How to make argument return his own name?

Question

My code

for i in list_tasks:
    globals()['icon1%s' % i].bind(on_press=lambda x: **print(#here I want to print his name)**

for example globals()['icon1%s' % i] is "icon1clock" I want it to print his own name (icon1clock) on press.

Can you give us an hint of what kind of object is icon1clock? — Alexandre Mahdhaoui, Aug 31 '21 at 19:27
icon1clock is a button, I need him to get the name of himself — יואב שלו, Aug 31 '21 at 19:27
First of all, don't use individual variables like `icon1clock`. Use a *dict* with keys, e.g., `icons['clock'] = ...`. — chepner, Aug 31 '21 at 19:28
Unfortunately I don't know the nature of your button object. But if it's an instance of class you could print its name like shown: `[globals()['icon1%s' % i].bind(on_press=lambda x: print(x.__name__)) for i in list_tasks] ` — Alexandre Mahdhaoui, Aug 31 '21 at 19:30
Is this a duplicate of this question? https://stackoverflow.com/questions/18425225/getting-the-name-of-a-variable-as-a-string — Adam Smooch, Aug 31 '21 at 19:33
@JohnGordon Because `i` is a free variable in the body of the lambda expression, not the value read from `list_tasks`. By the time the callback is called, `i` has probably gotten a new value. — chepner, Aug 31 '21 at 19:33
@John Because by the time the callback triggers, `i`’s value has changed. — deceze, Aug 31 '21 at 19:33

score 1 · Answer 1 · answered Aug 31 '21 at 19:32

The tricky part is defining a lambda expression that wraps another value, rather than a name. I wouldn't bother trying. Write a function that creates the callback for you; the callback can be a closure over a variable.

def make_callback(name):
    def _(x):
        print(name)
    return _


# *Not* icon1clock = ...
icons = {
    'clock': ..., 
}

for i in list_tasks:
    icons[i].bind(on_press=make_callback(i))

How to make argument return his own name?

1 Answers1