带条件的两个列表的乘积

Question

我试图找到一种简洁的方法，从两个列表中构造组合元组列表。考虑以下列表：

x = [1, 2, 3]
y = [0, 3, 4, 5]

目标是获得如下列表：

[(None, 0), (1, None), (2, None), (3, 3), (None, 4), (None, 5)]

因此，匹配对一起进行，如果一个列表中没有对应的元素，那么它应该由某个元素填充（这里，None）

zip和itertools.zip_longest在这里不起作用，而且itertools.product显然返回列表的所有组合，这需要大量的后期筛选才能获得正确的组合

我试过：

prods = product(x, y)
matches = [ (i,j) for prods if i==j ]
unmatches = []
for i, j in prods:
   if i != j and (i, None) not in unmatches:
      unmatches += [ (i, None) ]
   elif i != j and (None, j) not in unmatches:
      unmatches += [ (None, j) ]

然而，这返回：

matches 
# [(3,3)]
unmatches
# [ (1, None), (None, 3), (None, 4), (None, 5), (2, None), (3, None) ]

因为product如何返回产品（例如，没有(0, j)术语）

有简单的方法吗

Answer 1

list comprehension和set可能会帮助你

x = [1, 2, 3]
y = [0, 3, 4, 5]
x = set(x)
y = set(y)
u = x.intersection(y)
diff = x - y
diff_2 = y - x
t = [(i, None) for i in diff] + [(i, i) for i in u] + [(None, i) for i in diff_2]

print(t)

结果：

[(1, None), (2, None), (3, 3), (None, 0), (None, 4), (None, 5)]

Answer 2

列表理解是一种方法：

[(i if i in x else None, i if i in y else None) for i in range(1, max(x + y) + 1)]

给出：

[(None, 0), (1, None), (2, None), (3, 3), (None, 4), (None, 5)]