Question

我有一系列日期

@dates = qw(2/1/1989 2/1/1970 2/1/1970 2/1/1989 6/1/1970 12/1/1970);

我需要将它从最旧到最新排序，并在排序后返回未排序数组的索引。

输出应该是这样的

#sorted array
2/1/1970
2/1/1970
6/1/1970
12/1/1970
2/1/1989
2/1/1989

#indexes
1 2 4 5 0 3

Answer 1

my @dates = qw(2/1/1989 2/1/1970 2/1/1970 2/1/1989 6/1/1970 12/1/1970);

my @idx = map $_->[0],
  sort {
    # compare years
    $a->[3] <=> $b->[3] ||
    # compare months
    $a->[1] <=> $b->[1] ||
    # compare days
    $a->[2] <=> $b->[2] ||
    # compare index for stable sort for duplicate values
    $a->[0] <=> $b->[0] 
  }
  map [$_, split /\D/, $dates[$_] ],
  0 .. $#dates;

print "indexes @idx\n";
# sorted values
print "$_\n" for @dates[@idx];

输出

indexes 1 2 4 5 0 3
2/1/1970
2/1/1970
6/1/1970    
12/1/1970
2/1/1989
2/1/1989

Answer 2

使用Time::Piece和Schwartzian Transform

use strict;
use warnings;

use Time::Piece;

my @dates = qw(2/1/1989 2/1/1970 2/1/1970 2/1/1989 6/1/1970 12/1/1970);

my @idx = map { $_->[0] }
    sort { $a->[1] <=> $b->[1] }
    map { [$_, Time::Piece->strptime($dates[$_], '%m/%d/%Y') ] }
    (0..$#dates);

print "Indexes: @idx\n";

print "Dates: @dates[@idx]\n";

输出：

Indexes: 1 2 4 5 0 3
Dates: 2/1/1970 2/1/1970 6/1/1970 12/1/1970 2/1/1989 2/1/1989

也可以执行以下操作：

my @idx = sort { 
    Time::Piece->strptime($dates[$a], '%m/%d/%Y') <=> Time::Piece->strptime($dates[$b], '%m/%d/%Y')
} (0..$#dates);

排序日期数组并在perl中返回索引

2 个答案: