将嵌套列表排序为次低

Question

给出物理课上每个学生的名字和成绩，将它们存储在一个嵌套列表中，并打印出任何成绩最低的学生的名字。注意：如果同一年级有多个学生，请按字母顺序排列他们的名字，并在新的一行上打印每个名字。

输入：

5
Harry
37.21
Berry
37.21
Tina
37.2
Akriti
41
Harsh
39

输出：

>>Berry
>>Harry

我做到了

if __name__ == '__main__':
l2=[]
l1=[]
for i in range(int(input())):
    name = input()
    l1.append(name)
    score = float(input())
    l1.append(score)
    l2.append(l1)
print(l2)

但它输出：

>>[['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]

Answer 1

^{一个旧问题，但添加一个答案以便获得帮助}

这确实是来自hackerrank的一个面试问题。

下面是我对这个问题的简单解决方案-我只需遍历嵌套的成绩列表。在每次迭代中，将新的得分与lowest得分和第二低得分（slowest）进行比较，同时保持一个names列表，其中得分等于第二低得分。

注：names总是与slowest分数相关，如果slowest改变，我们也改变names。如果新分数等于slowest分数，则添加新名称。评论将帮助你进一步。。。

def second_lowests(grades):
    """
    returns list of names of students with second lowest score
    """
    # intialize the `lowest` and second lowest `slowest` score and names
    grades = iter(grades)
    lname, lowest = next(grades)
    slname, slowest = next(grades)
    if slowest < lowest:
        lowest, slowest = slowest, lowest
        names = [lname]
    elif slowest == lowest: # we don't know, if lowest can be second lowest!
        names = [lname, slname]
    else:
        names = [slname]

    for name, score in grades:
        if score == slowest:
            names.append(name)
            continue
        if score == lowest:
            continue
        if score < lowest:
            if slowest == lowest:
                pass
            else:
                names = [lname]
            lowest, slowest = score, lowest
            lname = name
        elif score < slowest:
            slowest = score
            names = [name]
        elif score > slowest and slowest == lowest:
            slowest = score
            names = [name]

    if slowest == lowest: # all have same score
        return []
    names.sort()
    return names


if __name__ == '__main__':
    nested_list = []
    for _ in range(int(raw_input())):
        name = raw_input()
        score = float(raw_input())
        nested_list.append([name, score])
    assert 2 <= len(nested_list) <= 5
    print ("\n".join(second_lowests(nested_list)))

我不知道这是否是一个更好的解决方案，但它通过了所有的测试用例。

Answer 2

if __name__ == '__main__':
    total = []
    for i in range(int(input())):
        name = input()
        score = float(input())
        name_score = list((score, name))
        total.append(name_score)
    total.sort()
    min_mark = total[0][0]
    count = 0
    for i in range(len(total)):
        if min_mark == total[i][0]:
            count = count+1
    if count >= 1:
        for j in range(count):
            total.pop(0)
    nd_min_mark = total[0][0]
    for i in range(len(total)):
        if nd_min_mark == total[i][0]:
            print(total[i][1])

Answer 3

要回答您的问题，可以将输入作为

n=int(input())
arr=[[input(),float(input())] for _ in range(0,n)]

我相信这是一个黑客等级的问题 这是解决办法

n=int(input())
arr=[[input(),float(input())] for _ in range(0,n)]
arr.sort(key=lambda x: (x[1],x[0]))
names = [i[0] for i in arr]
marks = [i[1] for i in arr]
min_val=min(marks)
while marks[0]==min_val:
    marks.remove(marks[0])
    names.remove(names[0])    
for x in range(0,len(marks)):
    if marks[x]==min(marks):
        print(names[x])