日付/時刻文字列を12時間形式で並べ替える

日付/時刻文字列を12時間形式で並べ替える

次の内容を含むログファイルがあります。

11-12-2014 - 03:03:59AM lat = 41.990516; lon = -93.430704<br>
11-12-2014 - 03:05:15AM lat = 41.001546; lon = -93.443352<br>
11-12-2014 - 03:11:50AM lat = 42.039054; lon = -93.442001<br>
11-12-2014 - 12:08:03AM lat = 41.937911; lon = -93.369249<br>
11-12-2014 - 12:11:29AM lat = 41.949656; lon = -93.329133<br>
11-12-2014 - 12:23:02AM lat = 42.025385; lon = -93.347026<br>
11-12-2014 - 12:29:10AM lat = 41.033341; lon = -93.380586<br>
11-12-2014 - 12:38:08AM lat = 41.036720; lon = -93.436851<br>
11-12-2014 - 12:45:20AM lat = 41.998129; lon = -93.400943<br>
11-12-2014 - 12:53:36AM lat = 41.961489; lon = -93.414624<br>

どうすれば24時間制に変換して正しくソートできますか?

ベストアンサー1

そしてperl

$ perl -MTime::Piece -anle '
  $F[2] = Time::Piece->strptime($F[2],"%r")->strftime("%H:%M:%S");
  push @out, [$F[2]."-".join("-", reverse(split("-",$F[0]))), join(" ",@F)];
  END {
    print for map  { $_->[1] }
              sort { $a->[0] cmp $b->[0] } @out;
}' file
11-12-2014 - 00:08:03 lat = 41.937911; lon = -93.369249<br>
11-12-2014 - 00:11:29 lat = 41.949656; lon = -93.329133<br>
11-12-2014 - 00:23:02 lat = 42.025385; lon = -93.347026<br>
11-12-2014 - 00:29:10 lat = 41.033341; lon = -93.380586<br>
11-12-2014 - 00:38:08 lat = 41.036720; lon = -93.436851<br>
11-12-2014 - 00:45:20 lat = 41.998129; lon = -93.400943<br>
11-12-2014 - 00:53:36 lat = 41.961489; lon = -93.414624<br>
11-12-2014 - 03:03:59 lat = 41.990516; lon = -93.430704<br>
11-12-2014 - 03:05:15 lat = 41.001546; lon = -93.443352<br>
11-12-2014 - 03:11:50 lat = 42.039054; lon = -93.442001<br>

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