You don't assign anything to $tokpos. The assignment is part of a comment - syntax highlighting in your editor should've told you. strict would've told you, too.

你沒有給$ tokpos分配任何東西。作業是評論的一部分 - 編輯器中的語法高亮應該告訴你。嚴格也會告訴你的。

Also, you should probably use $prevtokpos in the division: $tokpos is the new value that you haven't met before. To get the output for the last token, you have to process it outside the loop, as there's no $tokpos to replace it. To avoid code repetition, use a subroutine to do that:

此外,您應該在分部中使用$ prevtokpos:$ tokpos是您之前未遇到的新值。要獲取最后一個令牌的輸出,您必須在循環外處理它,因為沒有$ tokpos來替換它。為避免代碼重復,請使用子例程來執行此操作:

#!/usr/bin/perl
use warnings;
use strict;

my %SumHash;
my %ScoreHash;

sub output {
    my $token = shift;
    for (keys %ScoreHash) {
        my $normscore = $ScoreHash{$_} / $SumHash{$token};
        print "$_\t$ScoreHash{$_}\t$normscore\n";
    }
    undef %ScoreHash;
    undef %SumHash;
}

my $prevtokpos;
while (<DATA>){
    chomp;
    if (my ($tokpossense,$rawscore) = /^\s{4}(.+): (\d+)/) {
        my ($tok, $pos, $sensenr) = split /#/, $tokpossense;
        my $tokpos = "$tok\#$pos";
        if (defined $prevtokpos && $tokpos ne $prevtokpos) {
            output($prevtokpos);
        }

        $prevtokpos = $tokpos;
        $SumHash{$tokpos} += $rawscore;
        $ScoreHash{$tokpossense} = $rawscore;
    }
}
output($prevtokpos);

__DATA__
i#CL take#v#17 my#CL checks#n#1 to#CL the#CL bank#n#2 .#IT 
Context: i#CL <target>take#v</target> my#CL checks#n to#CL the#CL bank#n
  Scores for take#v
    take#v#1: 17
    take#v#10: 158
    take#v#17: 174
  Winning score: 174
Context: i#CL take#v my#CL <target>checks#n</target> to#CL the#CL bank#n .#IT
  Scores for checks#n
    check#n#1: 198
    check#n#2: 117
    check#n#3: 42
  Winning score: 198
Context: take#v my#CL checks#n to#CL the#CL <target>bank#n</target> .#IT
  Scores for bank#n
    bank#n#1: 81
    bank#n#2: 202
    bank#n#3: 68
    bank#n#4: 37
  Winning score: 202

You're confusing yourself by trying to print the results as soon as $tokpos changes. For one thing it's the values for $prevtokpos that are complete, but your trying to output the data for $tokpos; and also you're never going to display the last block of data because you require a change in $tokpos to trigger the output.

$ tokpos更改后,嘗試打印結果會讓您感到困惑。首先,$ prevtokpos的值是完整的,但是你試圖輸出$ tokpos的數據;而且你永遠不會顯示最后一個數據塊,因為你需要更改$ tokpos來觸發輸出。

It's far easier to accumulate all the data for a given file and then print it when the end of file is reached. This program works by keeping the three values $tokpos, $sense, and $rawscore for each line of the output in array @results, together with the total score for each value of $tokpos in %totals. Then it's simply a matter of dumping the contents of @results with an extra column that divides each value by the corresponding total.

累積給定文件的所有數據然后在到達文件末尾時打印它會容易得多。這個程序的工作原理是保持數組@results中輸出的每一行的三個值$ tokpos,$ sense和$ rawscore,以及%tokpos中每個值的總得分。%total。然后,只需將@results的內容轉儲為一個額外的列,將每個值除以相應的總數。

use strict;
use warnings;
use 5.014; # For non-destructive substitution

for my $file ( @ARGV ) {

    open my $fh, '<', $file or die $!;

    my (@results, %totals);

    while ( <$fh> ) {
        chomp;
        next unless my ($tokpos, $sense, $rawscore) = / ^ \s{4} ( [^#]+ \# [^#]+ ) \# (\d+) : \s+ (\d+)  /x;
        push @results, [ $tokpos, $sense, $rawscore ];
        $totals{$tokpos} += $rawscore;
    }

    print "** $file **\n";
    for my $item ( @results ) {
        my ($tokpos, $sense, $rawscore) = @$item;
        printf "%s\t%s\t%6.4f\n", $tokpos.$sense, $rawscore, $rawscore / $totals{$tokpos};
    }
    print "\n";
}

output

** tokpos.txt **
take#v#1  17  0.0487
take#v#10 158 0.4527
take#v#17 174 0.4986
check#n#1 198 0.5546
check#n#2 117 0.3277
check#n#3 42  0.1176
bank#n#1  81  0.2088
bank#n#2  202 0.5206
bank#n#3  68  0.1753
bank#n#4  37  0.0954