lambda in for loop only takes last value [duplicate]

Answer 1

Please read about minimal examples. Without reading your code, I believe you have run into a well known issue addressed in previous questions and answers that needs 2 lines to illustrate. Names in function bodies are evaluated when the function is executed.

funcs = [lambda: i for i in range(3)]
for f in funcs: print(f())

prints '2' 3 times because the 3 functions are identical and the 'i' in each is not evaluated until the call, when i == 2. However,

funcs = [lambda i=i:i for i in range(3)]
for f in funcs: print(f())

makes three different functions, each with a different captured value, so 0, 1, and 2 are printed. In your statement

__cMenu.add_command(label="{}".format(option),
    command=lambda: self.filter_records(column, option))

add option=option before : to capture the different values of option. You might want to rewrite as

lambda opt=option: self.filter_records(column, opt)

to differentiate the loop variable from the function parameter. If column changed within the loop, it would need the same treatment.

Answer 2

Closures in Python capture variables, not values. For example consider:

def f():
    x = 1
    g = lambda : x
    x = 2
    return g()

What do you expect the result of calling f() to be? The correct answer is 2, because the lambda f captured the variable x, not its value 1 at the time of creation.

Now if for example we write:

L = [(lambda : i) for i in range(10)]

we created a list of 10 different lambdas, but all of them captured the same variable i, thus calling L[3]() the result will be 9 because the value of variable i at the end of the iteration was 9 (in Python a comprehension doesn't create a new binding for each iteration; it just keeps updating the same binding).

A "trick" that can be seen often in Python when capturing the value is the desired semantic is to use default arguments. In Python, differently from say C++, default value expressions are evaluated at function definition time (i.e. when the lambda is created) and not when the function is invoked. So in code like:

L = [(lambda j=i: j) for i in range(10)]

we're declaring a parameter j and setting as default the current value of i at the time the lambda was created. This means that when calling e.g. L[3]() the result will be 3 this time because of the default value of the "hidden" parameter (calling L[3](42) will return 42 of course).

More often you see the sightly more confusing form

lambda i=i: ...

where the "hidden" parameter has the same name as the variable of which we want to capture the value of.

Answer 3

I know I am late, but I found a messy workaround which gets the job done (tested in Python 3.7)

If you use a double lambda (like I said, very messy) you can preserve the value, like so:

Step 1: Create the nested lambda statement:

send_param = lambda val: lambda: print(val)

Step 2: Use the lambda statement:

send_param(i)

The send_param method returns the inner most lambda (lambda: print(val)) without executing the statement, until you call the result of send_param which takes no arguments, for example:

a = send_param(i)
a()

Only the second line will execute the print statement.