Question

我正在尝试生成一个程序来检查2D数组的最高和值。然后它需要将每个值相互比较到输出，这是数组中每个位置的所有总和中的最大值。

我能够为一定数量的元素执行此操作，但我想根据用户输入的元素数量为无限数量的元素执行此操作。

Person = [123,321,213]

TotalMorn =[[3,5,6,4,5,3,7],
            [2,1,1,2,3,4,5],
            [3,4,5,6,7,6,5]]

TotalAft =[[8,7,6,2,6,7,5],
           [4,3,2,3,4,3,2],
           [1,1,1,1,2,3,1]]

def WeeklyTotal(K):
    Total = 0
    for j in range(7):
        TotalM = TotalMorn[K][j]
        TotslA = TotalAft[K][j]
        Total = TotalM + TotalA + Total
    return Total

Total1 = WeeklyTotal(0)
Total2 = WeeklyTotal(1)
Total3 = WeeklyTotal(2)

if Total1 > Total2 and Total1 > Total3:
    print 'Person', (Person[0]), 'Has produced the most profit with ', Total1
elif Total2 > Total1 and Total2 > Total3:
    print 'Peron', (Person[1]), 'Has produced the most profit with ', Total2
elif Total3 > Total1 and Total3 > Total2:
    print 'Person', (Person[2]), 'Has produced the most profit with ', Total3

Answer 1

在您的情况下，您需要将 person 绑定到他的 totalScore 。然后，您将查找最高分数值，因此您将拥有与此值相关联的最高分数的人，例如......

arr = [('John', 15), ('Sue', 13), ('Jack', 20), ('Tom', 5)]    
print max(arr, key = lambda x:x[1])
# outputs : ('Jack', 20)

在这里，您已将得分绑定到人到元组列表。您要求 max 函数使用每个元组的第二个元素查找最高值（这就是key = lambda x:x[1]的意思）。

在您的示例中，数据的格式如...

people = ['John', 'Sue', 'Jack', 'Tom']
scores = [15, 13, 20, 5]

...你可能想用一个函数来确定最高值的索引，这样做：

#   We set the highest score at the beginning as the first value, and save the index.
curr_score = scores[0]
saved_index = 0

#   We then iterate through scores, not looping through the first item, as its index
#   is already saved. We set the start of the index to `1`.
for index, score in enumerate(scores[1:], start=1):
    if score > curr_score:
        #   We save the new highest score.
        curr_score = score

        #   And keep its index.
        saved_index = index

#   At the end of the loop, we saved the maximum value index, and can use it
#   on `people` array.
print people[saved_index]
#   It will output the person with the highest score.

Answer 2

我假设您要做的是在数组WeeklyTotal中找到最大值？

maxIndex, maxValue = 0,0
for i in range(0,len(WeeklyTotal)-1):
    if WeeklyTotal[i] > maxValue :
        maxValue = WeeklyTotal[i]
        maxIndex = i
print 'Person ',Person[maxIndex],' Has produced the most profit with ',WeeklyTotal[maxIndex]

Answer 3

这是一个使用一些python习语的简短解决方案：

（感谢IMCoins提醒我max(a, key = lambda...）

people = [123,321,213]

morning =[[3,5,6,4,5,3,7],
            [2,1,1,2,3,4,5],
            [3,4,5,6,7,6,5]]

afternoon =[[8,7,6,2,6,7,5],
           [4,3,2,3,4,3,2],
           [1,1,1,1,2,3,1]]


def makeTotals(people, morning, afternoon):
    return [(p,sum(m+a)) for (p,m,a) in zip(people, morning, afternoon)]

allTotals = makeTotals(people, morning, afternoon)

print(max(allTotals, key = lambda x:x[1]))

输出：

(123, 74)

makeTotals函数zip向上排列数组，以便将每个人的早晨和下午分数对齐，(p,sum(m+a))创建tuple此人及其总数。< / p>

对max的调用使用key = lambda x:x[1]参数（感谢@IMCoins）提取上午和下午得分最大的元组。

Answer 4

一个班轮怎么样＆＃39; （带有换行符和添加的空格以便于阅读）

Person = [123,321,213]

TotalMorn =[[3,5,6,4,5,3,7],
            [2,1,1,2,3,4,5],
            [3,4,5,6,7,6,5]]

TotalAft =[[8,7,6,2,6,7,5],
           [4,3,2,3,4,3,2],
           [1,1,1,1,2,3,1]]


print('Person %s Has produced the most profit with %s'
      % max(zip(map(sum,
                    zip(map(sum, TotalMorn),
                        map(sum, TotalAft))),
                Person))[::-1])

Person 123 Has produced the most profit with 74

比较从2D数组中获取的无限数量的值，以查看哪个元素具有最大总和

4 个答案: