CODEWITHSUNDEEP
pythonpython-3.x
Joshua Boca
Joshua Boca

Reputation: 13

Python 3 Syntax Error

  method = input("Is it currently raining? ")
if method=="Yes" :
  print("You should take the bus.")
else: distance = input("How far in km do you want to travel? ")
if distance == > 2:
    print("You should walk.")
elif distance ==  < 10 :
  print("You should take the bus.")
else: 
  print("You should ride your bike.")

Nvm, i fixed it..for those who have the same problem and were on Grok Learning it was just an indention issue and I forgot to write int...

Upvotes: 1

Views: 1293

Answers (2)

Fredrik
Fredrik

Reputation: 940

So since you added a second question, I'll add a second answer :)

In Python 3, the input() function always returns a string, and you cannot compare strings and integers without converting things first (Python 2 had different semantics here).

>>> distance = input()
10
>>> distance
'10' <- note the quotes here
>>> distance < 10
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: str() < int()

To convert a string to an integer value, use int(string):

>>> distance = int(distance)
>>> distance
10 <- no quotes here
>>> distance < 10
False

(also note that your code snippet above has an indentation issue -- you'll end up on the "if distance < 2" line whether you answer "Yes" or not. To fix this, you have to indent everything that should be in the "else" branch in the same way.)

Upvotes: 2

Fredrik
Fredrik

Reputation: 940

You need to specify what to compare with for every comparison, so

elif distance <=2 and >=10

should be:

elif distance <=2 and distance >=10:

(there are more clever ways to do this, but the above is the quickest fix)

Upvotes: 2

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