Python 如何在另一个列表内容中重复一个列表的元素，直到第二个列表的长度满足？

Question

如何在另一个列表内容中重复一个列表的元素，直到第二个列表的长度满足？例如：

LA = [0,1,2]
LB = [(0,0),(1,0),(2,0),(3,0),(4,0),(5,0),(6,0)]
the end result should be:
LC = [(0,0,0),(1,0,1),(2,0,2),(3,0,0),(4,0,1),(5,0,2),(6,0,0)]

希望能一行完成

Answer 1

像这样尝试 enumerate() 以及列表理解 -

[elem + (LA[i % len(LA)],) for i, elem in enumerate(LB)]

Answer 2

您可以使用 itertools.cycle:

from itertools import cycle

LA = [0,1,2]
LB = [(0,0),(1,0),(2,0),(3,0),(4,0),(5,0),(6,0)]

LC = [(i, j, k) for (i, j), k in zip(LB, cycle(LA))]
print LC
# [(0, 0, 0), (1, 0, 1), (2, 0, 2), (3, 0, 0), (4, 0, 1), (5, 0, 2), (6, 0, 0)]

这是可行的，因为 zip 会生成项目，直到其中一个可迭代对象耗尽...但是 cycle 对象是 ineexhaustible，所以我们将保留 LA 的填充项目，直到 LB 用完。

Answer 3

#use list comprehension and get the element from LA by using the index from LB %3.
[v+(LA[k%3],) for k,v in enumerate(LB)]
Out[718]: [[0, 0, 0], [1, 0, 1], [2, 0, 2], [3, 0, 0], [4, 0, 1], [5, 0, 2], [6, 0, 0]]

Answer 4

这里有一个更 "explicit" 的版本，适用于任何长度的 LA。

LA = [0,1,2]
LB = [(0,0),(1,0),(2,0),(3,0),(4,0),(5,0),(6,0)]

i = 0
LC = []
for x,y in LB:
    try:
        z = LA[i]
    except IndexError:
        i = 0
        z = LA[i]
    LC.append((x,y,z))
    i += 1

print LC
[(0, 0, 0), (1, 0, 1), (2, 0, 2), (3, 0, 0), (4, 0, 1), (5, 0, 2), (6, 0, 0)]

Python 如何在另一个列表内容中重复一个列表的元素，直到第二个列表的长度满足？

Python How to repeat a list's elements in another list content until the length of the second list fulfilled?

python

list

repeat

python-2.7