Question

I know theres something wrong with the syntax here...
print ("this program will check if a number is even")
Number1 = input("please enter a number")
if Number1 % 2 == 0:
print ("the number is even!")
else
print("the number is odd!")

but im not sure what.

Answer 1

I see u have print as a function, what version of Python u have?

Answer 2

ermm 3.3?

Answer 3

-?

Answer 4

U will get problems with MOOC if u use 3 instead of 2 (are you on MOOC?)

Answer 5

yeah I am. Okay i will switch to 2. Is there anything wrong with my if statement?

Answer 6

AFAICS, 2.7.3 seems to work fine..

Let me try your code....

Answer 7

Is Number a class name?

Answer 8

i guess it would be print "this program will check if a number is even"
Number1 = raw_input("please enter a number")
if Number1 % 2 == 0:
print "the number is even!"
else
print"the number is odd!"
for python 2

Answer 9

Number might be, but im using Number1 so i didnt think it would effect it

Answer 10

what about the "else" does it need a : or anything?

Answer 11

I don't know that much about 3, seems that it migh be a problem to start with a capital if that is for Class

Answer 12

U should have else:

I am just trying to get it to run....

Answer 13

Put it in here http://www.codeskulptor.org/

Answer 14

Okay i have, what is this?

Answer 15

hmmm, seems to work in there

Answer 16

else has to be indented left....

Answer 17

Now I am running as .py file in cmd.exe

Answer 18

so not underneath the print, in the same column as if

Answer 19

After I enter a number

it complains about type error


please enter a number 5
Traceback (most recent call last):
File "F:\!Work\!Coursera\Week1\wolfy.py", line 2, in <module>
if Number1 % 2 == 0:
TypeError: not all arguments converted during string formatting

Let me try it without "Number"

Answer 20

You use raw_input which converts input to string, you should then convert the input to int with int(input) or directly use input() instead of raw_input

Answer 21

I tried aNumber1 and get "not defined"

Answer 22

I suggest this :

Number1 = input("please enter a number")
if Number1 % 2 == 0:
print "the number is even!"
else:
print"the number is odd!"

Answer 23

@FabienToune Yes....

Answer 24

okay thanks :)

Answer 25

probably be about 2 minutes until im stuck again :)

Answer 26

:-)

Answer 27

The skulptor is what I am using in the Coursera Python course....

Answer 28

why not idle?

Answer 29

It has a stripped down simple GUI

Answer 30

for simple game programming

Answer 31

Everyone is using same thing in the cloud, stops all the problems with version numbers and so on....

Answer 32

oh okay, cool

Answer 33

Their course started exact same time as MOOC so I thought I will try to do both at same time...

Answer 34

But discussion here is better, they have only standard type of forum...

Answer 35

which course?

Answer 36

https://class.coursera.org/interactivepython-2012-001/class/index

(maybe u have to register I think)

Answer 37

Back in an hour or so...

Answer 38

i will have a look. With the mooc assignments, we dont actually send them to anyone do we?

Answer 39

There is a timetable of sorts but it really doesn't matter if you fall behind a bit.

The other is quite strict (I am not sure I will keep to it in the end)

Answer 40

The indentation level of the else function is not the same of the if function
place else outside. It will work. Like this:

Number1 = input("please enter a number")
if Number1 % 2 == 0:
print ("the number is even!")
else:
print("the number is odd!")

Answer 41

There are two problems here that have already been mentioned separately. First, the 'else' line needs to be at the same indentation level as the 'if' line. Second, the 'else' line should read 'else:' the colon at the end of the line is a very important and forgettable bit of syntax in python.

From a purely stylistic standpoint, variable identifiers usually dont begin with a capital letter. Additionally the input begins immediately after the string that you provide for the prompt. something along the lines of 'number1 = input("please enter a number: ")' would be a bit cleaner.