Array @p is a multiline array, e.g. $p[1] is the second line.
Array @p是一个多行数组，例如$ p [1]是第二行。
This code will explain what I want:
这段代码将解释我想要的东西：
$size=@p; # line number of array @p
for($i=0; $i<$size; $i++)
{
@p{$i}= split(/ +/,$p[$i]);
}
I want the result should be like this:
我希望结果应该是这样的：
@p0 = $p[0] first line of array @p goes to array @p0;
@p1 = $p[1] second line of array @p goes to array @p1;
...
...
and so on.
等等。
But above code does not work, how can I do it?
但是上面的代码不起作用，我该怎么办呢？
2 个解决方案
#1
4
It is a bad idea to dynamically generate variable names.
动态生成变量名称是个坏主意。
I suggest the best solution here is to convert each line in your @p array to an array of fields.
我建议这里最好的解决方案是将@p数组中的每一行转换为字段数组。
Lets suppose you have a better name for @p, say @lines. Then you can write
让我们假设你有一个更好的@p名称，比如说@lines。然后你就可以写了
my @lines = map [ split ], <$fh>;
to read in all the lines from the file handle $fh and split them on whitespace. The first field of the first line is then $lines[0][0]. The third field of the first line is $lines[0][2] etc.
读入文件句柄$ fh中的所有行并将它们拆分为空格。第一行的第一个字段是$ lines [0] [0]。第一行的第三个字段是$ lines [0] [2]等。
#2
3
First, the syntax @p{$i} accesses the entry with the key $i in a hash %p, and returns it in list context. I don't think you meant that. use strict; use warnings; to get warned about undeclared variables.
首先，语法@p {$ i}在散列％p中使用键$ i访问该条目，并在列表上下文中返回它。我不认为你的意思。用严格;使用警告;警告未申报的变量。
You can declare variables with my, e.g. my @p; or my $size = @p;
您可以使用我的声明变量，例如我的@p;或我的$ size = @p;
Creating variable names on the fly is possible, but a bad practice. The good thing is that we don't need to: Perl has references. A reference to an array allows us to nest arrays, e.g.
可以动态创建变量名称，但这是一种不好的做法。好消息是我们不需要：Perl有参考。对数组的引用允许我们嵌套数组，例如
my @AoA = (
[1, 2, 3],
["a", "b"],
);
say $AoA[0][1]; # 2
say $AoA[1][0]; # a
We can create an array reference by using brackets, e.g. [ @array ], or via the reference operator \:
我们可以使用括号创建数组引用，例如[@array]，或通过引用运算符\：
my @inner_array = (1 .. 3);
my @other_inner = ("a", "b");
my @AoA = (\@inner_array, \@other_array);
But careful: the array references still point to the same array as the original names, thus
但要小心：数组引用仍然指向与原始名称相同的数组，因此
push @other_inner, "c";
also updates the entry in @AoA:
还会更新@AoA中的条目：
say $AoA[1][2]; # c
Translated to your problem this means that you want:
转换为您的问题这意味着您想要：
my @pn;
for (@p) {
push @pn, [ split /[ ]+/ ];
}
There are many other ways to express this, e.g.
还有许多其他方式来表达这一点，例如：
my @pn = map [ split /[ ]+/ ], @p;
or
要么
my @pn;
for my $i ( 0 .. $#p ) {
$pn[$i] = [ split /[ ]+/, $p[$i] ];
}
To learn more about references, read
要了解有关参考的更多信息，请阅读
- perlreftut,
- perlreftut，
- perldsc, and
- perldsc，和
- perlref.
- perlref。
#1
4
It is a bad idea to dynamically generate variable names.
动态生成变量名称是个坏主意。
I suggest the best solution here is to convert each line in your @p array to an array of fields.
我建议这里最好的解决方案是将@p数组中的每一行转换为字段数组。
Lets suppose you have a better name for @p, say @lines. Then you can write
让我们假设你有一个更好的@p名称，比如说@lines。然后你就可以写了
my @lines = map [ split ], <$fh>;
to read in all the lines from the file handle $fh and split them on whitespace. The first field of the first line is then $lines[0][0]. The third field of the first line is $lines[0][2] etc.
读入文件句柄$ fh中的所有行并将它们拆分为空格。第一行的第一个字段是$ lines [0] [0]。第一行的第三个字段是$ lines [0] [2]等。
#2
3
First, the syntax @p{$i} accesses the entry with the key $i in a hash %p, and returns it in list context. I don't think you meant that. use strict; use warnings; to get warned about undeclared variables.
首先，语法@p {$ i}在散列％p中使用键$ i访问该条目，并在列表上下文中返回它。我不认为你的意思。用严格;使用警告;警告未申报的变量。
You can declare variables with my, e.g. my @p; or my $size = @p;
您可以使用我的声明变量，例如我的@p;或我的$ size = @p;
Creating variable names on the fly is possible, but a bad practice. The good thing is that we don't need to: Perl has references. A reference to an array allows us to nest arrays, e.g.
可以动态创建变量名称，但这是一种不好的做法。好消息是我们不需要：Perl有参考。对数组的引用允许我们嵌套数组，例如
my @AoA = (
[1, 2, 3],
["a", "b"],
);
say $AoA[0][1]; # 2
say $AoA[1][0]; # a
We can create an array reference by using brackets, e.g. [ @array ], or via the reference operator \:
我们可以使用括号创建数组引用，例如[@array]，或通过引用运算符\：
my @inner_array = (1 .. 3);
my @other_inner = ("a", "b");
my @AoA = (\@inner_array, \@other_array);
But careful: the array references still point to the same array as the original names, thus
但要小心：数组引用仍然指向与原始名称相同的数组，因此
push @other_inner, "c";
also updates the entry in @AoA:
还会更新@AoA中的条目：
say $AoA[1][2]; # c
Translated to your problem this means that you want:
转换为您的问题这意味着您想要：
my @pn;
for (@p) {
push @pn, [ split /[ ]+/ ];
}
There are many other ways to express this, e.g.
还有许多其他方式来表达这一点，例如：
my @pn = map [ split /[ ]+/ ], @p;
or
要么
my @pn;
for my $i ( 0 .. $#p ) {
$pn[$i] = [ split /[ ]+/, $p[$i] ];
}
To learn more about references, read
要了解有关参考的更多信息，请阅读
- perlreftut,
- perlreftut，
- perldsc, and
- perldsc，和
- perlref.
- perlref。