如何选取年份

jason680

回复 1# sunzhiguolu


>> ...要精确打印出相隔10年之内行

何来精确??!!

1999121   =>1999/1/21
1999121   =>1999/12/1

1999121 2000111 有一年吗？

sunzhiguolu

回复 11# jason680
还是大神您考虑的周全, 我没有想到. 光想着怎么能够获取到正确的年份信息. 谢谢点醒梦中人.

   

sunzhiguolu

穷举了一个, 只能是玩玩了. 正如 Jason680 大神所说, 这日期咋能精确呀!

1. #!/usr/bin/perl
   
2. use strict;
   
3. use warnings;
   
4. 
   
5. my ($sRegx) = (qr/(0?[1-9]|1[0-2])([1-2]?[0-9]|0?[1-9]|3[01])/);
   
6. while (<DATA>){
   
7.     next if (/\A\s*\z/);
   
8.     my $sYears = $_;
   
9.     my @aYears = ([0..2], [0..2]);
   
10. 
    
11.     foreach my $sIdx (0 .. $#aYears){
    
12.         @{$aYears[$sIdx]}[1,2] = ($1, $2) if (/$sRegx\b/);
    
13.         $aYears[$sIdx][0] = substr ($_, 0, $-[1]);
    
14.         s/\d+\s*//;
    
15.     }
    
16. 
    
17.     foreach my $raYear (@aYears){
    
18.         if (length $raYear->[0] == 2){
    
19.             if ($raYear->[0] =~ /\A([01]\d+)/){
    
20.                 $raYear->[0] = "20$1";
    
21.                 next;
    
22.             }
    
23.             $raYear->[0] = "19$raYear->[0]";
    
24.         }
    
25.     }
    
26. 
    
27.     my ($rA, $rB) = splice (@aYears);
    
28.     next if (abs ($rA->[0] - $rB->[0]) > 10);
    
29.     next if ($rA->[0] - $rB->[0] == 10 and $rA->[1] > $rB->[1]);
    
30.     next if ($rB->[0] - $rA->[0] == 10 and $rB->[1] > $rA->[1]);
    
31.     next if ($rA->[0] - $rB->[0] == 10 and $rA->[1] == $rB->[1] and $rA->[-1] > $rB->[-1]);
    
32.     next if ($rB->[0] - $rA->[0] == 10 and $rB->[1] == $rA->[1] and $rB->[-1] > $rA->[-1]);
    
33.     print $sYears;
    
34. }
    
35. 
    
36. __DATA__
    
37. 1987913 19970912
    
38. 2000315 20100316
    
39. 
    
40. 2004410 20140410
    
41. 
    
42. 1999121 20050806
    
43. 20050806 1999121
    
44. 
    
45. 19901019 2016515
    
46. 
    
47. 99121 050806
    

复制代码

sunzhiguolu

输出结果:

1987913 19970912
2004410 20140410
1999121 20050806
20050806 1999121
99121 050806
050806 99121

sunzhiguolu

我将发生歧义的日期部分, 修改了下.
在进行日期比较的时候, 发现有好多冗余的代码, 不知道如何进行优化下, 还请大家指点, 谢谢大家...

1. #!/usr/bin/perl
   
2. use strict;
   
3. use warnings;
   
4. 
   
5. my ($sRegx) = (qr/(0?[1-9]|1[0-2])([12][0-9]|0?[1-9]|3[01])/);
   
6. while (<DATA>){
   
7.     next if (m/\A\s*\z/);
   
8.     my @aDate = ({},{});
   
9.     my $sYears = $_;
   
10. 
    
11.     foreach my $sIdx (0 .. $#aDate){
    
12.         ($aDate[$sIdx]{'M'}, $aDate[$sIdx]{'D'}) = ($1, $2) if (m/$sRegx\b/);
    
13.         $aDate[$sIdx]{'Y'} = substr ($_, 0, $-[1]);
    
14.         if (length $aDate[$sIdx]{'Y'} == 2){
    
15.             if ($aDate[$sIdx]{'Y'} =~ m/\A([01]\d+)/){
    
16.                 $aDate[$sIdx]{'Y'} = "20$1";
    
17.             }else{
    
18.                 $aDate[$sIdx]{'Y'} = "19$aDate[$sIdx]{'Y'}";
    
19.             }
    
20.         }
    
21.         s/\d+\s*//;
    
22.     }
    
23. 
    
24.     my ($Ya, $Yb) = splice (@aDate);
    
25.     next if (abs ($Ya->{'Y'} - $Yb->{'Y'}) > 10);
    
26.     next if ($Ya->{'Y'} - $Yb->{'Y'} == 10 and $Ya->{'M'} > $Yb->{'M'});
    
27.     next if ($Yb->{'Y'} - $Ya->{'Y'} == 10 and $Yb->{'M'} > $Ya->{'M'});
    
28.     next if ($Ya->{'Y'} - $Yb->{'Y'} == 10 and $Ya->{'M'} == $Yb->{'M'} and $Ya->{'D'} > $Yb->{'D'});
    
29.     next if ($Yb->{'Y'} - $Ya->{'Y'} == 10 and $Yb->{'M'} == $Ya->{'M'} and $Yb->{'D'} > $Ya->{'D'});
    
30.     print $sYears;
    
31. }
    
32. 
    
33. __DATA__
    
34. 19870913 1997912
    
35. 20000315 20100316
    
36. 
    
37. 2004410 20140410
    
38. 
    
39. 19990121 20050806
    
40. 20050806 19990121
    
41. 
    
42. 19901019 2016515
    
43. 
    
44. 991201 050806
    

复制代码

sunzhiguolu

就是在代码的 25~29 行处 (判断不符合条件的日期), 另外, 如果其他地方需要改进的话也请大家指点, 谢谢大家...

jason680

回复 15# sunzhiguolu

How about this easy way ...

while (<DATA>){
    next if (m/\A\s*\z/);

    my $sLine = $_;
    my @aDay;
    foreach(split){

      # 1997912 => 19970912
      #   99121 =>   990121
      s/^((..)+)(...)$/${1}0${3}/;
      
      #991201 => 19991201  , 99 >= 18 , 1918 ~ 1999
      #050806 => 20050806  , 05 <  18 , 2000 ~ 2017
      s/^(..)(....)$/19+($1<18)."$1$2"/e;
   
      push @aDay, $_;
    }
    #print "@aDay\n";
    if($aDay[0] < $aDay[1]){
      @aDay[0,1] = @aDay[1,0];
    }
    #      10 years(YYMMDD) => 100000   
    if( $aDay[0] <= $aDay[1] +  100000){  # ONLY for years(YY0000)
      print $sLine;
    }
}

   

sunzhiguolu

回复 17# jason680
开眼了, 还能这么做加法. 谢谢您指点, 学习了, 学了好几手!!!

   

sunzhiguolu

非常感谢 Jason680 大神指点, 重新修改了下.

1. #!/usr/bin/perl
   
2. use strict;
   
3. use warnings;
   
4. 
   
5. while (<DATA>){
   
6.     next if (m/\A\s*\z/);
   
7.     my @aDate = ({},{});
   
8.     my $sOffset = 0;
   
9.     my $sLine = $_;
   
10. 
    
11.     foreach (split){
    
12.         s/\A((?:\d{2})+)(\d{3})\z/${1}0$2/;
    
13.         s/\A(\d{2})(\d{4})\z/19+($1<18)."$1$2"/e;
    
14.         @{$aDate[$sOffset++]}{'Y','M','D'} = m/\A(\d{4})(\d{2})(\d{2})/g;
    
15.     }
    
16. 
    
17.     my ($Ya, $Yb) = sort {local $" = ""; "@$b{'Y','M','D'}" cmp "@$a{'Y','M','D'}"} @aDate;
    
18.     next if ($Ya->{'Y'} - $Yb->{'Y'} > 10);
    
19.     next if ($Ya->{'Y'} - $Yb->{'Y'} == 10 and $Ya->{'M'} > $Yb->{'M'});
    
20.     next if ($Ya->{'Y'} - $Yb->{'Y'} == 10 and $Ya->{'M'} == $Yb->{'M'} and $Ya->{'D'} > $Yb->{'D'});
    
21.     print $sLine;
    
22. }
    
23. 
    
24. __DATA__
    
25. 19870913 1997912
    
26. 20000315 20100316
    
27. 
    
28. 2004410 20140410
    
29. 
    
30. 19990121 20050806
    
31. 20050806 19990121
    
32. 
    
33. 19901019 2016515
    
34. 
    
35. 991201 050806
    

复制代码

RE_HASH

$>  cat aa.pl ; perl  aa.pl
sub guessDate
{
  $_ = shift;
  my ($y, $m, $d) = /^((?:19|20)*\d{2}|0?\d{1})([01]?\d)([0-2]?\d|3[01])$/;
  $y +=1900 if ($y < 100);
  $y +=100 if ($y < 1950);
  return sprintf "%4d%02d%02d", $y, $m, $d;
}
printf STDERR "DEBUG INFO:\n______________\n";
while (<DATA>)
{
  chomp;
  @A = map {guessDate($_)} split;

  print STDERR "$_\tConverted to :\t", "@A", ($A[0] + 100000 > $A[1])?"\t< 10 years":"\t> 10 years", "\n" ;
  ($A[0] + 100000 > $A[1]) ? push @{$O->{'10 Years -'}}, $_: push @{$O->{'10 years +'}}, $_  ;
}


printf "OUTPUT:\n______________\n%s", join "\n", @{$O->{'10 Years -'}};

__DATA__
19870913 1997912
20000315 20100316
0411 0411
2004410 20140410
89112 90112
19990121 20050806
20050806 19990121
01121 02122
19901019 2016515
991201 050806

DEBUG INFO:
______________
19870913 1997912        Converted to :  19870913 19970912       < 10 years
20000315 20100316       Converted to :  20000315 20100316       > 10 years
0411 0411       Converted to :  20040101 20040101       < 10 years
2004410 20140410        Converted to :  20040410 20140410       > 10 years
89112 90112     Converted to :  19891102 19901102       < 10 years
19990121 20050806       Converted to :  19990121 20050806       < 10 years
20050806 19990121       Converted to :  20050806 19990121       < 10 years
01121 02122     Converted to :  20011201 20021202       < 10 years
19901019 2016515        Converted to :  19901019 20160515       > 10 years
991201 050806   Converted to :  19991201 20050806       < 10 years
        Converted to :          < 10 years
OUTPUT:
______________
19870913 1997912
0411 0411
89112 90112
19990121 20050806
20050806 19990121
01121 02122
991201 050806