组合框绑定-管理数据

zyfwsgd6  于 2021-09-08  发布在  Java
关注(0)|答案(2)|浏览(296)

希望是一个快速的,但我知道它不会。我正在为自己编写一个小应用程序,我收集银币,未来我将做的不仅仅是银币,但现在我想记录我所拥有的,计划是使用组合框仅显示每个子组合框所需的数据
每个硬币的数据是(金属类型、国家、硬币类型、年份、纯度)如下所示,我有每个金属类型(金、银、铂、钯)的国家列表,每个国家至少有一枚硬币,并非所有国家都有所有金属类型,一些国家有更多。算上它,组合框事件将至少有54条if语句
有没有更合适的方法来管理这些数据,也许是字典?

{"Australia1": {"Silver": "Red Kangaroo", "Koala", "Kookaburra"}}                                   
{"Australia2": {"Gold": "Gold Nugget", "Dragon Rectangular Coin"}}
{"Australia3": {"Platinum": "coin1", "coin2"}}

等等

from tkinter import *
from tkinter import ttk

def draw_main_window():
    metals = ['Palladium',
              'Platinum',
              'Gold',
              'Silver']

    gold_country = ["Austria",
                    "Canada",
                    "China",
                    "Iran",
                    "Isle of Man",
                    "Israel",
                    "Kazakhstan",
                    "Malaysia",
                    "Malta",
                    "Mexico",
                    "New Zealand",
                    "Poland",
                    "Russia",
                    "Somalia",
                    "South Africa",
                    "Ukraine",
                    "United Kingdom",
                    "United States"]

    silver_country = ["Armenia",
                      "Australia",
                      "Austria",
                      "Canada",
                      "China",
                      "Congo (Republic)",
                      "Cook Islands",
                      "Isle of Man",
                      "Mexico",
                      "New Zealand",
                      "Niue/Fiji",
                      "Russia",
                      "Rwanda",
                      "Serbia",
                      "Somalia",
                      "South Africa",
                      "South Korea",
                      "Ukraine",
                      "United Kingdom",
                      "United States"]

    palladium_list = ["Australia",
                      "Canada",
                      "China",
                      "Portugal",
                      "Russia",
                      "United States"]

    platinum_list = ["Australia",
                     "Austria",
                     "Canada",
                     "Isle of Man",
                     "United Kingdom",
                     "United States"]

    def metal_click(event):
        metal_value = cmb_metal.get()

        if metal_value == "Palladium":
            cmb_country['values'] = palladium_list

        if metal_value == "Platinum":
            cmb_country['values'] = platinum_list

        if metal_value == "Gold":
            cmb_country['values'] = gold_country

        if metal_value == "Silver":
            cmb_country['values'] = silver_country

    def country_click(event):
        current_value = cmb_country.get()

        if current_value == "Australia":
            cmb_type['values'] = "Emu"

        if current_value == "Canada":
            cmb_type['values'] = "Palladium Maple Leaf"

    root = Tk()
    root.geometry("640x480")
    root.title("Silver-Inventory")

    Label(root, text="Metal").grid(column=0, row=0)
    Label(root, text="Country").grid(column=0, row=1)
    Label(root, text="Coin Type").grid(column=0, row=2)
    Label(root, text="Year").grid(column=0, row=3)
    Label(root, text="Metal").grid(column=0, row=4)
    Label(root, text="Fineness").grid(column=0, row=5)

    cmb_metal = ttk.Combobox(root, values=metals)
    cmb_metal.grid(column=1, row=0)
    cmb_metal.bind("<<ComboboxSelected>>", metal_click)

    cmb_country = ttk.Combobox(root)
    cmb_country.grid(column=1, row=1)
    cmb_country.bind("<<ComboboxSelected>>", country_click)

    cmb_type = ttk.Combobox(root)
    cmb_type.grid(column=1, row=2)

    root.mainloop()

draw_main_window()
zte4gxcn

zte4gxcn1#

您通常可以转换一长串的 if 语句转换为简单的字典查找。
例如,与此相反:

metal_value = cmb_metal.get()
if metal_value == "Palladium":
    cmb_country['values'] = palladium_list

if metal_value == "Platinum":
    cmb_country['values'] = platinum_list

if metal_value == "Gold":
    cmb_country['values'] = gold_country

if metal_value == "Silver":
    cmb_country['values'] = silver_country

... 可以使用字典在金属值和列表之间创建Map:

mapping = {
    "Palladium": palladium_list,
    "Platinum": platinum_list,
    "Gold": gold_country,
    "Silver": silver_country,
}
metal_value = cmb_metal.get()
cmb_country['values'] = mapping[metal_value]
hmae6n7t

hmae6n7t2#

我认为嵌套字典是一种很好的方法。我一直很喜欢@aaron hall's Vividict 类在他对问题的回答中显示了实现嵌套字典的最佳方法是什么?因为它使构建它们变得非常容易。这个概念来自perl,被称为“自活化”。
下面演示了如何使用它保存您拥有的数据类型:

from pprint import pprint

# Quick and dirty way to define a bunch of constant strings to make it

# unnecessary to quote them in the code every place they are used.

namespace = globals()
for name in ("Gold Silver Platinum "
             "Armenia Austria Australia Canada China Isle_of_Man "
             "Coin1 Coin2 Coin3 Coin4".split()):
    namespace[name] = name

class Vividict(dict):
    def __missing__(self, key):
        value = self[key] = type(self)()
        return value

data = {(Gold, Austria): [Coin1, Coin2],
        (Gold, Canada): [Coin1, Coin2, Coin3],
        (Gold, China): [Coin1, Coin2],
        (Silver, Armenia): [Coin1, Coin2],
        (Silver, Australia): [Coin1, Coin2],
        (Platinum, Australia): [Coin1],
        (Platinum, Austria): [Coin1, Coin2, Coin3],
        (Platinum, Canada): [Coin1, Coin2, Coin3, Coin4],
        (Platinum, Isle_of_Man): [Coin1, Coin2]}

d = Vividict()
for (metal, country), coins in data.items():
    d[metal][country] = coins

pprint(d)

下面是创建的嵌套字典:

{'Gold': {'Austria': ['Coin1', 'Coin2'],
          'Canada': ['Coin1', 'Coin2', 'Coin3'],
          'China': ['Coin1', 'Coin2']},
 'Platinum': {'Australia': ['Coin1'],
              'Austria': ['Coin1', 'Coin2', 'Coin3'],
              'Canada': ['Coin1', 'Coin2', 'Coin3', 'Coin4'],
              'Isle_of_Man': ['Coin1', 'Coin2']},
 'Silver': {'Armenia': ['Coin1', 'Coin2'], 'Australia': ['Coin1', 'Coin2']}}

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