so I have say two lists:
所以我说两个清单：
list1 = [1,2,3]
and
和
list2 = [4]
and I need to combine them to produce the following output:
我需要将它们组合起来以产生以下输出：
list3=[[1,4],[2,4],[3,4]]
itertools doesn't seem to have a method to accomplish this, the zip function ends when the second list does...
itertools似乎没有一个方法来实现这一点，zip函数在第二个列表时结束...
I'm sure there's a one liner out there, but I'm finding too much stuff about similar but not the same problems on here and google.
我确定那里有一个班轮，但是我发现这里和谷歌有类似但不一样的问题。
Thanks for any help!
谢谢你的帮助！
5 个解决方案
#1
6
You can iterate over the list and concatenate the list2 value and the element for the current iteration:
您可以迭代列表并连接list2值和当前迭代的元素：
list1 = [1,2,3]
list2 = [4]
new_list = [[a]+list2 for a in list1]
Output:
输出：
[[1, 4], [2, 4], [3, 4]]
Or, an alternative, although lower solution using map:
或者，另一种方法，尽管使用map的较低解决方案：
final_list = map(lambda x:[x, list2[0]], list1)
Output:
输出：
[[1, 4], [2, 4], [3, 4]]
#2
6
You can use itertools izip_longest (py2) or itertools zip longest (py3) too:
你也可以使用itertools izip_longest（py2）或者itertools zip longest（py3）：
import itertools
list(map(list,itertools.izip_longest([], list1, fillvalue=list2[0])))
Returns:
返回：
[[4, 1], [4, 2], [4, 3]]
#3
3
Do you need a third alternative?
你需要第三种选择吗？
>>> list(map(list,zip(list2 * len(list1), list1)))
[[4, 1], [4, 2], [4, 3]]
#4
0
You can also try using itertools.cycle():
您也可以尝试使用itertools.cycle（）：
>>> import itertools
>>> list1 = [1,2,3]
>>> list2 = [4]
>>> print([list(x) for x in zip(list1, itertools.cycle(list2))])
[[1, 4], [2, 4], [3, 4]]
#5
-1
Are you looking for something like this?
你在找这样的东西吗？
Without any external module or heavy code:
没有任何外部模块或重型代码：
print(list(map(lambda x:[x,list2[0]],list1)))
when data is :
当数据是：
list1 = [1,2,3]
list2 = [4]
output:
输出：
[[1, 4], [2, 4], [3, 4]]
As someone pointed out this is already given answer ,Here is another solution:
正如有人指出这已经给出答案，这是另一种解决方案：
list1 = [1,2,3]
list2 = [4]
print(list(zip(list1,list2*len(list1))))
output:
输出：
[(1, 4), (2, 4), (3, 4)]
#1
6
You can iterate over the list and concatenate the list2 value and the element for the current iteration:
您可以迭代列表并连接list2值和当前迭代的元素：
list1 = [1,2,3]
list2 = [4]
new_list = [[a]+list2 for a in list1]
Output:
输出：
[[1, 4], [2, 4], [3, 4]]
Or, an alternative, although lower solution using map:
或者，另一种方法，尽管使用map的较低解决方案：
final_list = map(lambda x:[x, list2[0]], list1)
Output:
输出：
[[1, 4], [2, 4], [3, 4]]
#2
6
You can use itertools izip_longest (py2) or itertools zip longest (py3) too:
你也可以使用itertools izip_longest（py2）或者itertools zip longest（py3）：
import itertools
list(map(list,itertools.izip_longest([], list1, fillvalue=list2[0])))
Returns:
返回：
[[4, 1], [4, 2], [4, 3]]
#3
3
Do you need a third alternative?
你需要第三种选择吗？
>>> list(map(list,zip(list2 * len(list1), list1)))
[[4, 1], [4, 2], [4, 3]]
#4
0
You can also try using itertools.cycle():
您也可以尝试使用itertools.cycle（）：
>>> import itertools
>>> list1 = [1,2,3]
>>> list2 = [4]
>>> print([list(x) for x in zip(list1, itertools.cycle(list2))])
[[1, 4], [2, 4], [3, 4]]
#5
-1
Are you looking for something like this?
你在找这样的东西吗？
Without any external module or heavy code:
没有任何外部模块或重型代码：
print(list(map(lambda x:[x,list2[0]],list1)))
when data is :
当数据是：
list1 = [1,2,3]
list2 = [4]
output:
输出：
[[1, 4], [2, 4], [3, 4]]
As someone pointed out this is already given answer ,Here is another solution:
正如有人指出这已经给出答案，这是另一种解决方案：
list1 = [1,2,3]
list2 = [4]
print(list(zip(list1,list2*len(list1))))
output:
输出：
[(1, 4), (2, 4), (3, 4)]