Question

我无法将子程序从Perl转换为PHP（我是Perl的新手）。整个子程序如下：

sub find_all_subsets {
  if (1 == scalar (@_)) {return [@_]}
  else {
    my @all_subsets = () ;
    my $last_item = pop (@_) ;
    my @first_subsets = find_all_subsets (@_) ;
    foreach my $subset (@first_subsets) {
      push (@all_subsets, $subset) ;
      my @ext_subset = @{$subset} ;
      push (@ext_subset, $last_item) ;
      push (@all_subsets, [@ext_subset]) ;
    }
    push (@all_subsets, [$last_item]) ;
    return (@all_subsets) ;
  }
}

我的问题是我真的不太了解Perl语法，所以我在PHP中编写这些@{$subset}，[@ext_subset]和[$last_item]时遇到了麻烦。

如果问题很愚蠢，谢谢，对不起。

Answer 1

[]是一个数组引用运算符，用于将数组转换为数组引用

@{}是一个数组解引用运算符，将数组引用转换为数组

@a = (1,2,3);
$a = [ @a ];
@b = @{$a};       # now @b ==> (1,2,3)

Answer 2

当我跑步时

use Data::Dumper;
$Data::Dumper::Terse = 1;
$Data::Dumper::Indent = 0; 
my @x = (1,2,3,4);
my @y = find_all_subsets(@x);
foreach my $subset (@y) {
  print Dumper($subset), "\n";
}

（+原始脚本）输出

[1]
[1,4]
[1,3]
[1,3,4]
[1,2]
[1,2,4]
[1,2,3]
[1,2,3,4]
[2]
[2,4]
[2,3]
[2,3,4]
[3]
[3,4]
[4]

请注意输出中的[]，我们稍后会再回来。

function find_all_subsets (array $x) {
  if ( 1>= count($x) ) { // the >= differs from the original script, use == or === if you want to keep it "more original"
    return array($x);
  } 
  else {
    $all_subsets = array();
    $last_item = array_pop($x);
    $first_subsets = find_all_subsets($x) ;
    foreach ($first_subsets as $subset) {
      array_push($all_subsets, $subset);
      array_push($subset, $last_item);
      array_push($all_subsets, $subset);
    }
    array_push ($all_subsets, array($last_item));
    return $all_subsets;
  }
}

$x = array(1,2,3,4);
$y = find_all_subsets($x);
foreach($y as $subset) {
  echo '(', join(',', $subset), ")\n";
}

产生

(1)
(1,4)
(1,3)
(1,3,4)
(1,2)
(1,2,4)
(1,2,3)
(1,2,3,4)
(2)
(2,4)
(2,3)
(2,3,4)
(3)
(3,4)
(4)

到目前为止一切顺利。现在回到[]。 Data :: Dumper选择了这个[]而不是（）因为它不是一个数组而是一个数组引用（如果我没有使用正确的术语，那就是我; perl真的不是我的强项）。让我们改变perl测试脚本并查看所有这些参考资料的效果。

use Data::Dumper;
$Data::Dumper::Terse = 1;
$Data::Dumper::Indent = 0; 
$x2 = 2;
my @x = (1,\$x2,3,4);
my @y = find_all_subsets(@x);
$x2 = 99;
foreach my $subset (@y) {
  print Dumper($subset), "\n";
}

，输出变为

[1]
[1,4]
[1,3]
[1,3,4]
[1,\99]
[1,\99,4]
[1,\99,3]
[1,\99,3,4]
[\99]
[\99,4]
[\99,3]
[\99,3,4]
[3]
[3,4]
[4]

你看，我在调用find_all_subsets（）之后更改了$ x2 ，但仍然在结果中使用了新值，Data :: Dumper标记了“值” “作为参考（\ 99而不是简单的99）。你的php脚本中也需要这个功能吗？

Answer 3

这样的事情：

@{$subset}    # Return array reference as an ordinary array, not applicable in PHP.
[@ext_subset] # Turn an array into an array reference, not applicable in PHP.
[$last_item]  # Create an array with $last_item as the only element, equal to array($last_item) in PHP.

Answer 4

这是正确的（在php中）吗？

function find_all_subsets($array) {
    if (count($array) == 1) {
        return $array;
    }
    else {
        $all_subsets = array();
        $last_item = array_pop($array);
        $first_subsets = find_all_subsets($array);
        foreach ($first_subsets as $key => $val) {
            array_push($all_subsets, $val);
            $ext_subset[] = $val;
            array_push($ext_subset, $last_item);
            array_push($all_subsets, $ext_subset);
        }
        array_push($all_subsets, $last_item);
        return $all_subsets;
    }
}

Answer 5

如果你只是需要找到所有子集，它就像这样

function subsets($ary) {
    $out = array();
    $len = count($ary);
    for($i = 0; $i < (1 << $len); $i++) {
        for($p = array(), $j = 0; $j < $len; $j++)
            if($i & (1 << $j))
                $p[] = $ary[$j];
        $out[] = $p;
    }
    return $out;
}

我如何在PHP中添加这些：@ {$ subset}，[@ext_subset]，[$ last_item]？

5 个答案: