如何确定是否选择了列表中的第 2 项或第 1 项？

Question

我有这几点

[(3, 4), (3, 6), (3, 8), (4, 5), (4, 7), (5, 1), (5, 5), (7, 3), (7, 5), (8, 5)]

我对它们进行了一些操作，最终得到了这个列表

[(6.0, 3.605551275463989), (6.324555320336759, 5.385164807134504), (7.211102550927978, 7.280109889280518), (5.0990195135927845, 5.0), (5.830951894845301, 6.708203932499369), (5.0, 4.0), (4.123105625617661, 5.656854249492381), (2.23606797749979, 6.324555320336759), (2.23606797749979, 7.211102550927978), (1.4142135623730951, 8.06225774829855)]

列表中的第一项，例如(6.0, 3.605551275463989)对应于第一个点，例如(3,4 )。 我在列表的第一项中找到了两项之间的最小值，并提出了这个：

[3.605551275463989, 5.385164807134504, 7.211102550927978, 5.0, 5.830951894845301, 4.0, 4.123105625617661, 2.23606797749979, 2.23606797749979, 1.4142135623730951]

我现在想要做的是，例如，如果最小值是第 2 项，就像第一种情况一样，我想写如下： (3,4) M2 ，M2 表示它是第 2 项物品。 我该怎么做呢？ 我什至不知道从哪里开始解决这个问题，因为我没有遇到过类似的事情。

Answer 1

你的意思是这样的？

points = [(3, 4), (3, 6), (3, 8), (4, 5),
          (4, 7), (5, 1), (5, 5), (7, 3), (7, 5), (8, 5)]

results = [(6.0, 3.605551275463989), (6.324555320336759, 5.385164807134504), (7.211102550927978, 7.280109889280518), (5.0990195135927845, 5.0), (5.830951894845301, 6.708203932499369), (5.0, 4.0), (4.123105625617661, 5.656854249492381), (2.23606797749979, 6.324555320336759),
           (2.23606797749979, 7.211102550927978), (1.4142135623730951, 8.06225774829855)]

assert(len(points) == len(results))

for i, r in enumerate(results):
    if r[1] < r[0]:
        num = 2
    else:
        num = 1
    print(f'{points[i]}, M{num}')

Answer 2

a = [(6.0, 3.605551275463989), (6.324555320336759, 5.385164807134504), (7.211102550927978, 7.280109889280518), (5.0990195135927845, 5.0), (5.830951894845301, 6.708203932499369), (5.0, 4.0), (4.123105625617661, 5.656854249492381), (2.23606797749979, 6.324555320336759), (2.23606797749979, 7.211102550927978), (1.4142135623730951, 8.06225774829855)]
b = [(3, 4), (3, 6), (3, 8), (4, 5), (4, 7), (5, 1), (5, 5), (7, 3), (7, 5), (8, 5)]
for i in range(len(a)):
    j = list(a[i])
    print(f'{b[i]}-->{max(j)},M{j.index(max(j))+1}')

我将每个元组转换为一个列表并使用列表方法......

Output

(3, 4)-->6.0,M1
(3, 6)-->6.324555320336759,M1
(3, 8)-->7.280109889280518,M2
(4, 5)-->5.0990195135927845,M1
(4, 7)-->6.708203932499369,M2
(5, 1)-->5.0,M1
(5, 5)-->5.656854249492381,M2
(7, 3)-->6.324555320336759,M2
(7, 5)-->7.211102550927978,M2
(8, 5)-->8.06225774829855,M2

Answer 3

使用zip()并行循环所有 3 个列表。 然后只需将第二个列表中的项目中的选定值与第三个列表中的相应值进行比较，即可知道是打印M1还是M2 。

l1 = [(3, 4), (3, 6), (3, 8), (4, 5), (4, 7), (5, 1), (5, 5), (7, 3), (7, 5), (8, 5)]
l2 = [(6.0, 3.605551275463989), (6.324555320336759, 5.385164807134504), (7.211102550927978, 7.280109889280518), (5.0990195135927845, 5.0), (5.830951894845301, 6.708203932499369), (5.0, 4.0), (4.123105625617661, 5.656854249492381), (2.23606797749979, 6.324555320336759), (2.23606797749979, 7.211102550927978), (1.4142135623730951, 8.06225774829855)]
l3 = [3.605551275463989, 5.385164807134504, 7.211102550927978, 5.0, 5.830951894845301, 4.0, 4.123105625617661, 2.23606797749979, 2.23606797749979, 1.4142135623730951]

for i1, (m1, m2), selected in zip(l1, l2, l3):
    if selected == m1:
        which = "M1"
    else:
        which = "M2"
    print(i1, which)

Answer 4

这个怎么样

lst1 = [(3, 4), (3, 6), (3, 8), (4, 5), (4, 7), (5, 1), (5, 5), (7, 3), (7, 5), (8, 5)]
lst2 = [(6.0, 3.605551275463989), (6.324555320336759, 5.385164807134504), (7.211102550927978, 7.280109889280518), (5.0990195135927845, 5.0), (5.830951894845301, 6.708203932499369), (5.0, 4.0), (4.123105625617661, 5.656854249492381), (2.23606797749979, 6.324555320336759), (2.23606797749979, 7.211102550927978), (1.4142135623730951, 8.06225774829855)]

for i, pt in enumerate(lst2):
    argmin = pt.index(min(pt))
    print(lst1[i], 'M{}'.format(argmin + 1))

Output：

(3, 4) M2
(3, 6) M2
(3, 8) M1
(4, 5) M2
(4, 7) M1
(5, 1) M2
(5, 5) M1
(7, 3) M1
(7, 5) M1
(8, 5) M1