如何将列表中的值与嵌套列表的第一个值进行比较并返回嵌套列表结果？

Question

我有以下两个列表。

列表 1

(a,b,h,g,e,t,w,x)

列表二

((a,yellow),(h,green),(t,red),(w,teal))

我想返回以下内容

((a,yellow),(b,null),(h,green),(e,null),(t,red),(w,teal),(x,null))

for x in List_1:
     for y in list_2:
           if x == y
             print y
           else print x, "null"

您知道如何执行此操作吗？ 谢谢

Answer 1

列表理解中的简短 Python 列表理解：

[(i, ([j[1] for j in list2 if j[0] == i] or ['null'])[0]) for i in list1]

更长的版本：

def get_nested(list1, list2):
    d = dict(list2)
    for i in list1:
        yield (i, i in d and d[i] or 'null')
print tuple(get_nested(list1, list2))

Answer 2

试一试：

a = ('a', 'b', 'h', 'g', 'e', 't', 'w', 'x')
b = (('a', 'yellow'), ('h', 'green'), ('t', 'red'), ('w', 'teal'))
B = dict(b)
print [(x, B.get(x, 'null')) for x in a]

Answer 3

你的逻辑是正确的。您唯一需要的是形成一个列表，而不是直接打印结果。

如果您坚持使用嵌套循环（这是一项作业，对吧？），您需要这样的东西：

list1 = ["a", "b", "h", "g", "e", "t", "w", "x"]
list2 = [("a", "yellow"), ("h", "green"), ("t", "red"), ("w", "teal")]

result = [] # an empty list
for letter1 in list1:
  found_letter = False # not yet found
  for (letter2, color) in list2:
    if letter1 == letter2:
      result.append((letter2, color))
      found_letter = True # mark the fact that we found the letter with a color
  if not found_letter:
      result.append((letter1, 'null'))

print result

Answer 4

另一种方法

list1 = ["a", "b", "h", "g", "e", "t", "w", "x"]
list2 = [("a", "yellow"), ("h", "green"), ("t", "red"), ("w", "teal")]
print dict(((x, "null") for x in list1), **dict(list2)).items()