Question

我有一个元组列表，我需要在右侧找到具有最大值的元组。我还需要程序打印出具有此最大值的每个元组。我试过使用lambda，但是只返回左侧具有最大值的元组之一。具体来说，我尝试过：

Item.where(session: @user.sessions)

哪个返回max(TotalFriendsList,key=lambda x:x[1])

这是列表：

(1,3)

这是我需要的输出：

[(0, 2),
 (1, 3),
 (2, 3),
 (3, 3),
 (4, 2),
 (5, 3),
 (6, 2),
 (7, 2),
 (8, 3),
 (9, 1)]

谢谢！

Answer 1

现在，经过优化后，此列表将只遍历列表一次：

data = [(0, 2),
 (1, 3),
 (2, 3),
 (3, 3),
 (4, 2),
 (5, 3),
 (6, 2),
 (7, 2),
 (8, 3),
 (9, 1)]

maxVal = -1
result = None

for datum in data:
    val = datum[1]
    if val > maxVal:
        result = [datum]
        maxVal = val
    elif val == maxVal:
        result.append(datum)

print result

print

# and the way he wanted it printed...
print "[" + ",\n".join([str(v) for v in result]) + "]"

请注意，由于第一个条目包含“获胜元组”，因此该代码根本不会处理失败者。它只是为它们做一些比较，而比较几乎没有花费。因此，基本上它只需一次就可以完成构建结果列表的最少工作量。如果第一个元组不是赢家，那就没有效率了。

输出：

[(1, 3), (2, 3), (3, 3), (5, 3), (8, 3)]

[(1, 3),
(2, 3),
(3, 3),
(5, 3),
(8, 3)]

只有我按照那个人的要求做了;）（tic）-所有答案都正确，...对解决相同问题的不同方式具有很好的见识

Answer 2

您可以通过获取最大值，然后获取对应的元组来保持紧凑性：

l = [(0, 2),
     (1, 3),
     (2, 3),
     (3, 3),
     (4, 2),
     (5, 3),
     (6, 2),
     (7, 2),
     (8, 3),
     (9, 1)]

# pass a generator expression to `max`
greatest = max(item[1] for item in l)

# "filter" `l`, keeping only tuples that have the greatest value as their second element
result = [item for item in l if item[1] == greatest]
print(result)
# [(1, 3), (2, 3), (3, 3), (5, 3), (8, 3)]

Answer 3

>>> lst = [(0, 2),
...  (1, 3),
...  (2, 3),
...  (3, 3),
...  (4, 2),
...  (5, 3),
...  (6, 2),
...  (7, 2),
...  (8, 3),
...  (9, 1)]
>>> mx = max(lst, key=lambda x: x[1])
>>> filter(lambda x: x[1] == mx[1], lst)
[(1, 3), (2, 3), (3, 3), (5, 3), (8, 3)]

查找具有多个相同值的元组的最大值

3 个答案: