Ruby - 以随机顺序返回一个数组

Question

在Ruby中以随机顺序返回数组的最简单方法是什么？任何可以在IRB会话中使用的很好和简短的东西

[1,2,3,4,5].random()
# or 
random_sort([1,2,3,4,5])

Answer 1

array.shuffle

Answer 2

如果你没有[] .shuffle,[] .sort_by {rand}就像sepp2k所指出的那样工作..sort_by暂时用某种东西替换每个元素,以便进行排序,在本例中是一个随机数.

[] .sort {rand-0.5}但是,不会正常洗牌.如果对阵列进行随机排序,某些语言(例如某些Javascript实现)不能正确地重排数组,有时会产生相当大的后果.

JS分析(带图表!):http://www.robweir.com/blog/2010/02/microsoft-random-browser-ballot.html

Ruby也不例外!它有同样的问题.:)

#sort a bunch of small arrays by rand-0.5
a=[]
100000.times{a <<  [0,1,2,3,4].sort{rand-0.5}}

#count how many times each number occurs in each position
b=[]
a.each do |x|
    x.each_index do |i|
        b[i] ||=[]
        b[i][x[i]] ||= 0
        b[i][x[i]] += 1
    end
end
p b

=>

[[22336, 18872, 14814, 21645, 22333],
 [17827, 25005, 20418, 18932, 17818],
 [19665, 15726, 29575, 15522, 19512],
 [18075, 18785, 20283, 24931, 17926],
 [22097, 21612, 14910, 18970, 22411]]

每个元素应在每个位置出现约20000次.[] .sort_by(rand)给出了更好的结果.

#sort with elements first mapped to random numbers
a=[]
100000.times{a <<  [0,1,2,3,4].sort_by{rand}}

#count how many times each number occurs in each position
...

=>

[[19913, 20074, 20148, 19974, 19891],
 [19975, 19918, 20024, 20030, 20053],
 [20028, 20061, 19914, 20088, 19909],
 [20099, 19882, 19871, 19965, 20183],
 [19985, 20065, 20043, 19943, 19964]]

类似于[] .shuffle(这可能是最快的)

[[20011, 19881, 20222, 19961, 19925],
 [19966, 20199, 20015, 19880, 19940],
 [20062, 19894, 20065, 19965, 20014],
 [19970, 20064, 19851, 20043, 20072],
 [19991, 19962, 19847, 20151, 20049]]