使用循环按下一个创建的单选按钮并调用其命令时出现KeyError

Question

我在一个循环中创建多个单选按钮，并通过lambda为它们分配一个带有一个参数的函数(这些函数也是使用循环和字典创建的(，但当我按下其中一个单选按钮时，我会得到以下错误：

Exception in Tkinter callback
Traceback (most recent call last):
File "/home/shirin/anaconda3/lib/python3.8/tkinter/__init__.py", line 1892, in __call__
return self.func(*args)
File "/tmp/ipykernel_7981/822586079.py", line 99, in <lambda>
lst_r["r{0}".format(f+j)] = Radiobutton(text = algorithms[f+j], var = v, value = f+j, height=2, width=30, indicator = 0, command = lambda: d['use_algo'+str(f+j)](algorithms[f+j]))
KeyError: 'use_algo35'

对于算法文件，我创建了4.py文件，其中包含以下代码：

def print_array(array):
array2 = []
for i in array:
array2.append(i) 
print array2
array = arr
print_array(array)

这里有一个和我一样的最小代码；运行此代码会返回KeyError:"use_algo5"：

from tkinter import *
root = Tk()
row_num = 0
font = ('Verdana', 10)
algorithms = ['algorithm1','algorithm2','algorithm3','algorithm4']
def enter():
try:
class entries:
def __init__(self, root):
self.variables = []
n = 4
for i in range(n):
self.variables.append(StringVar())
self.labels = []
self.entries = []
c, d, e = 0,0,0

for ii in range(int(n/4)):
for jj in range(4):
num = c+(jj+1)
self.labels.append(Label(root , text = num))
self.labels[-1].grid(padx=0, pady=0, row=ii+d+1, column=jj)
self.entries.append(Entry(root, font = font, textvariable =self.variables[c+(jj)], width = 30))
self.entries[-1].grid(padx=0, pady=0, row=ii+d+2, column=jj)
d+=2   
c+=4
e = ii+1+d

label_algo = Label(font = font, text = 'Choose an algorithm:')
label_algo.grid(row = int(n/4)+e+2 , column = 0, padx = 29, pady = 15)
row_num = int(n/4)+e+2

def use_algo(algo_name):   ### radio button functions:
arr = []
for i in range(int(len(self.entries))):
arr.append(self.entries.get())
arr = exec(open("/home/shirin/Desktop/Python Project/Algorithms/" + algo_name + ".py").read())
label_output.configure(text = 'The array is {}'.format(arr))
d = {f'use_algo{k}': partial(use_algo, algo_name=algorithms[k]) for k in range(int(len(algorithms)))}
v = IntVar()
lst_r = {}
f = 0
for i in range(int(len(algorithms)/2)):    ### making radio buttons:
for j in range(2):
lst_r["r{0}".format(f+j)] = Radiobutton(text = algorithms[f+j], var = v, value = f+j, height=2, width=30, indicator = 0, command = lambda: d['use_algo'+str(f+j)](algorithms[f+j]))
lst_r["r{0}".format(f+j)].grid(row=row_num+1+i, column=j)
f+=2
label_output = Label(font = font)
label_output.grid(row = row_num+1+int(len(algorithms)/4)+1 , column = 0, padx = 29, pady = 15) 

T = entries(root)

except:
pass
label_num = Label(font = font, text = 'Press enter button')
label_num.grid(row = 0 , column = 0, padx = 30, pady = 15)
button_enter = Button(text = 'Enter', command = enter, width = 20)
button_enter.grid(row = 0 , column = 2, padx = 30)
root.mainloop()

Answer 1

当通过单击其中一个单选按钮执行lambda时，f+j是5，因为在创建这些单选按钮的for循环之后，f是4，j是1。

此外，通常在command选项中直接使用partial()，如下所示：

command=partial(use_algo, algorithms[f+j])

那么CCD_ 6根本不是必需的。

注意，use_algo()内部的for循环是不必要的，因为arr被：覆盖

arr = exec(open("/home/shirin/Desktop/Python Project/Algorithms/" + algo_name + ".py").read())