I'm trying to make a program in Python 3.5 that asks the user to enter a number from 1 to 9. The the program will also guess a number from 1 to 9. If the user guesses the same number correctly, then the program will print Correct . Otherwise, the program will print Wrong. I wrote a program. Everything went fine with my code. However, when I guessed a number correctly, the program suddenly wrote Wrong instead of Correct. What am I doing wrong?
Here is my code:
print('Enter a number from 1 to 9')
x = int(input('Enter a number: '))
import random
random = print(random.randint(1,9))
if int(x) != random:
print ('Wrong')
else:
print ('Correct')
You are saving the result of a print() call (and masking random). print() returns None, so it will always be unequal to an integer, and therefore always "wrong."
import random
print('Enter a number from 1 to 9')
x = int(input('Enter a number: '))
r = random.randint(1,9)
if x != r:
print('Wrong')
else:
print('Correct')
Also note that I've moved the import statement to the top, avoided a second int() cast on something you've already done that to, and removed the space between the print reference and its arguments.
Here is the mistake,
random = print(random.randint(1,9))
You need to do something like this.
random = random.randint(1,9)
print(random)
Also, you have already converted the input to int so, you can do just this.
if x != random:
As pointed out your mistake is the line
random = print(random.randint(1,9))
But why?
functions (like print() take something, do something (with it) and give something back.
Example:
sum(3,4) takes 3 and 4, may add them and returns 7.
print("Hello World") on the other hand takes "Hello world", prints it on the screen but has nothing useful to give back, and therefore returns None (Pythons way to say "nothing").
You then assign None to the name random and test if it equals your number, which it (of course) doesn't.
Related
So I basically wanna compare "Number" and "Guess" in the if statement, but no matter what it says they don't match (I get the else response, not included here). Even if I copy the random number they don't match.
Thanks in advance!
import time
def the_start():
points = 0
attempt = 1
print("Attempt:",attempt)
print("Your goal is to guess a number between 1 and 10 - Points:",points)
time.sleep(2)
attempt = attempt + 1
number = random.randint(0,10)
print(number)
guess = input("What is your guess? :")
time.sleep(2)
if guess == number:
points = points + 1
print("OMG YOU WERE RIGHT! Here, have some fake cheers! *cheer*")
time.sleep(5)
guess is a string. You need to do conversion of the string and handle error conditions. int() will convert a string to an integer, but it will throw an exception if the string is not purely numbers.
import random
print "Welcome to the number guesser program. We will pick a number between 1 and 100 and then ask you to pick a number between 1 and 100. If your number is 10 integers away from the number, you win!"
rand_num = random.randint(1, 100)
user_input = raw_input('Put in your guess:').isdigit()
number = int(user_input)
print number
if abs(rand_num - number) <= 10:
print "Winner"
else:
print "Loser"
Whenever I try to run this code, the computer always generates the number 1. And if I put in a number that is 10 (or less) integers away from one it will still display the else statement. I will have my gratitude to whoever can solve my predicament. I am new to python try to keep your answers simple please.
raw_input returns a string you entered as input. isdigit() checks whether a string is a digit or not and returns True or False.
In your code you're assigning the return value of isdigit to user_input So, you get either True or False. Then you convert it to a number, thus getting either one or zero. This is not what you want.
Let user_input be just a result of raw_input. Then check whether it is a valid number with user_input.isdigit(). If it's not a number, print an error message and exit (or re-ask for input), else convert user_input to an integer and go on.
The problem is product by this sentenc:
user_input = raw_input('Put in your guess:').isdigit().
The return value of isdigit() is True or False. When you enter 1 or any digit number, it will return True(True=1,False=0), so you will always get 1, if you enter digit. You can change like this:
import random
print "Welcome to the number guesser program.
We will pick a number between
1 and 100 and then ask you to pick a number
between 1 and 100. If your number is 10 integers
away from the number, you win!"
rand_num = random.randint(1, 100)
user_input= raw_input('Put in your guess:')
is_digit = user_input.isdigit()
if is_digit==True:
number = int(user_input)
print number
if abs(rand_num - number) <= 10:
print "Winner"
else:
print "Loser"
else:
print 'Your input is not a digit.'
Look at this line:
user_input = raw_input('Put in your guess:').isdigit()
The function raw_input gives you user input, but then you call isdigit and as consequence in your variable user_input you get result of isdigit. So, it has boolean value.
Then in
number = int(user_input) you cast it to integer type. In python true is 1 and false is 0. That's why you always get 1.
My if statement works but else doesn't can anyone help me?
this is my code. Btw if anyone knows how to ask for a retry after one time would be awesome!
import random
print('choose a number between 1 and 10,if you guess right you get 10 points if you guess wrong you lose 15points')
answer = input()
randint = random.randint(0,2)
print('the answer is ',randint)
if [answer == randint]:
print('gratz! you win 10points!')
else:
print('you lose 15points!')
Don't put brackets around your if statement. When you do that, you are creating a new list. Change it to this:
if answer == randint:
You could put parentheses around it if you wanted to, but not []. Your second problem is that random.randint() returns an integer, but input() returns a string (in Python3). You could say if int(answer) == randint: instead, or you could say if answer == str(randint):. Your third problem, as #cricket_007 pointed out is randint(0, 2) will return an integer between 0 and 2, not 1 and 10. Just change that line to randint = random.randint(1, 10).
If you want it to loop after each game, just add these lines around it like so:
import random
loop = true
while loop:
----insert your code here----
This will make it loop endlessly.
Hope this helps!
I am trying to write a simple progrm in Python but when my number is getting equal to number the else statement keeps repeating itself. Help
print('Tell me your name Ramesh')
import random
name=input()
print('Howdy, '+name)
print('Guess my number which is in between 1 to 20')
number=input()
mynumber = random.randint(1, 20)
count = 0
while(mynumber!=number):
if (int(number) > mynumber):
print('Howdy, Your number is too high. Plz renter')
number=input()
count=count+1
elif (int(number)< mynumber):
print('Howdy, Your number is too low. Plz renter')
number=input()
count=count+1
else :
print('Howdy you got it in '+ str(count)+' chances. The correct number is '+number)
I've only just started python myself and had the same issue.
Your if statements all convert the entered number to an int however your while doesn't. so it's comparing an int and string. Try:
while(mynumber != int(number))
or change the input to read
number = int(input("blah")).
I like this more even though it will throw an error if you enter text.
You may also need to change the while to read while(mynumber != int(number)) as well as adding in the break within the ELSE. You could also remove the else and just let the loop fall out if the numbers match and then show the success message.
I have a function below which is part of my big main function, what I want this function to do is that whenever called, I want the function to check if the user input is
a number or not. If it is a number it will return the number and break.
But if it is not a number I want it to loop again and again.when I try to
run it, it gives me unexpected an error:
unexpected eof while parsing
can any body help me what I am missing or how I should rearrange my code? thank you!
def getnumber():
keepgoing==True
while keepgoing:
number = input("input number: ")
result1 = number.isdigit()
if result1 == 1:
return number
break
elif keepgoing==True:
A neater and clearer what to do what you are already doing:
def getnumber():
while True:
number = input("Input number: ")
if number.isdigit():
return number
That's all you need, the extra variables are superfluous and the elif at the end makes no sense. You don't need to check booleans with == True or == 1, you just do if condition:. And nothing happens after return, your break will never be reached.
You don't need the last line:
elif keepgoing==True:
It's waiting for the rest of the file after the :.
Side note, it should be a single = in the first part, and can just be written simpler as well.
def getnumber():
while True:
number = input("input number: ")
result1 = number.isdigit()
if result1:
return number
Since you're inside the while loop, it'll keep executing. Using return will end the while loop, as will breaking and exiting the program. It will wait for input as well each time, though.
While assigning you have used keepgoing == True, I think it should be keepgoing=True
The following solution works on my machine, although I am running Python 2.7
def get_num():
while True: #Loop forever
number_str = raw_input("> Input a number: ") #Get the user input
if number_str.isdigit(): #They entered a number!
return int(number_str) #Cast the string to an integer and return it
I used raw_input rather than input, because raw_input gives you the exact string the user entered, rather than trying to evaluate the text, like input does. (If you pass the 12 to input, you'll get the number 12, but if you pass "12", you'll get the string '12'. And, if you pass my_var, you'll get whatever value was in my_var).
Anyway, you should also know that isdigit() returns whether or not the string has only digits in it and at least one character - that is not the same thing as isNumber(). For instance, "123".isdigit() is True, but "123.0".isdigit() is False. I also simplified your loop logic a bit.