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以 12 小时格式对日期时间字符串进行排序

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以 12 小时格式对日期时间字符串进行排序

我有一个日志文件，其内容如下：

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>

如何将其转换为 24 小时时间并正确排序？

答案1

和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>

答案2

您可以使用 GNUdate命令来完成此操作。它可以接受字符串并打印出相应的日期：

$ date -d "11/11/2014 04:12:03PM"
Tue Nov 11 16:12:03 CET 2014

但请注意，它不喜欢DD-MM-YYY：

$ date -d "11-11-2014"
date: invalid date ‘11-11-2014’

因此，首先sed对您的文件运行 a 以将所有内容替换-为/.然后，传递它read以将每个字段放入一个单独的变量中，进行转换和排序：

$ sed 's#-#/#g' file | while read date _ hour rest; do 
    echo "$(date -d "$date $hour" +"%F - %R:%S") $rest"
  done | sort -h
2014-11-12 - 00:08:03 lat = 41.937911; lon = /93.369249<br>
2014-11-12 - 00:11:29 lat = 41.949656; lon = /93.329133<br>
2014-11-12 - 00:23:02 lat = 42.025385; lon = /93.347026<br>
2014-11-12 - 00:29:10 lat = 41.033341; lon = /93.380586<br>
2014-11-12 - 00:38:08 lat = 41.036720; lon = /93.436851<br>
2014-11-12 - 00:45:20 lat = 41.998129; lon = /93.400943<br>
2014-11-12 - 00:53:36 lat = 41.961489; lon = /93.414624<br>
2014-11-12 - 03:03:59 lat = 41.990516; lon = /93.430704<br>
2014-11-12 - 03:05:15 lat = 41.001546; lon = /93.443352<br>

这适用于您的示例，但如果您还需要将二月 ( 02) 排序在 11 月 ( 11) 之前，则会失败。因此，一个技巧是将日期打印为自纪元以来的秒数，对其进行排序，然后将其删除：

$ sed 's#-#/#g' file | while read date _ hour rest; do 
  printf "%s\t%s %s\n" "$(date -d "$date $hour" +"%s")" "$date - $hour" "$rest"
done | sort | cut -f 2-
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>
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>

或者，以 24 小时格式打印日期：

$ sed 's#-#/#g' file | while read date _ hour rest; do 
    printf "%s\t%s %s\n" "$(date -d "$date $hour" +"%s")" \ 
    "$(date -d "$date $hour" +"%F - %R:%S")" "$rest"
  done | sort | cut -f 2-
2014-11-12 - 00:08:03 lat = 41.937911; lon = /93.369249<br>
2014-11-12 - 00:11:29 lat = 41.949656; lon = /93.329133<br>
2014-11-12 - 00:23:02 lat = 42.025385; lon = /93.347026<br>
2014-11-12 - 00:29:10 lat = 41.033341; lon = /93.380586<br>
2014-11-12 - 00:38:08 lat = 41.036720; lon = /93.436851<br>
2014-11-12 - 00:45:20 lat = 41.998129; lon = /93.400943<br>
2014-11-12 - 00:53:36 lat = 41.961489; lon = /93.414624<br>
2014-11-12 - 03:03:59 lat = 41.990516; lon = /93.430704<br>
2014-11-12 - 03:05:15 lat = 41.001546; lon = /93.443352<br>
2014-11-12 - 03:11:50 lat = 42.039054; lon = /93.442001<br>

答案3

使用 python 和dateutil模块 ( pip install dateutil)，您可以直接对日期时间对象进行排序：

#! /usr/bin/env python
import sys
from dateutil.parser import parse

lines = []
for line in open(sys.argv[1]):
    d, rest = line[:24], line[24:]
    lines.append((parse(d), rest))

for x in sorted(lines):
    print x[0], x[1],

从...开始python program.py inputfile

这可以在没有的情况下完成，dateutil但使用它的优点是您不必指定输入时间格式，只要它不模糊即可。

答案4

我会awk使用sort：

awk '{while("date +%T -d" $3|getline x){$3=x}}1' logfile | sort -t- -n -k3 -k1 -k2

首先让我们稍微修改一下日志以具有不同的日期：

11-12-2010 - 03:03:59AM lat = 41.990516; lon = -93.430704
11-12-1998 - 03:05:15AM lat = 41.001546; lon = -93.443352
11-12-2030 - 03:11:50AM lat = 42.039054; lon = -93.442001
11-12-2014 - 12:08:03AM lat = 41.937911; lon = -93.369249
11-12-2014 - 12:11:29AM lat = 41.949656; lon = -93.329133
11-11-2014 - 12:23:02AM lat = 42.025385; lon = -93.347026
11-12-2011 - 12:29:10AM lat = 41.033341; lon = -93.380586
11-12-2011 - 12:38:08AM lat = 41.036720; lon = -93.436851
10-12-2014 - 12:45:20AM lat = 41.998129; lon = -93.400943
11-12-2014 - 12:53:36AM lat = 41.961489; lon = -93.414624

结果将是：

11-12-1998 - 03:05:15 lat = 41.001546; lon = -93.443352
11-12-2010 - 03:03:59 lat = 41.990516; lon = -93.430704
11-12-2011 - 00:29:10 lat = 41.033341; lon = -93.380586
11-12-2011 - 00:38:08 lat = 41.036720; lon = -93.436851
10-12-2014 - 00:45:20 lat = 41.998129; lon = -93.400943
11-11-2014 - 00:23:02 lat = 42.025385; lon = -93.347026
11-12-2014 - 00:08:03 lat = 41.937911; lon = -93.369249
11-12-2014 - 00:11:29 lat = 41.949656; lon = -93.329133
11-12-2014 - 00:53:36 lat = 41.961489; lon = -93.414624
11-12-2030 - 03:11:50 lat = 42.039054; lon = -93.442001

这里的技巧是用作-字段分隔符，并首先在第三个字段（年）上排序，然后在第一个（月）上排序，然后在第二个（日）上排序。所有排序都是数字（-n选项）。

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