问题：如何将列表合并为元组列表？
实现以下目标的Python方法是什么？
# Original lists:
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
# List of tuples from 'list_a' and 'list_b':
list_c = [(1,5), (2,6), (3,7), (4,8)]
的每个成员list_c都是一个元组，其第一个成员是from list_a，第二个成员是from list_b。
What is the Pythonic approach to achieve the following?
# Original lists:
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
# List of tuples from 'list_a' and 'list_b':
list_c = [(1,5), (2,6), (3,7), (4,8)]
Each member of list_c is a tuple, whose first member is from list_a and the second is from list_b.
回答 0
在Python 2中：
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> zip(list_a, list_b)
[(1, 5), (2, 6), (3, 7), (4, 8)]
在Python 3中：
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> list(zip(list_a, list_b))
[(1, 5), (2, 6), (3, 7), (4, 8)]
In Python 2:
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> zip(list_a, list_b)
[(1, 5), (2, 6), (3, 7), (4, 8)]
In Python 3:
>>> list_a = [1, 2, 3, 4]
>>> list_b = [5, 6, 7, 8]
>>> list(zip(list_a, list_b))
[(1, 5), (2, 6), (3, 7), (4, 8)]
回答 1
在python 3.0中，zip返回一个zip对象。您可以调用以获得清单list(zip(a, b))。
In python 3.0 zip returns a zip object. You can get a list out of it by calling list(zip(a, b)).
回答 2
您可以使用地图lambda
a = [2,3,4]
b = [5,6,7]
c = map(lambda x,y:(x,y),a,b)
如果原始列表的长度不匹配，这也将起作用
You can use map lambda
a = [2,3,4]
b = [5,6,7]
c = map(lambda x,y:(x,y),a,b)
This will also work if there lengths of original lists do not match
回答 3
Youre looking for the builtin function zip.
回答 4
我不确定这是否是pythonic方式，但是如果两个列表具有相同数量的元素，这似乎很简单：
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
list_c=[(list_a[i],list_b[i]) for i in range(0,len(list_a))]
I am not sure if this a pythonic way or not but this seems simple if both lists have the same number of elements :
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
list_c=[(list_a[i],list_b[i]) for i in range(0,len(list_a))]
回答 5
我知道这是一个古老的问题，已经得到回答，但是由于某些原因，我仍然想发布此替代解决方案。我知道很容易找出哪个内置函数可以完成您所需的“魔术”，但是知道您可以自己完成该操作也不会有什么害处。
>>> list_1 = ['Ace', 'King']
>>> list_2 = ['Spades', 'Clubs', 'Diamonds']
>>> deck = []
>>> for i in range(max((len(list_1),len(list_2)))):
while True:
try:
card = (list_1[i],list_2[i])
except IndexError:
if len(list_1)>len(list_2):
list_2.append('')
card = (list_1[i],list_2[i])
elif len(list_1)<len(list_2):
list_1.append('')
card = (list_1[i], list_2[i])
continue
deck.append(card)
break
>>>
>>> #and the result should be:
>>> print deck
>>> [('Ace', 'Spades'), ('King', 'Clubs'), ('', 'Diamonds')]
I know this is an old question and was already answered, but for some reason, I still wanna post this alternative solution. I know it’s easy to just find out which built-in function does the “magic” you need, but it doesn’t hurt to know you can do it by yourself.
>>> list_1 = ['Ace', 'King']
>>> list_2 = ['Spades', 'Clubs', 'Diamonds']
>>> deck = []
>>> for i in range(max((len(list_1),len(list_2)))):
while True:
try:
card = (list_1[i],list_2[i])
except IndexError:
if len(list_1)>len(list_2):
list_2.append('')
card = (list_1[i],list_2[i])
elif len(list_1)<len(list_2):
list_1.append('')
card = (list_1[i], list_2[i])
continue
deck.append(card)
break
>>>
>>> #and the result should be:
>>> print deck
>>> [('Ace', 'Spades'), ('King', 'Clubs'), ('', 'Diamonds')]
回答 6
您在问题陈述中显示的输出不是元组而是列表
list_c = [(1,5), (2,6), (3,7), (4,8)]
检查
type(list_c)
考虑到您想要结果作为list_a和list_b中的元组，请执行
tuple(zip(list_a,list_b))
The output which you showed in problem statement is not the tuple but list
list_c = [(1,5), (2,6), (3,7), (4,8)]
check for
type(list_c)
considering you want the result as tuple out of list_a and list_b, do
tuple(zip(list_a,list_b))
回答 7
一种不使用的替代方法zip：
list_c = [(p1, p2) for idx1, p1 in enumerate(list_a) for idx2, p2 in enumerate(list_b) if idx1==idx2]
万一不仅要获取元组1st与1st，2nd与2nd …而且要获取2个列表的所有可能组合，可以使用
list_d = [(p1, p2) for p1 in list_a for p2 in list_b]
One alternative without using zip:
list_c = [(p1, p2) for idx1, p1 in enumerate(list_a) for idx2, p2 in enumerate(list_b) if idx1==idx2]
In case one wants to get not only tuples 1st with 1st, 2nd with 2nd… but all possible combinations of the 2 lists, that would be done with
list_d = [(p1, p2) for p1 in list_a for p2 in list_b]
