在两个列表中偏移正数和负数，然后在Python中获得其余数字

发布于 2025-01-24 09:44:59 字数 267 浏览 0 评论 0原文

我有两个列表，包括一系列正数或负数，两个列表可能具有不等的长度，例如：

lst1 = [2, 3, 3, 5, 6, 6, 6, 8, 10]
lst2 = [-1, -2, -2, -3, -6, -10, -11]

我想要得到的结果是：

lst_ofs = [-1, -2, 3, 5, 6, 6, 8, -11]

在两个列表中具有同样绝对值的正数和负数，被数量抵消，有什么简单的方法吗？

原文

I have two lists consisting of a series of positive or negative numbers, and two lists may have unequal lengths, like this:

lst1 = [2, 3, 3, 5, 6, 6, 6, 8, 10]
lst2 = [-1, -2, -2, -3, -6, -10, -11]

The result I want to get is:

lst_ofs = [-1, -2, 3, 5, 6, 6, 8, -11]

Positive and negative numbers with equal absolute value in the two lists are offset by quantity, is there any easy way to do this?

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我最亲爱的 2025-01-31 09:44:59

该解决方案将每个数字的正和负实例的数量计算为最大值。必要时在每次迭代上扩展输出列表。

lst1 = [2, 3, 3, 5, 6, 6, 6, 8, 10]
lst2 = [-1, -2, -2, -3, -6, -10, -11]

lst_ofs = []
for i in range(max(max(lst1), -min(lst2)) + 1):
    n = lst1.count(i) - lst2.count(-i)
    if abs(n) > 0:
        lst_ofs.extend([int(i * n / abs(n))] * abs(n))
print(lst_ofs)

输出：

[-1, -2, 3, 5, 6, 6, 8, -11]

This solution counts the number of the positive and negative instances of each number up to the maximum. The output list is extended on each iteration as necessary.

lst1 = [2, 3, 3, 5, 6, 6, 6, 8, 10]
lst2 = [-1, -2, -2, -3, -6, -10, -11]

lst_ofs = []
for i in range(max(max(lst1), -min(lst2)) + 1):
    n = lst1.count(i) - lst2.count(-i)
    if abs(n) > 0:
        lst_ofs.extend([int(i * n / abs(n))] * abs(n))
print(lst_ofs)

Output:

[-1, -2, 3, 5, 6, 6, 8, -11]

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醉南桥 2025-01-31 09:44:59

result = lst1 + lst2
# keep checking for pairs until none are left
while True:
    for v in result:
        if -v in result:
            # if v and -v are in the list, remove them both
            # and restart the for loop
            result.remove(v)
            result.remove(-v)
            break
    else:
        # if the for loop made it all the way to the end,
        # break the outer loop
        break

这不是最有效的解决方案，因为循环的每次删除两个值时启动。但这应该起作用。

result = lst1 + lst2
# keep checking for pairs until none are left
while True:
    for v in result:
        if -v in result:
            # if v and -v are in the list, remove them both
            # and restart the for loop
            result.remove(v)
            result.remove(-v)
            break
    else:
        # if the for loop made it all the way to the end,
        # break the outer loop
        break

This isn't the most efficient solution, because the for loop starts over each time two values are removed. But it should work.

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