Python作业

Question

我一直在写一个记录作业的软件，用python写的。我会给出一个显示所有作业的小片段。

def delHomework(homework,element,HLabel,HDelete):
    del homework.get(element)
    HLabel.destroy()
    HDelete.destroy()

row = 0 #keeps track on which height everything is inserted
for i in homeworks: #homeworks is a list of all homework's
    HLabel = Label(text=i) #displays the homework
    HLabel.grid(column=0,row=row)

    HDelete = Button(text="delete",command=lambda: 
        delHomework(homework,i,HLabel,HDelete)
    #is the button to delete the homework

    HDelete.grid(column=1,row=row)

    row += 1

问题是，即使它全部显示正确，当您尝试删除任何作业时，最后一个作业被删除而不是关联的作业，因为 lambda 在该迭代中引用的是变量的最新状态，而我可以弄清楚如何让它发挥作用。希望这个问题有道理。

Answer 1

假设 HDelete 的括号缺失只是复制粘贴问题。

实际问题是，在函数被调用之前，不会为 lambda 函数计算 i 的值，当它被调用时，它使用 i 的最新值，在这种情况下，这将是 homworks 列表中 i 的最后一个值。

一个非常简单的例子-

>>> lst = [1,2,3,4,5,6]
>>> newlst = []
>>> for i in lst:
...     newlst.append(lambda : print(i))
...
>>> newlst[1]()
6
>>> newlst[2]()
6

您应该将元素作为默认参数传递给 lambda。示例 -

HDelete = Button(text="delete",command=lambda ele = i, hl = HLabel, hd = HDelete: 
    delHomework(homework,ele,hl,hd))

Example/Demo 展示它的工作原理 -

>>> lst = [1,2,3,4,5,6]
>>> newlst = []
>>> for i in lst:
...     newlst.append(lambda ele = i: print(ele))
...
>>> newlst[2]()
3
>>> newlst[1]()
2

Answer 2

如以下问题的已接受答案中所述：What do (lambda) function closures capture?

你可以这样做：

HDelete = Button(text="delete",command=(lambda i, l, d: lambda:
    delHomework(homeworks, i, l, d))(i, HLabel, HDelete)

或类似这样的内容：

def createDeleteCommand(homeworks, i, l, d):
    return lambda: delHomework(homeworks, i, l, d)

HDelete = Button(text="delete",command=
     createDeleteCommand(homeworks, i, HLabel, HDelete))

Python作业

Python Homework

python

closures