使用列表推导来简化转换元组元组中的列表列表的函数

Question

i want to change this function using a list comprehension in order that it will have 1 or to lines. the function transfor a list containing list in a tuple containing tupples inside

def lol(lista):
        novotuplo = ()
            for i in range(len(lista)):
                novotuplo += (tuple(lista[i]),)

            return novotuplo

Answer 1

This should work:

lista = [[1,2], [3,4], [5,6]]
print(tuple(tuple(i) for i in lista))
# ((1, 2), (3, 4), (5, 6))

A more explicit explanation of the above list comprehension would be create a tuple of all elements in lista converted to tuple .

Answer 2

我认为应该这样做。

novotuplo = tuple(tuple(item) for item in lista)

Answer 3

If you want it in a functional form, this is one way. You do not need an index here because lst in lista iterates over the elements directly.

lista = [[1],[2],[3]]
def lol(lista):
    novotuplo = tuple(tuple(lst) for lst in lista)
    return novotuplo

print (lol(lista))
# ((1,), (2,), (3,))

Answer 4

您可以将项目map到tuple()构造函数：

tuple(map(tuple, lista))