python - Unorderable types in "if" predicate -

def determinelettergrade():     global lettergrade, testavg;     if (testavg <= str(100)) , (testavg >= str(90)):         lettergrade = "a";     else:         if (testavg <= 89.99) , (testavg >= 87):             lettergrade = "b+";         else:             if (testavg <= 86.99) , (testavg >= 80):                 lettergrade = "b";             else:                 if (testavg <= 79.99) , (testavg >= 77):                     lettergrade = "c+";                 else:                     if (testavg <= 76.99) , (testavg >= 70):                         lettergrade = "c";                     else:                         if (testavg <= 69.99) , (testavg >= 67):                             lettergrade = "d+";                         else:                             if (testavg <= 66.99) , (testavg >= 60):                                 lettergrade = "d";                             else:                                 if (testavg <= 59.99) , (testavg >= 0):                                     lettergrade = "f"     # end determinelettergrade function 

midterm project, creating functions, function determinelettergrade. using if statements determine letter grade test average. when go run it, error message:

if (testavg <= str(100)) , (testavg >= str(90)): typeerror: unorderable types: function() <= str() 

the error says on tin: interpreter can't figure out larger when hand function , string. means testavg name of function elsewhere in code. need call it, appropriate arguments, , have return proper value. call once, , save local variable average_score = testavg() (pass arguments needed). can make comparison... shouldn't casting other values strings. leave them integers.

also, should restructure code replace repeated else..if blocks structure uses elif keyword.