Question

有一些元组列表，使用列表理解可以很容易地将此列表格式化为列表列表。但是，如何使用for循环来做同样的事情？

some_tuples = [(1, 2, 3), (4, 5, 6), (7, 8, 9)]
[[x for x in a_tuple] for a_tuple in some_tuples]

输出将是：

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

我尝试使用for循环

a_tuple = []
for i in some_tuples:
    for x in i:
        a_tuple.append([x])
a_tuple

输出将是：

[[1], [2], [3], [4], [5], [6], [7], [8], [9]]

如何在不使用列表理解的情况下获取[[1,2,3], [4,5,6], [7,8,9]]？

Answer 1

使用list()将元组转换为列表：

some_tuples = [(1, 2, 3), (4, 5, 6), (7, 8, 9)]

a_tuple = []
for i in some_tuples:
    a_tuple.append(list(i))

print(a_tuple)

结果：

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Answer 2

不使用列表推导的最简单方法是将内部元组映射到列表：

some_tuples = [(1, 2, 3), (4, 5, 6), (7, 8, 9)]
some_lists = map(list, some_tuples)  # list(map(list, some_tuples)) on Python 3.x
print(some_lists)  # [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Answer 3

你的清单理解：

print([[x for x in a_tuple] for a_tuple in some_tuples])

与：

相同

final_result=[]
for a_tuple in some_tuples:
    sub_result=[]
    for x in a_tuple:
        sub_result.append(x)
    final_result.append(sub_result)

print(final_result)

输出：

[[1, 2, 3], [4, 5, 6], [7, 8, 9]]

Python如何使用循环来填充列表中的子列表

3 个答案:

结果：