Python 如何在输出窗口中并排打印单独的多行ascii符号

免责声明：一般来说，我是一个相对较新的python用户和程序员

我正在为一组卡片构建一个类，对于

\uuuu str\uuuu

方法，我希望将当前在卡片组中的卡片的ascii符号返回为十三行四列。稍后，当我实际使用这个类进行游戏时，在显示玩家的手时，我将需要类似的逻辑。我希望找到一种方法来做到这一点，其中列的数量是可变的，行的数量取决于列的数量和列表的长度（或者简单地说，当卡片用完时停止）。这样，我的

\uuuu str\uuuu

返回就可以了，它有4列，玩家的手可以控制不同的列数

因为我只想理解这样做的逻辑，所以我将问题简化为下面的代码。我已经做了很多研究，但是我还没有找到一个我能理解或者不使用导入库的例子。我已经学会在print语句后使用逗号来防止强制换行，但即使使用该工具，我也无法找到使用for和while循环的方法。我还将粘贴来自最终用例的一些代码。这只是许多没有成功的例子，可能很可怕，但这正是我所处的位置

简化用例：

# Each string in each list below would actually be one line of ascii art for
# the whole card, an example would be '|{v}   {s}   |'

deck = [['1','2','3','4'],
    ['5','6','7','8'],
    ['9','10','11','12'],
    ['a','b','c','d'],
    ['e','f','g','h'],
    ['i','j','k','l']]

# expected output in 3 columns:
#
#   1   5   9
#   2   6   10
#   3   7   11
#   4   8   12
#
#   a   e   i
#   b   f   j
#   c   g   k
#   d   h   l
#
# expected output in 4 columns:
#
#   1   5   9   a
#   2   6   10  b
#   3   7   11  c
#   4   8   12  d
#
#   e   i
#   f   j
#   g   k
#   h   l

def __str__(self):

    # WORKS empty list to hold ascii strings
    built_deck = []

    # WORKS fill the list with ascii strings [card1,card2,card3,card4...]
    for card in self.deck:
        built_deck.append(self.build_card(str(card[0]),str(card[1:])))

    # WORKS transform the list to [allCardsRow1,allCardsRow2,allCardsRow3,allCardsRow4...]
    built_deck = list(zip(*built_deck))

    # mark first column as position
    position = 1

    # initialize position to beginning of deck
    card = 0

    # Try to print the table of cards ***FAILURE***
    for item in built_deck:
        while position <= 4:
            print(f'{item[card]}\t',)
            card += 1
            continue
        position = 1
        print(f'{item[card]}')
        card += 1
    #return built_deck

最终用例：

# Each string in each list below would actually be one line of ascii art for
# the whole card, an example would be '|{v}   {s}   |'

deck = [['1','2','3','4'],
    ['5','6','7','8'],
    ['9','10','11','12'],
    ['a','b','c','d'],
    ['e','f','g','h'],
    ['i','j','k','l']]

# expected output in 3 columns:
#
#   1   5   9
#   2   6   10
#   3   7   11
#   4   8   12
#
#   a   e   i
#   b   f   j
#   c   g   k
#   d   h   l
#
# expected output in 4 columns:
#
#   1   5   9   a
#   2   6   10  b
#   3   7   11  c
#   4   8   12  d
#
#   e   i
#   f   j
#   g   k
#   h   l

def __str__(self):

    # WORKS empty list to hold ascii strings
    built_deck = []

    # WORKS fill the list with ascii strings [card1,card2,card3,card4...]
    for card in self.deck:
        built_deck.append(self.build_card(str(card[0]),str(card[1:])))

    # WORKS transform the list to [allCardsRow1,allCardsRow2,allCardsRow3,allCardsRow4...]
    built_deck = list(zip(*built_deck))

    # mark first column as position
    position = 1

    # initialize position to beginning of deck
    card = 0

    # Try to print the table of cards ***FAILURE***
    for item in built_deck:
        while position <= 4:
            print(f'{item[card]}\t',)
            card += 1
            continue
        position = 1
        print(f'{item[card]}')
        card += 1
    #return built_deck

def\uuuu str\uuuuuu（自）：
#使用空列表来保存ascii字符串
已建甲板=[]
#WORKS使用ascii字符串填充列表[card1、card2、card3、card4…]
对于self.deck中的卡：
build_-deck.append（self.build_-card（str（card[0]），str（card[1:]））
#WORKS将列表转换为[allCardsRow1、allCardsRow2、allCardsRow3、allCardsRow4…]
已建甲板=列表（zip（*已建甲板））
#将第一列标记为位置
位置=1
#将位置初始化为甲板的起始位置
卡=0
#尝试打印卡片表***失败***
对于内置甲板上的物品：
而position的诀窍是要认识到，你正在做的是对你的卡片矩阵进行连续的转置，然后打印它们，而你执行操作的矩阵的大小就是你想要显示的项目的数量。我们可以使用python中的
zip
获得转置

def display(deck, number_of_columns):
    col = 0
    while col < len(deck):
        temp = deck[col:col+number_of_columns]
        temp = zip(*temp)
        for x in temp:
            for y in x:
                print(y, end=" ")
            print()
        col += number_of_columns
display(deck, 3)
print()
display(deck, 4)