) = splice (@list,0,3)) {
print "$m $n $o\n";
}
此输出:
one two three
four five six
seven eight nine
答案 4 :(得分:1)
不容易。通过将元素作为数组引用推送到数组上,你最好使@d
成为一个三元素元组的数组:
foreach my $line (<>)
push @d, [ split /,/, $line ];
(除非你真的应该使用CPAN中的一个CSV模块。