在python中使用大量嵌套语句时改进代码

Question

最近，我在学习了一些关于Matlab的课程（主要集中在使用数学计算和绘图）之后开始使用python3，我仍然是一个初学者。 不过，我觉得创建一个计算器来计算pokemon的数量会很有趣，你可以用pokemon中可用的糖果来进化pokemon，然后使用Tkinter作为GUI。程序运行良好，效果良好。然而，我正在努力改进我的编码，这段代码包括许多嵌套的if和while语句，以及其他广泛的变量和参数使用。我想知道如何将这段代码改进为更“pythonic”的代码，或者这是否是创建这样一个程序的正确方法。 先谢谢你

    from tkinter import *
    from tkinter import messagebox


    def EvolveCalc():
        title = "Pokemon evolve calculator"
        Poke = var.get()
        Candies_total = Candies.get()
        PAmount = Poke_amount.get()
        if Poke == 'Pokemon':
            messagebox.showinfo(title,
                        "Please select a Pokemon from the dropdown menu")
        elif not PAmount.isdigit():
            messagebox.showinfo(title, "Please only enter numbers")
        elif not Candies_total.isdigit():
            messagebox.showinfo(title, "Please only enter numbers")
        else:
            Candies_total = int(Candies.get())
            PAmount = int(Poke_amount.get())
            Evolve = int((Candies_total - 1) / (Pokedex[Poke] - 1))
            Candies_needed = (PAmount * (Pokedex[Poke] - 1)) + 1
            if PAmount < Evolve:
                n = 0
                while Candies_needed <= Candies_total:
                    n = n + 1
                    PAmount = PAmount + 1
                    Candies_needed = ((PAmount) * (Pokedex[Poke] - 1)) + 1
                    Candies_total = Candies_total + 3
                    Evolve2 = int((Candies_total - 1) / (Pokedex[Poke] - 1))
                String1 = '''You have enough candies too evolve {0} {1},
                but you only have {2} {1} in storage and thus can only
                evolve {2} {1} If you catch {3} more {1} you can now
                evolve {4} {1}.'''
                messagebox.showinfo(title, String1.format(Evolve, Poke,
                                                          Poke_amount.get(),
                                                          n, Evolve2))
            else:
                m = 0
                while Candies_total <= Candies_needed:
                    m = m + 1
                    PAmount = PAmount - 1
                    Candies_needed = ((PAmount) * (Pokedex[Poke] - 1)) + 1
                    Candies_total = Candies_total + 1
                    Evolve2 = int((Candies_total - 1) / (Pokedex[Poke] - 1))
                String2 = 'Transfer {0} {1} so you can evolve a total of {2} {1}.'
                messagebox.showinfo(title, String2.format(m, Poke, Evolve2))


    root = Tk()

    Pokedex = {'pidgey': 12,
               'caterpie': 12,
               'weedle': 12
               }

    root.title("Pokemon evolve calculator")

    var = StringVar(root)
    var.set('Pokemon')
    choices = Pokedex.keys()
    Pokemon = OptionMenu(root, var, *choices)
    Pokemon.grid(column=0, row=1)

    L1 = Label(root, text='Candies')
    L1.grid(column=1, row=0)

    Candies = Entry(root)
    Candies.grid(column=1, row=1)

    L2 = Label(root, text='Pokemon amount in storage')
    L2.grid(column=2, row=0)

    Poke_amount = Entry(root)
    Poke_amount.grid(column=2, row=1)

    Calculate = Button(root, text='Calculate', command=EvolveCalc)
    Calculate.grid(column=1, row=2)

    root.mainloop()

Answer 1

1. 使用小写和下划线分隔的函数，使EvolveCalc成为evolve_calc
2. 使用小写和可读变量，使Poke和Candies_total变成poke和total_candies

3. 这段代码可以简化

   elif not PAmount.isdigit():
        messagebox.showinfo(title, "Please only enter numbers")
   elif not Candies_total.isdigit():
       messagebox.showinfo(title, "Please only enter numbers")
   

   至

   elif not PAmount.isdigit() or not Candies_total.isdigit():
       messagebox.showinfo(title, "Please only enter numbers")
   

4. 你的Pokedex可以格式化，这样更可读

   Pokedex = {'pidgey': 12,
              'caterpie': 12,
              'weedle': 12
              }
   

   至

   Pokedex = {
     'pidgey': 12,
     'caterpie': 12,
     'weedle': 12
   }
   

5. 我无法运行你的代码，因为它的格式不正确。第一个else块完成后，空格关闭。

其他建议：

• 阅读PEP8
• 查看代码示例
• 在函数中编写可重用的小代码
• 添加注释来解释你的一些数学知识