如何将两个列表合并到python中的一列列中？

Question

假设我有两个列表：

t1 = ["abc","def","ghi"] 
t2 = [1,2,3] 

我如何可以合并使用它蟒蛇，使输出列表将是：

t = [("abc",1),("def",2),("ghi",3)] 

，我已经尝试过的程序是：

t1 = ["abc","def"] 
t2 = [1,2]   
t = [ ] 
for a in t1: 
     for b in t2: 
       t.append((a,b)) 
print t 

输出是：

[('abc', 1), ('abc', 2), ('def', 1), ('def', 2)] 

我不想重复条目。

Answer 1

在Python 2.x中，你可以使用zip：

>>> t1 = ["abc","def","ghi"] 
>>> t2 = [1,2,3] 
>>> zip(t1, t2) 
[('abc', 1), ('def', 2), ('ghi', 3)] 
>>> 

然而，在Python 3.x中，zip返回一个zip对象（这是一个iterator），而不是一个列表。这意味着，你必须明确地将它们放在list结果转换成一个列表：

>>> t1 = ["abc","def","ghi"] 
>>> t2 = [1,2,3] 
>>> zip(t1, t2) 
<zip object at 0x020C7DF0> 
>>> list(zip(t1, t2)) 
[('abc', 1), ('def', 2), ('ghi', 3)] 
>>> 

Answer 2

使用ZIP：

>>> t1 = ["abc","def","ghi"] 
>>> t2 = [1,2,3] 
>>> list(zip(t1,t2)) 
[('abc', 1), ('def', 2), ('ghi', 3)] 
# Python 2 you do not need 'list' around 'zip' 

如果你不想重复的项目，你不在乎有关订单，用一套：

>>> l1 = ["abc","def","ghi","abc","def","ghi"] 
>>> l2 = [1,2,3,1,2,3] 
>>> set(zip(l1,l2)) 
set([('def', 2), ('abc', 1), ('ghi', 3)]) 

如果你想以uniquify：

>>> seen=set() 
>>> [(x, y) for x,y in zip(l1,l2) if x not in seen and (seen.add(x) or True)] 
[('abc', 1), ('def', 2), ('ghi', 3)]