列表列表的元素

发布于 2025-01-31 23:41:07 字数 336 浏览 1 评论 0原文

我有数字的列表（元组元组？）

[[1，2，3]，[4，5，6]，[7，8，9]]

我想转换在：

（123,456,789）

我对此代码做的，您能否向我建议一种更好的方法？

listacomb = []
for list in permdue:
    for number in list:
        numerix += "".join(str(number))
    listacomb.append(numerix)
    numerix=""
return listacomb

原文

I have a list of list(tuple of tuple?) of number

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

I want to transform in:

(123,456,789)

I did with this code, can you suggest to me a better way to do it?

listacomb = []
for list in permdue:
    for number in list:
        numerix += "".join(str(number))
    listacomb.append(numerix)
    numerix=""
return listacomb

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帅气尐潴 2025-02-07 23:41:07

您可以做一个列表理解：

data = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
output = tuple([int("".join(map(str, inner_list))) for inner_list in data])
print(output)

输出：

(123, 456, 789)

参考：

列表综合上的Python文档

You can do a list comprehension:

data = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
output = tuple([int("".join(map(str, inner_list))) for inner_list in data])
print(output)

Output:

(123, 456, 789)

References:

Python documentation on List comprehensions

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小鸟爱天空丶 2025-02-07 23:41:07

这是做到这一点的数学方法。我们可以向后迭代每个子派，并构建所需的数字。

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

for list in permdue:
    list = reversed(list)
    m = 1
    numerix = 0
    for number in list:
        numerix += number * m
        m = m * 10
    listacomb.append(numerix)

print(listacomb)  # [123, 456, 789]

此方法避免了性能惩罚弦解决方案。例如，通过服用[1，2，3]和Computing 3 + 20 + 100来获得123来起作用。

Here is a math approach to doing this. We can iterate each sublist backwards and build out the desired number.

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

for list in permdue:
    list = reversed(list)
    m = 1
    numerix = 0
    for number in list:
        numerix += number * m
        m = m * 10
    listacomb.append(numerix)

print(listacomb)  # [123, 456, 789]

This approach avoids the performance penalizing string solution. It works, for example, by taking [1, 2, 3] and computing 3 + 20 + 100, to obtain 123.

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明媚殇 2025-02-07 23:41:07

可以在没有这样的反转的情况下完成：

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

for p in permdue:
    n = 0
    for p_ in p:
        n = n * 10 + p_
    listacomb.append(n)
print(listacomb)

输出：

[123, 456, 789]

It can be done without reversal like this:

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

for p in permdue:
    n = 0
    for p_ in p:
        n = n * 10 + p_
    listacomb.append(n)
print(listacomb)

Output: