如何在Python中初始化字典，其中key是一个对象？

Question

我正在尝试重新表述发现here的实现。这就是我到目前为止所知道的：

import csv
import math
import random

training_set_ratio = 0.67

training_set = []
test_set = []


class IrisFlower:
    def __init__(self, petal_length, petal_width, sepal_length, sepal_width, flower_type):
        self.petal_length = petal_length
        self.petal_width = petal_width
        self.sepal_length = sepal_length
        self.sepal_width = sepal_width
        self.flower_type = flower_type

    def __hash__(self) -> int:
        return hash((self.petal_length, self.petal_width, self.sepal_length, self.sepal_width))

    def __eq__(self, other):
        return (self.petal_length, self.petal_width, self.sepal_length, self.sepal_width) \
               == (other.petal_length, other.petal_width, other.sepal_length, other.sepal_width)


def load_data():
    with open('dataset.csv') as csvfile:
        rows = csv.reader(csvfile, delimiter=',')
        for row in rows:
            iris_flower = IrisFlower(float(row[0]), float(row[1]), float(row[2]), float(row[3]), row[4])
            if random.random() < training_set_ratio:
                training_set.append(iris_flower)
            else:
                test_set.append(iris_flower)


def euclidean_distance(flower_one: IrisFlower, flower_two: IrisFlower):
    distance = 0.0
    distance = distance + math.pow(flower_one.petal_length - flower_two.petal_length, 2)
    distance = distance + math.pow(flower_one.petal_width - flower_two.petal_width, 2)
    distance = distance + math.pow(flower_one.sepal_length - flower_two.sepal_length, 2)
    distance = distance + math.pow(flower_one.sepal_width - flower_two.sepal_width, 2)
    return distance


def get_neighbors(test_flower: IrisFlower):
    distances = []
    for training_flower in training_set:
        dist = euclidean_distance(test_flower, training_flower)
        d = dict()
        d[training_flower] = dist
        print(d)
    return


load_data()
get_neighbors(test_set[0])

目前，打印以下代码块中的语句：

def get_neighbors(test_flower: IrisFlower):
    distances = []
    for training_flower in training_set:
        dist = euclidean_distance(test_flower, training_flower)
        d = dict()
        d[training_flower] = dist
        print(d)
    return

将具有类似以下内容的输出

{<__main__.IrisFlower object at 0x107774fd0>: 0.25999999999999945}

这是可以的。但我不想先创建字典，然后再追加键值，如下所示：

d = dict()
d[training_flower] = dist

所以这就是我正在尝试的：

d = dict(training_flower = dist)

但是，看起来dist方法使用的不是实例，而是字符串，因为我看到的打印内容如下所示：

{'training_flower': 23.409999999999997}
{'training_flower': 16.689999999999998}

如何在一条语句中使用对象作为键来创建字典？

Answer 1

要使用非有效关键字的键直接创建字典，请使用{}语法，如下所示：

代码：

d = {training_flower: 'a_value'}

测试代码：

training_flower = 'a key'
d = {training_flower: 'a_value'}
print(d)

结果：

{'a key': 'a_value'}

Answer 2

在编写d = dict(training_flower=dist)代码片段中，"training_flower“是dict函数的关键字参数，而不是对象。它等同于编写d = {'training_flower': dist}。使用对象作为键创建字典的唯一方法是使用后一种语法：

d = {training_flower: dist}

Answer 3

使用对象作为键初始化字典(edit：，Stephen示例中的字符串无论如何都是一个对象)

class Flower:

    def __repr__(self):
        return 'i am flower'


flower1 = Flower()
d = {flower1: 4}
print(d)

输出

{i am flower: 4}

这是我在这里的第一篇文章，我知道我来晚了，如果这是一个重复的解决方案，很抱歉。只是为了展示它对一个对象的工作。

我会支持斯蒂芬的答案，但我现在还不能。