Syntax error in python; unexpected indent

Question

The aim of this code is to grade the score which is between 0.0 and 1.0. below 0.6 is F, >= 0.6 is D, >= 0.7 is C, >= 0.8 is B and >= 0.9 is A; and I'm getting an error.

inp = raw_input ('Enter score between 0.0 & 1.0 here: ')
if inp == > 0.9:
 print A 
elif inp == > 0.8:
 print B 
elif inp == > 0.7:
 print C
elif inp == >0.6:
 print D 
else inp < 0.6:
 print F

In your code you have '== >' several times : what is that? Do you mean '>=' ? — user4322779
– user4322779, Commented Aug 17, 2015 at 20:58
Yes, you are doing it wrong. Where did you get the idea to do == >? Also, you'll need to change inp into a float and actually print strings like 'A' instead of unknown variables like A. That said, your current code's indentation appears correct. — TigerhawkT3
– TigerhawkT3, Commented Aug 17, 2015 at 21:16
@TigerhawkT3 idk... was doing kinda hit and try on this one... first days of coding here.. about float thing... can you elaborate it..?? Also i too noticed that i had to make A into strings to make it work.. thank you anyways — Sarmad Khalil
– Sarmad Khalil, Commented Aug 17, 2015 at 21:19
As @TigerhawkT3 mentions in a comment in one of the answers, it would be very helpful for yourself if you look through some basic Python documentation and/or tutorials for the appropriate syntax. Official Python tutorial. — vk1011
– vk1011, Commented Aug 17, 2015 at 21:27

chepner · Accepted Answer · 2015-08-17 21:29:30Z

2

inp = float(raw_input('Enter score between 0.0 & 1.0 here: '))  ## This line takes the raw input in from command prompt. float('') casts the raw_input string to a float type number. 
if inp >= 0.9:
    print "A"     
elif inp >= 0.8:
    print "B" 
elif inp >= 0.7:
    print "C"
elif inp >=0.6:
    print "D" 
else:
    print "F"

Rewrite your code as above. You don't need to have a logical statement for "else" and you don't have 2 equal signs for greater than or equals. Also, remember to convert your string input to integer.

Use "input" instead of raw_input if your on python version 3 or higher.

inp = float(input ('Enter score between 0.0 & 1.0 here: '))

edited Aug 17, 2015 at 21:29

chepner

538k77 gold badges594 silver badges746 bronze badges

answered Aug 17, 2015 at 21:06

bud

4857 silver badges22 bronze badges

Sign up to request clarification or add additional context in comments.

8 Comments

Sarmad Khalil Over a year ago

still didnt work. got the error "syntax error: invalid syntax (score.py, line 2)

bud Over a year ago

Sorry. Forgot to switch the => to >= to sign. Try this above.

bud Over a year ago

What version of python are you using?. raw_input is not correct if you are using version 3 or higher. The other thing you could try, is to completely start a new python file.

Sarmad Khalil Over a year ago

oooooooh!!! it worked!!! thanks a billion!!! If u have time, can you please explain the above code...?? especially the first line with float..??

TigerhawkT3 Over a year ago

For more information on basic functions and tasks, I suggest reading the official Python tutorial.

|

MaxTheG · Accepted Answer · 2015-08-17 20:53:21Z

1

Python knows where to end functions using indents/spaces. For example,

if(1 == 1):
    print "I indented here"

The code below causes an error because Python sees nothing is in the if statement

 if(1 == 1):
 print "I indented here"

answered Aug 17, 2015 at 20:53

MaxTheG

579 bronze badges

2 Comments

Sarmad Khalil Over a year ago

sorry but i m getting another error now. if you read the code, please tell me what is wrong. the error says... " syntax error: invalid syntax (score.py, line 2) "

MaxTheG Over a year ago

Could you post your code again or edit the existing code?

Collectives™ on Stack Overflow

Syntax error in python; unexpected indent

2 Answers 2

8 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

8 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related