Detecting if a random integer is odd or even using an if statement












-3















I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.



from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")


I need to know how to do the if statments and to make it detect if it's even or odd.






python python-3.x







share|improve this question




















  • 1





    Possible duplicate of stackoverflow.com/questions/21837208/…

    – sla3k
    Nov 20 '18 at 17:33






  • 2





    Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

    – Dennis Kuypers
    Nov 20 '18 at 17:35


















-3















I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.



from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")


I need to know how to do the if statments and to make it detect if it's even or odd.






python python-3.x







share|improve this question




















  • 1





    Possible duplicate of stackoverflow.com/questions/21837208/…

    – sla3k
    Nov 20 '18 at 17:33






  • 2





    Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

    – Dennis Kuypers
    Nov 20 '18 at 17:35
















-3












-3








-3








I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.



from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")


I need to know how to do the if statments and to make it detect if it's even or odd.






python python-3.x







share|improve this question
















I am trying to make a dice game and I want to make it so it will detect if the number on the dice is odd or even. This is what I tried and it didn't work.



from random import randrange



import random



import time



Even = [2, 4, 6]



Odd = [1, 3, 5]



x = random.randint(1, 6)



print("Rolling Dice...")



print("Your number is...." + str(x))



if str(x) == Even:



  print("It is an Even number!")



if str(x) == Odd:



  print("It is an Odd number!")


I need to know how to do the if statments and to make it detect if it's even or odd.






python python-3.x




python python-3.x






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 20 '18 at 17:32









jonrsharpe

78k11106215




78k11106215










asked Nov 20 '18 at 17:25









RichardTRichardT

31




31








  • 1





    Possible duplicate of stackoverflow.com/questions/21837208/…

    – sla3k
    Nov 20 '18 at 17:33






  • 2





    Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

    – Dennis Kuypers
    Nov 20 '18 at 17:35
















  • 1





    Possible duplicate of stackoverflow.com/questions/21837208/…

    – sla3k
    Nov 20 '18 at 17:33






  • 2





    Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

    – Dennis Kuypers
    Nov 20 '18 at 17:35










1




1





Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33





Possible duplicate of stackoverflow.com/questions/21837208/…

– sla3k
Nov 20 '18 at 17:33




2




2





Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35







Just use a modulo 2 if something % 2 == 0 # this is even and check for the remainder. Odd numbers will have the remainder 1 and even numbers zero.

– Dennis Kuypers
Nov 20 '18 at 17:35














2 Answers
2






active

oldest

votes


















1














For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:



from random import randrange



import random



import time



x = random.randint(1, 6)



print("Rolling Dice...")



time.sleep(2)



print("Your number is....{}".format(x))



if x % 2 == 0:



  print("It is an Even number!")



else:



  print("It is an Odd number!")





share|improve this answer


























  • This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

    – G. Anderson
    Nov 20 '18 at 17:45








  • 1





    Yes ofcourse, @G.Anderson, I have edited the answer.

    – Preetkaran Singh
    Nov 20 '18 at 17:49



















0














if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")






share|improve this answer























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1














    For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:



    from random import randrange
    

    
import random
    

    
import time
    

    
x = random.randint(1, 6)
    

    
print("Rolling Dice...")
    

    
time.sleep(2)
    

    
print("Your number is....{}".format(x))
    

    
if x % 2 == 0:
    

    
  print("It is an Even number!")
    

    
else:
    

    
  print("It is an Odd number!")





    share|improve this answer


























    • This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

      – G. Anderson
      Nov 20 '18 at 17:45








    • 1





      Yes ofcourse, @G.Anderson, I have edited the answer.

      – Preetkaran Singh
      Nov 20 '18 at 17:49
















    1














    For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:



    from random import randrange
    

    
import random
    

    
import time
    

    
x = random.randint(1, 6)
    

    
print("Rolling Dice...")
    

    
time.sleep(2)
    

    
print("Your number is....{}".format(x))
    

    
if x % 2 == 0:
    

    
  print("It is an Even number!")
    

    
else:
    

    
  print("It is an Odd number!")





    share|improve this answer


























    • This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

      – G. Anderson
      Nov 20 '18 at 17:45








    • 1





      Yes ofcourse, @G.Anderson, I have edited the answer.

      – Preetkaran Singh
      Nov 20 '18 at 17:49














    1












    1








    1







    For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:



    from random import randrange
    

    
import random
    

    
import time
    

    
x = random.randint(1, 6)
    

    
print("Rolling Dice...")
    

    
time.sleep(2)
    

    
print("Your number is....{}".format(x))
    

    
if x % 2 == 0:
    

    
  print("It is an Even number!")
    

    
else:
    

    
  print("It is an Odd number!")





    share|improve this answer















    For checking number being odd or even, you may not need the Even, Odd lists, you might check out the following code:



    from random import randrange
    

    
import random
    

    
import time
    

    
x = random.randint(1, 6)
    

    
print("Rolling Dice...")
    

    
time.sleep(2)
    

    
print("Your number is....{}".format(x))
    

    
if x % 2 == 0:
    

    
  print("It is an Even number!")
    

    
else:
    

    
  print("It is an Odd number!")






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 20 '18 at 17:49

























    answered Nov 20 '18 at 17:40









    Preetkaran SinghPreetkaran Singh

    114111




    114111













    • This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

      – G. Anderson
      Nov 20 '18 at 17:45








    • 1





      Yes ofcourse, @G.Anderson, I have edited the answer.

      – Preetkaran Singh
      Nov 20 '18 at 17:49



















    • This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

      – G. Anderson
      Nov 20 '18 at 17:45








    • 1





      Yes ofcourse, @G.Anderson, I have edited the answer.

      – Preetkaran Singh
      Nov 20 '18 at 17:49

















    This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

    – G. Anderson
    Nov 20 '18 at 17:45







    This could be further simplified with the second condition being else: instead of another if, but it's definitely more pythonic than what's already there

    – G. Anderson
    Nov 20 '18 at 17:45






    1




    1





    Yes ofcourse, @G.Anderson, I have edited the answer.

    – Preetkaran Singh
    Nov 20 '18 at 17:49





    Yes ofcourse, @G.Anderson, I have edited the answer.

    – Preetkaran Singh
    Nov 20 '18 at 17:49













    0














    if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")






    share|improve this answer




























      0














      if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")






      share|improve this answer


























        0












        0








        0







        if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")






        share|improve this answer













        if x % 2 == 0: print("It is an Even number!") else: print("It is an Odd number!")







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 20 '18 at 17:58









        DharmeshDharmesh

        8501212




        8501212






























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