如何理解python lambda中的闭包? [英] how to understand closure in python lambda?
问题描述
我想创建5个按钮,每个按钮绑定一个命令以打印此索引。
,但总是打印5。
I want to make 5 buttons and each buttons bind a commend to print this index. but always print 5.
我的代码就像这样,
for i in range(5):
make_button = Tkinter.Button(frame, text ="make!", command= lambda: makeId(i))
def makeId(i):
print(i)
它总是打印5, ?
感谢很多~~
It always print 5 , then how can i do ? thanks a lot~~
推荐答案
lambda执行时,lambdas中的变量解析。此时,对于所有按钮i = 5。要纠正此问题,请执行以下操作:
Resolution of variables in lambdas is done when lambda is executed. At this time, for all buttons i=5. To rectify this issue do as follows:
make_button = Tkinter.Button(frame, text ="make!", command= lambda i=i: makeId(i))
这将创建i作为lambda中的局部变量。这个局部变量将从循环中保持i的正确值。局部变量可以具有任何名称,不一定 i
,例如 command = lambda a = i:makeId(a))
。
This creates i as a local variable in a lambda. This local variable will hold correct value of i from the loop. the local variable can have any name, not necessarily i
, e.g. command= lambda a=i: makeId(a))
.
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