如何在对象列表中搜索最小索引

Question

我想在对象列表中搜索对象属性中存在的最小数字。

使用列表解析和min 函数很容易。当我想找到对象的索引时，问题就来了。

class School:
    def __init__(self, name, num_pupils, num_classrooms):
        self.name = name
        self.num_pupils = num_pupils
        self.num_classrooms = num_classrooms

    def students_per_class(self):
        return self.num_pupils / self.num_classrooms

    def show_info(self):
        print(f"{self.name} has {self.students_per_class():.2f} students per class.")

    def string_checker(question):
        valid = False
        while not valid:
            try:
                response = str(input(question))
                if all(character.isalpha() or character.isspace() for character in response):
                    valid = True
                    return response
                else:
                    print("Enter a string containing no numbers of special characters. ")
            except ValueError:
                print("Enter a string containing no numbers of special characters. ")

    def num_checker(question):
        valid = False
        while not valid:
            try:
                response = int(input(question))
                if (response):
                    valid = True
                    return response
                else:
                    print("Enter an integer containing no letters or special characters. ")

        except ValueError:
            print("Enter an integer containing no letters or special characters. ")

    def new_school():
        school_name = string_checker("School Name: ")
        num_pupils = num_checker("Number of Pupils: ")
        num_classrooms = num_checker("Number of Classrooms: ")
        return School(school_name, num_pupils, num_classrooms)


if __name__ == "__main__":
    schools = []
    school = School("Darfield High School", 900, 37)
    schools.append(school)
    school.show_info()

    for i in range(1):
        schools.append(new_school())

    for school in schools:
        school.show_info()

    print(min(school.students_per_class() for school in schools)) 
    # This works fine and prints the value of the lowest number
    print(schools.index(min(school.students_per_class() for school in schools))) 
    # This doesn't work as it tries to find the index of the value returned
    # from the min function as it should.

Answer 1

@Tomas Ordonez

你的答案在这里：

def minInt(instanceList):
    sorted_instanceList = sorted(instanceList, key=lambda instance: instance.int)
    minIndex = instanceList.index(sorted_instanceList[0])
    return minIndex

Answer 2

我想您可能正在寻找np.argmin 函数，如果您提供一个列表作为输入，它将返回最小元素的索引。

在这种情况下是：

import numpy as np 
min_ind = np.argmin([school.students_per_class() for school in schools])
print(schools[min_ind])

Answer 3

如果您希望能够直接比较、排序、查找项目的最小值/最大值，您可以使用“dunder”方法。这些方法允许您重载类的内置函数。

例如，如果您有这样的课程

class Item:
    def __init__(self, value):
        self.value = value

    def __lt__(self, other):
        return self.value < other.value

    def __eq__(self, other):
        return self.value == other.value

然后您可以创建两个实例并像这样直接比较它们，

A = Item(1)
B = Item(2)

print(A < B) # Prints True

或者如果你有一个项目列表

items = [A, B]

然后你可以通过去获得最小的项目

min_item = min(items)

或者通过去它的索引

min_item_index = items.index(min(items))

尽管仅引用最小项可能就足够了。

Answer 4

您可以使用min 的key 参数按索引搜索：

index = min(range(len(schools)), key=lambda i: schools[i].students_per_class())
print(schools[index])

键提供将被比较的值，而不是实际的序列。在这里，我的序列是range(len(schools))，它只是所有元素的索引。但我没有找到最小索引，而是让我们找到每个索引i 的最小值schools[i].students_per_class()。

Answer 5

使用enumerate 遍历列表，同时跟踪索引：

min(((i, school.students_per_class()) for i, school in enumerate(schools)), key=lambda x: x[1])

(i, school.students_per_class()) 是一个元组，使用min() 的key 参数，我们要求Python 找到school.students_per_class() 的最小值，随后返回它的索引。

了解 lambda here。