ruby​​ - 如何重新排列基於另一個數組的數組？

Question

我正在寫一個'洗牌程序'來和Ruby一起玩。認爲這是個人的功課，我自己分配了解更多:)

，我想輸出是在這裏：

----Triple Cut Deck on 3rd and 5th cards--------- 
    -- reset 
Number: 1, Position: 3, Suit: Clubs, Card: 3 
Number: 2, Position: 4, Suit: Clubs, Card: 4 
Number: 3, Position: 1, Suit: Clubs, Card: 5 
Number: 4, Position: 2, Suit: Clubs, Card: 6 
Number: 5, Position: 5, Suit: Clubs, Card: Ace 
Number: 6, Position: 6, Suit: Clubs, Card: 2 

，但我得到：

----Triple Cut Deck on 3rd and 5th cards--------- 
    -- reset 
Number: 1, Position: 3, Suit: Clubs, Card: 3 
Number: 2, Position: 4, Suit: Clubs, Card: 4 
Number: 3, Position: 5, Suit: Clubs, Card: 5 
Number: 4, Position: 5, Suit: Clubs, Card: 5 
Number: 5, Position: 6, Suit: Clubs, Card: 6 
Number: 6, Position: 6, Suit: Clubs, Card: 6 

基本上，我想得到的是重新命名的牌，以便'Ace，2,3,4,5,6「將他們的卡順序從」1,2,3,4,5「改變爲」5,6,3,4,1 ，2「，換句話說，底部的兩張牌（順序排列），底部的兩張牌在頂部和中間保持不變，這是一個3路切割的版本

我很難讓這個數組'重新排序'正常工作。 眼下卡「秩」和card_position越來越弄亂了如上所示，用重複，等等

class Card 
    RANKS = %w(Ace 2 3 4 5 6 7 8 9 10 J Q K) 
    SUITS = ['Clubs', 'Diamonds', 'Hearts', 'Spades'] 
    SCORES = [1..54] 
    attr_accessor :rank, :suit, :card_position 

    def initialize(id, rank='', suit='', card_position=0) 
    self.card_position = id 
    self.rank = RANKS[(id % 14)-1] 
    self.suit = SUITS[(id/14)] 
    end 
end 

class Deck 
    DECK_SIZE = 6 
    attr_accessor :cards 
    def initialize 
    self.cards = (1..DECK_SIZE).to_a.collect { |id| Card.new(id) } 
    @deck = cards 
    end 

    def process_cards 

    puts "\n----Triple Cut Deck on 3rd and 5th cards---------" 
    self.triple_cut_deck(3, 5, true) 
    self.show_deck 

    end 

    def show_deck 
    @deck.sort_by!(&:card_position).each_with_index do |card, index| 
     puts 'Number: ' + (index+1).to_s + ", Position: #{card.card_position.to_s}, Suit: #{card.suit.to_s}, Card: #{card.rank.to_s}" 
    end 
    end 

    def triple_cut_deck(top_cut, bottom_cut, reset_deck=false) 
    reset_the_deck(reset_deck) 

    top_cut-= 1 
    bottom_cut-= 1 
    deck_array_size = DECK_SIZE-1 

    @new_deck = [] 
    @new_deck[0..1] = @deck[4..5] 
    @new_deck[2..3] = @deck[2..3] 
    @new_deck[4..5] = @deck[0..1] 

    DECK_SIZE.times do |card| 
     @deck[card].card_position= @new_deck[card].card_position 
     @deck[card].card_position= @new_deck[card].card_position 
     @deck[card].card_position= @new_deck[card].card_position 
    end 
    end 


    def reset_the_deck(reset_deck) 
    puts reset_deck == true ? " -- reset" : 'no-reset' 
    initialize if (true && reset_deck) 
    end 

end 

Answer 1

這是你想要的嗎？

a = [1,2,3,4,5,6] 

n = 2 
b = a[-n, n] + a[n..-(n+1)] + a[0,n] 

p a # => [1,2,3,4,5,6] 
p b # => [5,6,3,4,1,2] 

Answer 2

不大可能

是最快的解決方案，但可以在兩個陣列zip一起（陣列排序鍵第一）並對結果進行排序，如下所示：

a = [ 8, 4, 2, 7, 5 ] 
b = [ 5, 7, 0, 3, 3 ] 

a.zip(b).sort.transpose.last 
# => [0, 7, 3, 3, 5] 

Answer 3

您試圖將位置保留爲數組中的方法和實際位置。這是很多簿記。切割和三倍或四倍切割基本上是相同的。

Ranks = %w(Ace 2 3 4 5 6 7 8 9 10 J Q K) 
Suits = ['Clubs', 'Diamonds', 'Hearts', 'Spades'] 
Card = Struct.new(:rank, :suit) 
deck = Suits.product(Ranks).map{|suit, rank| Card.new(rank, suit) } 
hand = deck[0, 6] 
def cut(hand, *at) #returns a new hand, multi-cutted 
    indices = (at.map{|i| i-1}+[0, hand.size]).sort 
    res = indices.each_cons(2).map{|i1,i2| hand[i1..i2-1] } 
    res.reverse.flatten 
end 
p cut(hand, 3, 5) 

輸出：

[#<struct Card rank="Ace", suit="Clubs">, #<struct Card rank="2", suit="Clubs">, 
    #<struct Card rank="3", suit="Clubs">, #<struct Card rank="4", suit="Clubs">, 
    #<struct Card rank="5", suit="Clubs">, #<struct Card rank="6", suit="Clubs">] 

    [#<struct Card rank="5", suit="Clubs">, #<struct Card rank="6", suit="Clubs">, 
    #<struct Card rank="3", suit="Clubs">, #<struct Card rank="4", suit="Clubs">, 
    #<struct Card rank="Ace", suit="Clubs">, #<struct Card rank="2", suit="Clubs">]