Xen*_*non 0 python grouping min
这是我的代码,它使得来自多个成员的小组,然后对每个小组进行评分,然后将它们相加以得到这种情况下4组的总分.
import random
def run(members, n_groups):
participants = list(range(1,members+1))*n_groups
random.shuffle(participants)
my_groups = list(zip(*[iter(participants)]*members))
print(my_groups)
def get_rating(group):
return (len(set(group)))
score = ((sum(get_rating(g) for g in my_groups)))
print(score)
return score
members = 4
n_groups = 4
print(min(run(members, n_groups) for _ in range(10)))
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输出:
[(3, 3, 4, 1), (4, 2, 3, 1), (2, 3, 2, 4), (1, 1, 4, 2)]
13
[(3, 1, 1, 4), (2, 3, 2, 4), (1, 4, 1, 2), (3, 2, 4, 3)]
12
[(2, 4, 4, 1), (3, 3, 1, 3), (3, 1, 2, 2), (4, 4, 2, 1)]
11
[(3, 1, 4, 1), (3, 1, 2, 1), (2, 2, 4, 4), (3, 2, 3, 4)]
11
[(1, 1, 3, 1), (2, 2, 3, 1), (3, 2, 4, 4), (4, 4, 3, 2)]
11
[(3, 2, 1, 2), (4, 1, 3, 4), (4, 2, 3, 1), (1, 2, 4, 3)]
14
[(2, 4, 3, 1), (3, 4, 4, 2), (1, 1, 2, 1), (3, 3, 4, 2)]
12
[(3, 3, 1, 2), (2, 1, 3, 3), (2, 4, 4, 2), (4, 1, 1, 4)]
10
[(1, 4, 3, 2), (4, 2, 3, 3), (3, 1, 2, 1), (4, 4, 1, 2)]
13
[(2, 4, 1, 3), (3, 2, 2, 3), (1, 4, 1, 3), (1, 4, 2, 4)]
12
10
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所以每个组都得到一个评级,然后最低分显示在底部,有没有办法可以重新打印获得该分数的组?因为如果我尝试使用100,000,那么我很难检查每一个以找到它来自哪个组!
我试过了:
import random
def run(my_groups, score):
participants = list(range(1,members+1))*n_groups
random.shuffle(participants)
my_groups = list(zip(*[iter(participants)]*members))
print(my_groups)
def get_rating(group):
return (len(set(group)))
score = ((sum(get_rating(g) for g in my_groups)))
print(score)
return score
members = 4
n_groups = 4
print(min(run(members, n_groups) for _ in range(10), key=lambda x:x[1]))
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但是,对于括号运行,这会返回"生成器表达式必须括号,如果不是唯一的参数"吗?
改变run回归my_groups, score.
然后改变
min(run(members, n_groups) for _ in range(10))
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至
min((run(members, n_groups) for _ in range(10)), key=lambda x:x[1])
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这使得它min返回整个(my_groups, score)元组,但只比较score(元组的第二个元素)来找到最小值.
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