随机填充数组 - perl

Question

我需要知道如何在 perl 中随机填充数组。例如：我想声明一个包含 10 个小于 60 的元素的数组。有人可以解释我该怎么做或给我发送任何指南吗？我将不胜感激。

Answer 1

你可以简单地做：

my @r = map int(rand(60)), 0..9;
say Dumper\@r;

Answer 2

您可以利用 perl 的散列键随机排序功能。这将在每次运行时随机填充一个包含 10 个元素的数组：

use warnings;
use strict; 

my @nums = (1 .. 60);

my %data;

$data{$_}++ foreach @nums;

my $count = 0;
my @random;
foreach (keys %data){
    $count++;
    push @random, $_ if $count <= 10;
}

Answer 3

我假设你的意思是 "ten non-negative integers less than 60"。

有重复的可能性：

my @rands = map { int(rand(60)) } 1..10;

例如，

$ perl -E'say join ",", map { int(rand(60)) } 1..10;'
0,28,6,49,26,19,56,32,56,16       <-- 56 is repeated

$ perl -E'say join ",", map { int(rand(60)) } 1..10;'
15,57,50,16,51,58,46,7,17,53

$ perl -E'say join ",", map { int(rand(60)) } 1..10;'
13,57,26,47,30,14,47,55,39,39     <-- 47 and 39 are repeated

没有重复的可能性：

use List::Util qw( shuffle );

my @rands = (shuffle 0..59)[0..9];

例如，

$ perl -MList::Util=shuffle -E'say join ",", (shuffle 0..59)[0..9];'
13,50,8,21,11,24,28,51,55,38

$ perl -MList::Util=shuffle -E'say join ",", (shuffle 0..59)[0..9];'
1,0,58,46,47,49,52,33,5,13

$ perl -MList::Util=shuffle -E'say join ",", (shuffle 0..59)[0..9];'
19,43,45,49,23,53,2,38,59,35

随机填充数组 - perl

Random flling the array - perl

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random

perl