在Python字典中找到最大的键

问题描述

我需要帮助，在python中找到一种方法来获取多维python词典中的最大N个项目.例如:

I need help finding a way in python to get the max N items in a multi-dimensional python dictionary. For example:

 things = {
          "car": { "weight": 100 },
          "apple": { "weight": 1 },
          "spanner": { "weight": 10 }
          }

在这种情况下，我想在词典中找到2个权重最高的项目，特别是这些项目的键.因此，在这种情况下，它应该返回["car", "spanner"]

In this case, I would want to find the 2 highest-weighted items in the dictionary, specifically the keys of these items. So in this case, it should return ["car", "spanner"]

注意:这是我第一次尝试遗传算法，因此我可能做得不正确.完全没有.

因为我是英国人，所以我正在寻找我能想象的最好的茶，所以我正在编写一个python程序，该程序生成10个随机的茶，然后使用自然选择在这10个茶中排名前5位，依此类推上.

As I am British, I am searching for the best cup of tea I can imagine, so I am writing a python program that generates 10 random cups of tea, then uses natural selection to find the top 5 in that ten and so on.

将一杯茶建模为python字典，具有5个键:

A cup of tea is modelled as a python dictionary, with 5 keys:

{
 "brew_time": Some Number,
 "milk": Some Number,
 "sweeteners": Some Number,
 "fitness": Some Number (This is what I'm interested in),
 "name": Some randomly generated name (Doesn't really matter)
}

我的程序将吐出的一杯茶看起来像这样:

A cup of tea my program will spit out will look something like this:

{'brew_time': 2.0, 'milk': 0.5, 'sweeteners': 3.0, 'name': 'bold cup', 'fitness': 0}

然后生成10杯茶，存储在teas变量中.这是该输出的示例:

It then generates 10 cups of tea, stored in the teas variable. This is an example of an output of that:

{0: {'brew_time': 2.0, 'milk': 0.4, 'sweeteners': 1.0, 'name': 'unafraid brew', 'fitness': 0}, 1: {'brew_time': 3.0, 'milk': 0.5, 'sweeteners': 3.0, 'name': 'fire-eating blend', 'fitness': 0}, 2: {'brew_time': 2.0, 'milk': 0.6, 'sweeteners': 2.0, 'name': 'fearless drink', 'fitness': 0}, 3: {'brew_time': 2.0, 'milk': 0.9, 'sweeteners': 3.0, 'name': 'fire-eating blend', 'fitness': 0}, 4: {'brew_time': 2.0, 'milk': 0.8, 'sweeteners': 2.0, 'name': 'fire-eating cuppa', 'fitness': 0}, 5: {'brew_time': 3.0, 'milk': 0.3, 'sweeteners': 1.0, 'name': 'fire-eating drink', 'fitness': 0}, 6: {'brew_time': 4.0, 'milk': 0.7, 'sweeteners': 2.0, 'name': 'dauntless medley', 'fitness': 0}, 7: {'brew_time': 3.0, 'milk': 0.3, 'sweeteners': 2.0, 'name': 'dauntless cuppa', 'fitness': 0}, 8: {'brew_time': 3.0, 'milk': 0.9, 'sweeteners': 2.0, 'name': 'epic drink', 'fitness': 0}, 9: {'brew_time': 2.0, 'milk': 0.4, 'sweeteners': 2.0, 'name': 'gusty drink', 'fitness': 0}}

我现在正在尝试编写一个名为selection()的函数，该函数将从字典中删除5种最不适合的茶. (我可以使用rank_tea()函数来设置茶的适应度，该函数采用一个数组并设置所有茶的适应度，该数字是介于0到1之间的数字，代表茶的质量)

I'm now trying to code a function called selection() that will remove the 5 least fit teas from the dictionary. (The fitness of a tea is set by me, using the rank_tea() function, which takes an array and sets all the teas fitnesses, which is a number between 0 - 1 that represents the quality of the tea)

这是我到目前为止所获得的，但是它不起作用:

This is what I've got so far, but it doesn't work:

def selection():
    teaCopy = teas.copy()
    fitnesses = []
    for i in range(0, len(teaCopy)):
        fitnesses.append(teas[i]["fitness"])

    print(fitnesses)

    max_fitnesses_indicies = sorted(range(len(fitnesses)), key=lambda x: fitnesses[x])
    print(max_fitnesses_indicies)

    len_array = []
    print(len_array)
    for i in range(0, len(teas)):
        len_array.append(i)

    to_be_del = list( set(max_fitnesses_indicies) - set(len_array) )
    print(to_be_del)

完整的代码.很抱歉，我只是没有回答这个问题而已.不想错过任何事情.

This is the full code. Sorry for the length of the question, I just didn't want to miss anything.

任何帮助将不胜感激

推荐答案

您可以简单地使用:

>>> sorted(things.keys(),key=lambda x:things[x]['weight'],reverse=True)
['car', 'spanner', 'apple']

获得按重量排序的物品清单(此处按相反顺序，以便较重的物品先被排序).因此，如果您致电:

To obtain a list of items sorted by their weight (here in reversed order such that the more heavy things are sorted first). So if you call:

>>> sorted(things.keys(),key=lambda x:things[x]['weight'],reverse=True)[:2]
['car', 'spanner']

您得到两个最重的东西.但这将以 O(n log n) 运行.如果您希望获得的 k个值 k 很小(与总数相比).您可以使用heapq:

you get the two heaviest. But this will run in O(n log n). In case the number of values k you wish to obtain is small (compared to the total number). You can use heapq:

from heapq import nlargest

result = nlargest(k,things.keys(),key=lambda x:things[x]['weight'])

据我所知-以 O(n log k) ( k 选择).

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