Q:

Python tkinter label cannot be called

I am making a simple tkinter GUI with Python and I need to have about 10 or more labels to place.  I place the labels using the grid method but after a certain amount it comes up with an error saying "TypeError: Label object is not callable".  Here is the code for the labels and the grid placing:
UnitPointLabel = ttk. Label(root, text = unit_points)
UnitLabel = ttk. Label(root, text = unit)
HQ1 = ttk. Label(root, text = "HQ:")
HQ2 = ttk. Label(root, text = "HQ:")
Elite1 = ttk. Label(root, text = "Elite:")
Elite2 = ttk. Label(root, text = "Elite:")
Elite3 = ttk. Label(root, text = "Elite:")
Troop1 = ttk. Label(root, text = "Troop:")
Troop2 = ttk. Label(root, text = "Troop:")
Troop3 = ttk. Label(root, text = "Troop:")
Troop4 = ttk. Label(root, text = "Troop:")
Troop5 = ttk. Label(root, text = "Troop:")
Troop6 = ttk. Label(root, text = "Troop:")
Fast_Attack1 = ttk. Label(root, text = "Fast Attack:")
Fast_Attack2 = ttk. Label(root, text = "Fast Attack:")
Heavy_Support1 = ttk. Label(root, text = "Heavy Support:")
Heavy_Support2 = ttk. Label(root, text = "Heavy Support:")
Heavy_Support3 = ttk. Label(root, text = "Heavy Support:")

UnitPointLabel.grid(row = 3, column = 7)
HQ1.grid(row = 3, column = 5)
HQ2.grid(row = 4, column = 5)
Troop1.grid(row = 5, column = 5)
Troop2.grid(row = 6, column = 5)
Troop3.grid(row = 7, column = 5)
Troop4.grid(row = 8, column = 5)
Troop5.grid(row = 9, column = 5)
Troop6.grid(row = 10, column = 5)
Fast_Attack1.grid(row = 11, column = 5)
Fast_Attack2.grid(row = 12, column = 5)

Stops working here: Heavy_Support1(row = 3, column = 6)
Heavy_Support2(row = 2, column = 6)
Heavy_Support3(row = 3, column = 6)

As you can see I tried moving the rows and columns around but its still not working.  Anyone know why this is happening?  

A:

First rule of debugging: assume the error message is telling the truth.  What does it say?  It says "Label object is not callable".  That means you have some object that is an instance of a Label object, and you are trying to call it.  For example, "foo=Label(...); ...; foo()". 
So, looking at your code, what label objects are you trying to call?  Take a look at these three lines:
Heavy_Support1(row = 3, column = 6)
Heavy_Support2(row = 2, column = 6)
Heavy_Support3(row = 3, column = 6)

Looking back up through your code I see that Heavy_Support1 (et al.)  are all instances of Label, and you are indeed trying to call it.  
To make a long story short, it looks like you forgot to add .grid on those lines.  They should be:
Heavy_Support1.grid(row = 3, column = 6)
Heavy_Support2.grid(row = 2, column = 6)
Heavy_Support3.grid(row = 3, column = 6)

