如何使用&x201C；列表”；使用python添加一个新列，查找元素之间的对应关系？；

问题描述如下：

现在我们有了

list1

和

list2

。它们的行数和列数不是固定的，但它们的行数相同，因此我们可以将

list2

添加到

list1

作为新列。但是，请注意，我们使用的列表本质上是一维数组，除了列表中的元素是元组。。。它们看起来像二维数组，但实际上不是。它们本质上是一维列表

示例问题：

list1= [("A","the first",1,"one"),("B","the second",2,"two"),("C","the third",3,"three"),("D","the 4th",4,"four")]
list2 = [("C","cat","animal" ),("B","banana","fruit"),("A","apple","fruit") ,("D","do","verb")  ]

样本输出：

[("A","the first",1,"one","apple","fruit"),
("B","the second",2,"two","banana","fruit" ),
("C","the third",3,"three","cat","animal" ),
("D","the 4th",4,"four","do","verb")]

请注意，

list1

和

list2

中的列数不是固定的，行数必须相等。您可以使用

list1

中的列0作为合并lsit1和List2的参考。您需要去掉重复的列

我知道这个简单的算法在任何情况下都可以编写。如何使整个算法简洁、巧妙

from collections import defaultdict

list1= [("A","the first",1,"one"),("B","the second",2,"two"),("C","the third",3,"three"),("D","the 4th",4,"four")]
list2 = [("C","cat","animal" ),("B","banana","fruit"),("A","apple","fruit") ,("D","do","verb") ]

d = defaultdict(lambda: list())

for column, *rest_cols in list1+list2:
    d[column] += [c for c in rest_cols if c not in d[column]]  # can also made with |= set(rest_cols) and d as defaultdict of set() but the order won't maintained

output = [(k, *v) for k, v in d.items()]  # converting back to the required list of tuples
print(output)

给出：

[('A', 'the first', 1, 'one', 'apple', 'fruit'), ('B', 'the second', 2, 'two', 'banana', 'fruit'), ('C', 'the third', 3, 'three', 'cat', 'animal'), ('D', 'the 4th', 4, 'four', 'do', 'verb')]

给出：

[('A', 'the first', 1, 'one', 'apple', 'fruit'), ('B', 'the second', 2, 'two', 'banana', 'fruit'), ('C', 'the third', 3, 'three', 'cat', 'animal'), ('D', 'the 4th', 4, 'four', 'do', 'verb')]

这是一个简单的zip（）应用程序，添加了一个在索引级别不匹配列表的扭曲：

list1 = [("A","the first",1,"one"),("B","the second",2,"two"),("C","the third",3,"three"),("D","the 4th",4,"four")]
list2 = [("C","cat","animal" ),("B","banana","fruit"),("A","apple","fruit") ,("D","do","verb")  ]

result = [ a+b[1:] for a,b in zip(sorted(list1),sorted(list2)) ]

for row in result:print(row)

('A', 'the first', 1, 'one', 'apple', 'fruit')
('B', 'the second', 2, 'two', 'banana', 'fruit')
('C', 'the third', 3, 'three', 'cat', 'animal')
('D', 'the 4th', 4, 'four', 'do', 'verb')

但是，对于这种“键控”结构，我建议使用字典而不是元组列表。

这是一个简单的zip（）应用程序，添加了一个扭曲，列表在索引级别上没有顺序匹配：

list1 = [("A","the first",1,"one"),("B","the second",2,"two"),("C","the third",3,"three"),("D","the 4th",4,"four")]
list2 = [("C","cat","animal" ),("B","banana","fruit"),("A","apple","fruit") ,("D","do","verb")  ]

result = [ a+b[1:] for a,b in zip(sorted(list1),sorted(list2)) ]

for row in result:print(row)

('A', 'the first', 1, 'one', 'apple', 'fruit')
('B', 'the second', 2, 'two', 'banana', 'fruit')
('C', 'the third', 3, 'three', 'cat', 'animal')
('D', 'the 4th', 4, 'four', 'do', 'verb')

然而，对于这种“键控”结构，我建议使用字典而不是元组列表