Is there a way to simplify this code?

Question

s = input('How do you like, placeholder, on a scale from 1-10?')
print(s)
if s=="1":
print('custom entry here')
if s=="2":
print('custom entry here')
if s=="3":
print('custom entry here')
if s=="4":
print('custom entry here')
if s=="5":
print('custom entry here')
if s=="6":
print('custom entry here')
if s=="7":
print('custom entry here')
if s=="8":
print('custom entry here')
if s=="9":
print('custom entry here')
if s=="10":
print('custom entry here')
else:
print('That is not a valid entry')

Answer 1

responses=(
'custom 1',
'custom 2', # etc
)

s = int(input('How do you like, placeholder, on a scale from 1-10?'))
try:
print("{}\n{}".format(s, responses[s-1]))
except IndexError:
print('That is not a valid entry')

Answer 2

dont python but if it were c or vb i would use a switch/case statement.. but i hear tell python doesnt have one..

I found this. it might be of help.

http://blog.simonwillison.net/post/57956755106/switch

oh and this:
http://stackoverflow.com/questions/60208/replacements-for-switch-statement-in-python

Answer 3

Hi Kolby,
yes, there is!

s = input('How do you like, placeholder, on a scale from 1-10?')
print(s)

if int(s) in range(1,11):
print('custom entry here')

if s!="10":
print('That is not a valid entry')

These lines produce the same output, as your code does.

However I am not sure, your code does what you expect it to do.
I guess, you want to print “...custom...“ if the input is an integer between 1 and 10. And if it is not, you want to print “...not valid...“, right?
If you replace all 'if' by 'elif' in your code, except for the first one, your code will work this way.

Also for this, there will be a shorter version, similar to the one I posted :-)

Best and have fun! nmc

Answer 4

If messages are different for each scale, you can use something like this:

scale = input('Please enter a scale from 1-10?')
print(scale)

scale_messages = {
'1': 'custom message 1',
'2': 'custom message 2',
'3': 'custom message 3',
'4': 'custom message 4',
'5': 'custom message 5',
'6': 'custom message 6',
'7': 'custom message 7',
'8': 'custom message 8',
'9': 'custom message 9',
'10': 'custom message 10',
}

message = scale_messages.get(
scale, 'That is not a valid scale')

print(message)

Answer 5

Thank you

Answer 6

So I changed the code to:s = input('How do you like, placeholder, on a scale from 1-10? ')
print(s)
if int(s)in range(1,11):
print('Custom entey here')
if s > 10:
print('That is not a valid entry')
I thought that a > would be more appropriate than != because in the case you wrote in it only works for the specified number. But now I get a Type error: unorderable types, str() > int()

Answer 7

Mayur Chaudhari almost gave you the implementation of what I suggested.
However, there are still three errors in this code snippet.

Answer 8

s = input("How do you like,placeholder,on a scale from 1-10");
for s in range(1,10):
print("custom entry here")
else:
print("Invalid Entry")

Answer 9

Hi Kolby,
you are on the correct way with your revised code.
You got this error because your variable s is still a string (which you can't compare with an integer).
You can handle this in two ways:
1. cast the string s into an integer, as in the first if-statement.
or
2. replace your second if-statement by using else. Doing so, you will also detect invalid user input values that are less than 1, such as 0.

Option two is the 'cleanest' way.

Answer 10

Thank you the code now works

Answer 11

its easy by using loop here