Python Traceback (most recent call last) error

Question

I have started coding for about a week, and while practicing to create a shape calculator, I encountered such an error:

Traceback (most recent call last):
File "python", line 4
if option = 'C':
          ^
SyntaxError: invalid syntax

The code is as follows:

print "The Calculator has been launched"
option = raw_input ("What shape is your object?     Enter C for circle or T 
for Triangle.")
if option = 'C': 
    radius = float (raw_input ("What is the radius of your circle?") )
    area_1 = 3.14159 * ( radius ** 2)
    print area_1 

elif option = 'T':
    base = float (raw_input ("What is the base of the triangle?"))
    height = float (raw_input ("What is the corresponding height of the 
    triangle?"))
    area_2 = (base * height) * 1/2
    print area 
else :
    print "Please, enter a valid shape"

I would be very grateful if someone could explain the cause of the error.

Thanks!

`if option == c` the `=` is assignment operator and `==` is for value equality. — ᴀʀᴍᴀɴ, Jun 29 '18 at 13:09
That worked thanks! Yet, now I encountered this problem `Traceback (most recent call last): File "python", line 6 area_1 = 3.14159 * radius**2 ^ IndentationError: unindent does not match any outer indentation level` — Lazimsiz, Jun 29 '18 at 16:02

score 4 · Accepted Answer · answered Jun 29 '18 at 13:09

4

When comparing you must use a ==. The = is only used for assignment.

So in your example the line should be

if option == 'C':

answered Jun 29 '18 at 13:09

ltd9938

1,444
1
15
29

Thanks for the feedback! == seems to have fixed the issue partially, but now I have this error. Any ideas on how it can be fixed? Traceback (most recent call last): File "python", line 6 area_1 = 3.14159 * radius**2 ^ IndentationError: unindent does not match any outer indentation level – Lazimsiz Jun 29 '18 at 16:00
That means the indentation isn't lining up. Double check the indentation for that line. https://stackoverflow.com/questions/14979224/indentation-error-in-python – ltd9938 Jun 29 '18 at 16:02

score 2 · Answer 2 · answered Jun 29 '18 at 13:12

Yes it is actually a very basic error that everyone does in the beginning :)
The = operator does not mean the same thing in code as it does in math. Here it means that you want to assign a value to a variable (you can also think of it as the := operator that you can see in maths or other coding languages).

The operator you need to compare two elements is == which returns a boolean value: either True or False

score 1 · Answer 3 · answered Jun 29 '18 at 13:19

It is worth to mention that this code will be difficult for user to work beacue of nescessity of inputting big letters (Shitf + letter). To avoid that, just use lower() method.

if option.lower() == "c":
   do_something()

Now, user can input both big or small letter ("c" or "C"), and program will be no differ to that. Of course the nesscesity of using "==" in any comparing is necessity.

Anshuman Sharma · Answer 4 · 2018-06-29T13:16:56.407

0

you can also use 'is'

if option is 'C':

Don't use the equality "==" symbol to compare objects to None Use "is" instead

"etc" is None  # => False
None is None  # => True
# negate with not
not True  # => False
not False  # => True

# Equality is ==
1 == 1  # => True
2 == 1  # => False

# Inequality is !=
1 != 1  # => False
2 != 1  # => True

edited Jun 29 '18 at 13:16

answered Jun 29 '18 at 13:11

Anshuman Sharma

57
4

what does this have to do with the ops issue? – L_Church Jun 29 '18 at 13:18
`is` might work in this particular case, but it is bad advice. `==` is the appropriate operator to compare strings. – khelwood Jun 29 '18 at 15:10

Python Traceback (most recent call last) error

4 Answers4