查找两个列表之间具有最大差异的元素索引

Question

有两个列表

l1 = [1, 2.3, 1.4, 1.1]

l2 = [2.1, 0.9, 1.1, 3.2]

并且我想在列表中找到列表之间差异最大的元素。

即：

l1 = [1, 2.3, 1.4, 1.1] 应该 return 0 对于最小数字的索引 0

l2 = [2.1, 0.9, 1.1, 3.2] 应该 return 3 对于最大数

的索引 3

Answer 1

试试这个：

l1 = [1, 2.3, 1.4, 1.1]
l2 = [2.1, 0.9, 1.1, 3.2]


def max_diff(l1, l2):
    min_l1 = min(l1)
    max_l1 = max(l1)
    min_l2 = min(l2)
    max_l2 = max(l2)
    max_diff1 = max_l2 - min_l1
    max_diff2 = max_l1 - min_l2
    if max_diff1 > max_diff2:
        return l1.index(min_l1), l2.index(max_l2), max_diff1
    else:
        return l1.index(max_l1), l2.index(min_l2), max_diff2


l1_index, l2_index, max_diff = max_diff(l1, l2)
print(l1_index, l2_index, max_diff)  # 0 3 2.2

请注意，这会计算最大差异，并且可以 return (min(l1), max(l2)) 或 (max(l1), min(l2))，具体取决于列表。

Answer 2

这里有一个更简洁的方法：

l1 = [1, 2.3, 1.4, 1.1]
l2 = [2.1, 0.9, 1.1, 3.2]

for k, v in {min: ['lo', l1], max: ['hi', l2]}.items():
    print(f'{v[0]}={k(range(len(v[1])), key=v[1].__getitem__)}')

输出：

lo=0
hi=3

Answer 3

如果有多个结果：

l1 = [1, 2.3, 3.2, 1]
l2 = [1, 1, 1.1, 3.2]


def find_all(list1, list2):
    def return_all(func, lst):
        return [pos for pos, val in enumerate(lst) if val == func(lst)]
    return [return_all(max, list1), return_all(min, list2)]


def output(res, order):
    print(f"max position list {order[0]}: {res[order[0] - 1]}")
    print(f"min position list {order[1]}: {res[order[1] - 1]}")


if max(l1) - min(l2) > max(l2) - min(l1):
    results = find_all(l1, l2)
    output(results, (1, 2))
elif max(l1) - min(l2) < max(l2) - min(l1):
    results = find_all(l2, l1)
    output(results[::-1], (2, 1))
else:
    results = find_all(l1, l2)
    output(results, (1, 2))
    print ('And:')
    results = find_all(l2, l1)
    output(results[::-1], (2, 1))

输出：

    max position list 1: [2]
    min position list 2: [0, 1]
    And:
    max position list 2: [3]
    min position list 1: [0, 3]

查找两个列表之间具有最大差异的元素索引

Find elements' indices with maximum difference between two lists

python

list