PyQt QPushButton clicked connect lambda can't get correct value [duplicate] - pyqt5

Question

This question already has answers here:
tkinter creating buttons in for loop passing command arguments
(3 answers)
Closed 2 years ago.
Problemset:
Context Menu should show filter variables dynamically and execute a function with parameters defined inside the callback.
Generic descriptions show properly, but function call is always executed with last set option.
What I have tried:
#!/usr/bin/env python
import Tkinter as tk
import ttk
from TkTreectrl import MultiListbox
class SomeClass(ttk.Frame):
def __init__(self, *args, **kwargs):
ttk.Frame.__init__(self, *args, **kwargs)
self.pack(expand=True, fill=tk.BOTH)
self.grid_rowconfigure(0, weight=1)
self.grid_columnconfigure(0, weight=1)
self.View=MultiListbox(self)
__columns=("Date","Time","Type","File","Line","-","Function","Message")
self.View.configure(columns=__columns, expandcolumns=(0,0,0,0,0,0,0,1))
self.View.bind("", self.cell_context)
self.View.grid(row=0, column=0, sticky=tk.NW+tk.SE)
self.__recordset = []
self.__recordset_filtered = False
#Some dummy values
self.__recordset.append(["Date", "Time", "INFO", "File", "12", "-", "Function", "Message Info"])
self.__recordset.append(["Date", "Time", "DEBUG", "File", "12", "-", "Function", "Message Info"])
self.__recordset.append(["Date", "Time", "WARNING", "File", "12", "-", "Function", "Message Info"])
self.__refresh()
def cleanView(self):
self.View.delete(0, tk.END)
def __refresh(self):
self.cleanView()
for row in self.__recordset:
self.View.insert(tk.END, *row)
def filter_records(self, column, value):
print("Filter Log Recordset by {column} and {value}".format(**locals()))
# Filter functionality works as expected
# [...]
def cell_context(self, event):
__cMenu=tk.Menu(self, tearoff=0)
if self.__recordset_filtered:
__cMenu.add_command(label="Show all", command=lambda: filter_records(0, ""))
else:
column=2
options=["INFO", "WARNING", "DEBUG"]
for i in range(len(options)):
option=options[i]
__cMenu.add_command(label="{}".format(option), command=lambda: self.filter_records(column, option))
# Also tried using for option in options here with same result as now
__cMenu.post(event.x_root, event.y_root)
if __name__=="__main__":
root=tk.Tk()
app=SomeClass(root)
root.mainloop()
The current output i get is:
Filter Log Recordset by 2 and DEBUG
No matter which of the three options i choose. I assume it has sth to do with the garbage collection that only the last option remains but i cannot figure out how to avoid this.
Any help is recommended.

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.