排序代码Python中的语法

Question

我应该做的是

Write a function top_k that takes a list of students and an integer k and returns a list of the names of the k students with the highest scores in alphabetical order. If there are students in the range (k + 1, · · · k + i) who have the same score as the kth student, include them in the list as well.

当我输入

top_k(student_list1, 5) 
>>> [('eugene', 'A', 21), ('simon', 'A', 21), ('john', 'A', 15), ('tiffany', 'A', 15), ('dave', 'B', 12)]
top_k(student_list2, 3)
>>>[('eugene', 'A', 21), ('simon', 'A', 21), ('john', 'A', 15), ('tiffany', 'A', 15)]

在哪里

^{pr2}$

我的代码是

def top_k(students, k):
    toptobottom = sorted(students, key = lambda x: x[2], reverse = True)
    takeout= toptobottom[:k+1]
    result = []
    for takethat in takeout:
        if takethat[2] == 21:
        result.append(takethat)
            result.sort( key= lambda x: x[0])
        elif takethat[2] ==15:
            result.append(takethat)
    return result

我不知道该放在哪里结果.排序（键..）。它应该和elif在同一行，还是应该有一个缩进？ 谢谢！在

Answer 1

您可以只对输入数组排序一次，然后选择正确的学生。在

def top_k(students, k):
    toptobottom = sorted(students, key = lambda x: (-x[2], x[0]))
    length = len(toptobottom)
    if k >= 0 and k < length:
        result = toptobottom[:k]
        i = k
    else:
        result = toptobottom
        i = 0
        k = 1
    while i < length and toptobottom[i][2] == toptobottom[k-1][2]:
        result.append(toptobottom[i])
        i += 1
    return result

Answer 2

除了你的缩进问题，你的代码是专门为输入数据编写的（即在你的函数中硬编码的最高分）。虽然这可能会让你完成练习，但这不是一个好的实践，你应该更全面地思考你要解决的问题。在

在伪代码中：

def top_k(students, k):
    create a blank output list
    for each student in the list of students sorted in reverse order by score:
        if either ((we have fewer than k students in the output so far) or 
                   (the student equals the score of the last student in output)):
            add the student to the output
        otherwise:
            stop looping
    return the output sorted by name

Answer 3

这绝不是一个答案，所以我真的不想把它标记成这样，但是我想插上几句话，它们太长了，不能发表评论，从长远来看可能对你有帮助。在

关于你的问题：

def top_k(students, k):
    toptobottom = sorted(students, key = lambda x: x[2], reverse = True)
    takeout= toptobottom[:k+1]
    result = []
    for takethat in takeout:
        if takethat[2] == 21:
            result.append(takethat)  # I indented this
            result.sort( key= lambda x: x[0])
        elif takethat[2] ==15:
            result.append(takethat)
    return result

我想我帮你解决了你的问题，但我还没有分析你的逻辑。我看到了硬编码的分数，就停止了阅读，因为这真是个坏主意。永远不要那样做，永远不要。尤其是当你在为算法课做作业的时候。在

第二件事是，我不认为你应该完全忽略一些python资源，python's standard library resource for sorting is enormous而且有很多库可以帮助你。在

第三，我们花了大量的精力为python的库和built in types that I really think you should check out提供某种完整性

第四，我根据你的例子准备了一些小片段。当然，我并没有把它变成一个函数，因为我想让你想想你在做什么。不管怎样，我想你会发现这个很有帮助的。在

^{pr2}$

输出如下：

correct sorted order
('eugene', 'A', 21)
('simon', 'A', 21)
('john', 'A', 15)
('tiffany', 'A', 15)
('dave', 'B', 12)
('jane', 'B', 10)
('charles', 'C', 9)
('ben', 'C', 8)
('freddy', 'D', 4)
('jimmy', 'F', 1)
grouped together
{1: [('jimmy', 'F', 1)], 4: [('freddy', 'D', 4)], 8: [('ben', 'C', 8)], 9: [('charles', 'C', 9)], 10: [('jane', 'B', 10)], 12: [('dave', 'B', 12)], 15: [('john', 'A', 15), ('tiffany', 'A', 15)], 21: [('eugene', 'A', 21), ('simon', 'A', 21)]}

let's get 'five' elements from the list, like in your example
[('eugene', 'A', 21), ('simon', 'A', 21), ('john', 'A', 15), ('tiffany', 'A', 15), ('dave', 'B', 12)]

let's get 'three' elements from the list, like in your example
[('eugene', 'A', 21), ('simon', 'A', 21), ('john', 'A', 15), ('tiffany', 'A', 15)]

你认为你可以综合所有这些，并以更好的理解完成你的任务？这里有一些复杂的事情在进行，比如dict、list和list理解，但是我认为这是一个很好的特性汇编，允许python将几乎所有的list问题切碎。在