2013-12-17 21 views
3

只需在一副紙牌中使用2-9。在Ruby中創建列表的更有效方式

什麼是首選和/或最有效的方式來獲取在Ruby中創建的套牌?

下面是我的問題,還是我該怎麼做?

deck = [] 
suits = ["spades", "diamonds", "clubs", "hearts"] 

for x in suits 
    for y in 2..9 
    w = y.to_s 
    deck.push(w+" of "+x) 
    end 
end 

回答

2
suits = ["spades", "diamonds", "clubs", "hearts"] 
deck = suits.product((2..9).to_a).map { |x,y| "#{y} of #{x}" } 

如果希望所有52張牌:

suits = ["spades", "diamonds", "clubs", "hearts"] 
faces = { 1 => 'Ace', 11 => 'Jack', 12 => 'Queen', 13 => 'King' } 
deck = suits.product((1..13).to_a).map { |x,y| "#{faces[y]||y} of #{x}" } 
# => Ace of spades, 2 of spades, ..., King of spades, Ace of diamonds, etc. 
1

如何:

suits.flat_map {|s| (2..9).map{|r| "#{r} of #{s}"}} 
2

嗯,你可以只使用地圖:

suits = ["spades", "diamonds", "clubs", "hearts"] 
deck = suits.map { |d| (2..9).map { |x| "#{x} of #{d}" } }.flatten 

但效率卻大致相同。

4

我喜歡product瓦特/塊

a, suits = [], ["spades", "diamonds", "clubs", "hearts"] 
suits.product((2..9).to_a) {|t,n| a << "#{n} of #{t}"} 
0
suits, values, deck = %w(spades hearts diamonds clubs), [*2..9], values.product(suits).map { |card| card * ' of ' } 
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