将列表转换为元组列表，每个元组中的项目重复值

Question

我有几个N维数组，如下所示。 如何将这些数组转换为元组？

num = ['1' '2']
letter1 = ['a' 'b' 'c' 'd' 'e' 'f']
letter2 = ['aa' 'bb' 'cc' 'dd' 'ee' 'ff']

我想要的结果是：

tuple1 = [(1,a),(1,b),(1,c),(1,d),(1,e),(1,f)]
tuple2 = [(2,aa),(2,bb),(2,cc),(2,dd),(2,ee),(2,ff)]

我有

tuple1 = tuple(num[0], letter1)
tuple2 = tuple(num[1], letter2)

但是我得到了一个TypeError: tuple() takes at most 1 argument (2 given)错误TypeError: tuple() takes at most 1 argument (2 given)

你的方法是什么？ 任何方法将不胜感激。 谢谢。

Answer 1

您可以使用带有itertools.repeat zip来实现此目的：

>>> from itertools import repeat
>>> num = ['1', '2']
>>> letter1 = ['a', 'b', 'c', 'd', 'e', 'f']

>>> tuple1 = list(zip(repeat(num[0]), letter1))
>>> tuple1
[('1', 'a'), ('1', 'b'), ('1', 'c'), ('1', 'd'), ('1', 'e'), ('1', 'f')]

PS： 注意逗号,在上述名单元素之间。 不使用逗号分隔的字符串被视为单个字符串。

---

从上面的答案中得到启发，这里是单个go中获取tuple1和tuple2的解决方案的列表理解版本：

letter2 = ['aa', 'bb', 'cc', 'dd', 'ee', 'ff'] # rest of the variables same as the above solution

tuple1, tuple2 = [list(zip(repeat(a), b)) for a, b in zip(num, (letter1, letter2))]

这些变量将以下列表作为值：

>>> tuple1
[('1', 'a'), ('1', 'b'), ('1', 'c'), ('1', 'd'), ('1', 'e'), ('1', 'f')]
>>> tuple2
[('2', 'aa'), ('2', 'bb'), ('2', 'cc'), ('2', 'dd'), ('2', 'ee'), ('2', 'ff')]

Answer 2

您可以使用iter和next ：

num = ['1', '2']
num = iter(num)
letter1 = ['a', 'b', 'c', 'd', 'e', 'f']
letter2 = ['aa', 'bb', 'cc', 'dd', 'ee', 'ff']
tuple1, tuple2 = [list(zip(i, [next(num)]*len(i))) for i in [letter1, letter2]]

输出：

[('a', '1'), ('b', '1'), ('c', '1'), ('d', '1'), ('e', '1'), ('f', '1')]
[('aa', '2'), ('bb', '2'), ('cc', '2'), ('dd', '2'), ('ee', '2'), ('ff', '2')]

无iter ：

tuple1, tuple2 = [[(i, h) for i in c] for c, h in zip([letter1, letter2], ['1', '2'])]

Answer 3

已经有一些好的答案，但我将使用列表理解来分享另一种选择。 它看起来像更多的代码，但适用于更长的nums和lists （只要它们的长度仍然相同）：

letter1 = ['a', 'b', 'c', 'd', 'e', 'f']
letter2 = ['aa', 'bb', 'cc', 'dd', 'ee', 'ff']
nums = [1, 2]

lists = [letter1, letter2]
tuples = [[(i, x) for x in l] for i, l in zip(nums, lists)]

tuple1, tuple2 = tuples

结果：

In [1]: tuple1
Out[1]: [(1, 'a'), (1, 'b'), (1, 'c'), (1, 'd'), (1, 'e'), (1, 'f')]

In [2]: tuple2
Out[2]: [(2, 'aa'), (2, 'bb'), (2, 'cc'), (2, 'dd'), (2, 'ee'), (2, 'ff')]

Answer 4

已经有了很好的答案，我只想给你提示你可以调整一下你的逻辑

num = ['1','2']
letter1 = ['a','b' ,'c', 'd', 'e', 'f']
letter2 = ['aa' ,'bb', 'cc', 'dd' ,'ee' ,'ff']
tuple1 = tuple(zip(*[[int(num[0])]*len(letter1),letter1]))
tuple2 = tuple(zip(*[[int(num[1])]*len(letter2), letter2]))

print(tuple1)
print(tuple2)

输出：

((1, 'a'), (1, 'b'), (1, 'c'), (1, 'd'), (1, 'e'), (1, 'f'))
((2, 'aa'), (2, 'bb'), (2, 'cc'), (2, 'dd'), (2, 'ee'), (2, 'ff'))