如何在嵌套列表理解中打印新行？

Question

是python的新版本。我想使用列表理解来显示两个骰子的可能性。我得到了输出。但是，当y达到6（y == 6）时是否可以打印新行？

我的编码：

d1=(1,2,3,4,5,6) 
d2=(1,2,3,4,5,6) 
print([(x,y) for x in d1 for y in d2])

我的输出：

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

我的期望输出：

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), 
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), 
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), 
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), 
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), 
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

Answer 1

打印具有可选参数end通常这是一个新行，但是您可以对其进行修改，以使其成为空格或您想要的任何内容。您可以使用它来重新创建所需的输出：

d1=(1,2,3,4,5,6)
d2=(1,2,3,4,5,6)
# this is bad style and should be avoided because of side effects (see comments)
([print((x,y), end='\n' if y==len(d2) else ' ') for x in d1 for y in d2]) 

更好的无副作用：

for tup in [(x,y) for x in d1 for y in d2]:
    print(tup, end='\n' if tup[1]==len(d2) else ' ')

输出：

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

而且，如果您真的想要一个真实的列表输出（带有所有特殊字符（“，”，“ [”，“]”），那么您还可以做更多的工作：

def print_nice_list(d1,d2):
    print("[",end="")
    for tup in [(x,y) for x in d1 for y in d2]:
        end = ", "
        if tup[1] == len(d2) and tup[0]==len(d1):
            end = "]"
        elif tup[1] == len(d2):
            end = ",\n"
        print(tup, end=end )

print_nice_list(d1,d2) 

输出：

[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]

编辑：根据评论重新设计答案。

Answer 2

如果仅用于可视化，则应使用pprint：

>>> import pprint
>>> lst = [(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]
>>> pprint.pprint(lst, compact=True, width=50)
[(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
 (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
 (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
 (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
 (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
 (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)]