python中的快速排序和选择排序方法

Question

class University:
        def __init__(self, name, numberoffaculity, numberofstudent):
            self.name = name
            self.numberoffaculity = numberoffaculity
            self.numberofstudent = numberofstudent


n = int(input())  # Number of universities that user wants to input
universities = []
for i in range(n):
    name = input()
    numberoffaculity = int(input())
    numberofstudent = int(input())
    uni = University(name, numberoffaculity, numberofstudent)
    universities.append(uni)

# Insert your code below this line


# Insert your code above this line

for uni in universities:
    print("%-10s %-10d %-10d"
          % (uni.name, uni.numberoffaculity, uni.numberofstudent))

我的目标是:程序获得数字"numberoffaculity“1或2，如果为1，则通过quicksort方法从大到小对列表进行排序和打印；如果是2E 212，则将根据E 115从E 216/code>通过E 117选择排序>E 218方法对列表进行排序和打印。

Answer 1

这应该能起作用。首先，我用一个键函数添加了quick_sort和selection_sort函数。

然后我根据用户输入调用它们：

def quick_sort(array,key = lambda x:x):
    if len(array) < 2:
        return array
    else:
        pivot = array[0]
        less = [i for i in array[1:] if key(i) <= key(pivot)]
        greater = [i for i in array[1:] if key(i) > key(pivot)]
        return quick_sort(less,key) + [pivot] + quick_sort(greater,key)


def selection_sort(lst,key = lambda x:x):
    empty_lst = []
    x = len(lst) - 1
    while x >= 0:
        for i in range(len(lst)):
            if key(lst[i]) <= key(lst[0]):
                lst[0], lst[i] = lst[i], lst[0]
                # this part compares the number in first index and numbers after the first index.
        g = lst.pop(0)
        empty_lst.append(g)
        x -= 1
    return empty_lst


user_input = int(input())
if user_input==1:
    universities = quick_sort(universities,key = lambda x:x.numberoffaculity)[::-1]
elif user_input==2:
    universities = selection_sort(universities,key = lambda x:x.numberofstudent)
else:
    print("invalid sort input , expected 1 or 2 and got %s"%user_input)