创建一个没有Lambda函数的Python计算器？

Question

我正在尝试用Tkinter创建一个简单的计算器来教孩子们如何编写基本程序。我已经制作了一个使用lambda函数的函数，但是我正在试图弄清楚如何在没有lambda的情况下制作一个。有人能帮忙吗？以下是我目前掌握的代码：

# start coding the buttons
button1 = Button(gui, text=' 1 ', fg='black', bg='gray65', 
                command=lambda: press(1), height=1, width=7) 
button1.grid(row=2, column=0) 

button2 = Button(gui, text=' 2 ', fg='black', bg='gray65', 
                command=lambda: press(2), height=1, width=7) 
button2.grid(row=2, column=1) 

button3 = Button(gui, text=' 3 ', fg='black', bg='gray65', 
                command=lambda: press(3), height=1, width=7) 
button3.grid(row=2, column=2) 

button4 = Button(gui, text=' 4 ', fg='black', bg='gray65', 
                command=lambda: press(4), height=1, width=7) 
button4.grid(row=3, column=0) 

button5 = Button(gui, text=' 5 ', fg='black', bg='gray65', 
                command=lambda: press(5), height=1, width=7) 
button5.grid(row=3, column=1) 

button6 = Button(gui, text=' 6 ', fg='black', bg='gray65', 
                command=lambda: press(6), height=1, width=7) 
button6.grid(row=3, column=2) 

button7 = Button(gui, text=' 7 ', fg='black', bg='gray65', 
                command=lambda: press(7), height=1, width=7) 
button7.grid(row=4, column=0) 

button8 = Button(gui, text=' 8 ', fg='black', bg='gray65', 
                command=lambda: press(8), height=1, width=7) 
button8.grid(row=4, column=1) 

button9 = Button(gui, text=' 9 ', fg='black', bg='gray65', 
                command=lambda: press(9), height=1, width=7) 
button9.grid(row=4, column=2) 

button0 = Button(gui, text=' 0 ', fg='black', bg='gray65', 
                command=lambda: press(0), height=1, width=7) 
button0.grid(row=5, column=0) 

plus = Button(gui, text=' + ', fg='black', bg='gray65', 
            command=lambda: press("+"), height=1, width=7) 
plus.grid(row=2, column=3) 

minus = Button(gui, text=' - ', fg='black', bg='gray65', 
            command=lambda: press("-"), height=1, width=7) 
minus.grid(row=3, column=3) 

multiply = Button(gui, text=' * ', fg='black', bg='gray65', 
                command=lambda: press("*"), height=1, width=7) 
multiply.grid(row=4, column=3) 

divide = Button(gui, text=' / ', fg='black', bg='gray65', 
                command=lambda: press("/"), height=1, width=7) 
divide.grid(row=5, column=3)  

Decimal= Button(gui, text='.', fg='black', bg='gray65', 
            command=lambda: press('.'), height=1, width=7) 
Decimal.grid(row=5, column='1')

Answer 1

在使用lambda的任何地方，都可以使用def：

button1 = Button(gui, text=' 1 ', fg='black', bg='gray65', 
                command=lambda: press(1), height=1, width=7) 

变成：

def press_1():
    press(1)

button1 = Button(gui, text=' 1 ', fg='black', bg='gray65', 
                command=press_1, height=1, width=7) 

您可以编写一个函数来创建一个带有特定数字的按钮，从而减少按钮生成过程中的一些麻烦：

def make_button(num: int) -> Button:
    def press_num():
        press(num)

    return Button(gui, text=f' {num} ', fg='black', bg='gray65', 
                command=press_num, height=1, width=7)

然后你可以做：

button1 = make_button(1)
button2 = make_button(2)

等等。您还可以通过在嵌套循环中构建网格的某些部分来进一步了解这一点，例如：

for row in range(3):
    for col in range(3):
        button = make_button(row * 3 + col + 1)
        button.grid(row=row+2, column=col)