I'm trying to make a small program in Python that generates a random number (from the throw of a die) and asks the user to guess what the number is, and if the user guesses wrong, then depending on whether the guess is higher than the actual number or not the program is supposed to re-prompt them for another guess.
The problem that I'm having with the code is that the output is always "That's a little high. Try again!", even if I input all the possible values.
Can someone please help me figure out which portion of the code is generating the error?
import random
dice = random.randint(1,6)
answer = raw_input ("What do you think is the number?")
while (answer != dice):
if answer > dice:
print ("That's a little high. Try again!\n")
else:
print ("That's a little low. Try again!\n")
answer = raw_input ("What do you think is the number?")
print ("That's correct! Good job. \n")
Thank you!
raw_input returns a string, which is always bigger than a number from 1-6 as its numerical representation is taken in comparison. you need to cast the input to int
answer = int(raw_input ("What do you think is the number?"))
Related
I am trying to make a calculator and the user must input which operation they want to use, I have only done addition so far but you get the point.
welcome = str(input("Welcome to the calculator, which form of operation do you wish to choose? "))
Addition = "Add", "add", "Addition", "addition", "+"
for x in Addition:
if welcome == Addition:
print("Works")
I run the code and when I input any of the strings in the "Addition" variable, it does not print "Works"?.
I have tried without the for loop and it still does not print "Works"
I am an absolute beginner at python and I just made this stackoverflow account today. I don't know how to write the code in proper code form in this question.
You are probably looking for in we can also use .lower() on the input so we have fewer things to match.
welcome = str(input("Welcome to the calculator, which form of operation do you wish to choose? ")).lower()
addition = ['add','addition','+']
if welcome in addition:
print('worked')
I am trying to make a simple guessing game. When I run the code and print the number in advance to check if the program is working, it keeps on having the same wrong response. In other words, even if the guess variable is equal to the num variable, the program still returns "Incorrect!", and I can't figure out why. Thank you in advance. The code is pretty self-explanatory, so I'll post it here.
import random
num = random.randint(1, 6)
print(num)
guess = input(f'Guess a number: ')
if guess == num:
print(f'Correct!')
else:
print(f'Incorrect!')
You are comparing different types. num is an integer number while the guess is a string. You need to convert it to an integer before comparing them.
Try using num == int(guess) instead.
I've been looking up on here and Google yet can't find the solution for what I'm looking for. "While" function keeps failing me as I don't know here I need to put it. I'm a total newbie so if you could kindly explain the solution, thank you:
number = raw_input("Number: ")
if int(number) % 2 == 0:
print "Even number"
else:
print "Odd number"
All I would like to do is to keep the function looping allowing the user to enter as many numbers as possible. It should only break if user puts in "stop" as the value.
Well if user types stop, program stops.So we need to check for that when we create our while loop like:
number_from_user = raw_input("Number: ")
while number_from_user != "stop":
try:
if int(number_from_user)% 2 == 0:
print "Even number"
else:
print "Odd number"
number_from_user = raw_input("Number:")
except ValueError:
print("Enter a number please")
number_from_user = raw_input("Number:")
And I suggest you make yourself familiar with while loops, which I found a video explaining and making an example "Guess the number game" using python. It should help you enough to make you able to solve your own problem with while loops.
https://www.youtube.com/watch?v=tu0zlBFRa_c
or:
https://www.youtube.com/watch?v=PoPxclbw6uc
And I suppose you're using python 2.7, in those videos they're using python 3+, you should type raw_input where they're typing input.
Edited: Added try and except.
I'm trying to learn how to program and I'm running into a problem....
I'm trying to figure out how to make sure someone inputs a number instead of a string. Some related answers I found were confusing and some of the code didn't work for me. I think someone posted the try: function, but it didn't work, so maybe I need to import a library?
Here's what I'm trying right now:
Code:
print "Hi there! Please enter a number :)"
numb = raw_input("> ")
if numb != str()
not_a_string = int(next)
else:
print "i said a number, not a string!!!"
if not_a_string > 1000:
print "You typed in a large number!"
else:
print "You typed in a smaller number!"
Also I have another question while I'm asking. How can I make it so it will accept both uppercase and lower case spellings? In my code below, if I were to type in "Go to the mall" but with a lowercase G it would not run the if statement because it only accepts the capital G.
print "What would you like to do: \n Go to the mall \n Get lunch \n Go to sleep"
answer = raw_input("> ")
if answer == "Go to the mall":
print "Awesome! Let's go!"
elif answer == "Get lunch":
print "Great, let's eat!"
elif answer == "Go to sleep":
print "Time to nap!"
else:
print "Not what I had in mind...."
Thanks. ^^
Edit: I'm also using python 2.7 not 3.0
You can do something like this:
while True: #infinite loop
ipt = raw_input(' Enter a number: ')
try:
ipt = int(ipt)
break #got an integer -- break from this infinite loop.
except ValueError: #uh-oh, didn't get an integer, better try again.
print ("integers are numbers ... didn't you know? Try again ...")
To answer your second question, use the .lower() string method:
if answer.lower() == "this is a lower case string":
#do something
You can make your string comparisons really robust if you want to:
if answer.lower().split() == "this is a lower case string".split():
In this case, you'll even match strings like "ThIs IS A lower Case\tString". To get even more liberal in what you accept, you'd need to use a regular expression.
(and all this code will work just fine on python2.x or 3.x -- I usually enclose my print statements in parenthesis to make it work for either version).
EDIT
This code won't quite work on python3.x -- in python3, you need to change raw_input into input to make it work. (Sorry, forgot about that one).
First,you should ask only one question per post.
Q1: use built-in .isdigit()
if(numb.isdigit()):
#do the digit staff
Q2:you can use string.lower(s) to solve the capital issue.
you may try
numb = numb.strip()
if numb.isdigit() or (numb[0] in ('+', '-') and numb[1:].isdigit():
# process numb
I am working through some Looping exercises, While statements in particular. Here are the instructions:
2.2) Modify the program so that it asks users whether they want to guess again each time. Use two variables, number for the number and answer for the answer to the question whether they want to continue guessing. The program stops if the user guesses the correct number or answers "no". (In other words, the program continues as long as a user has not answered "no" and has not guessed the correct number.)
Here is my code:
#!usr/bin/env python
#
#While statement
number = 24
while number != 24:
answer = raw_input("Guess my lucky number! Do you want to keep guessing?")
if number == 24:
print "You got it! That is great!"
elif answer == "no":
print "Thank you for playing."
else:
print "That is not the right answer! Try again."
When I run the module in IDLE, the end quote of That is great!" - becomes red and says invalid syntax. In terminal if I run $ python while.py nothing loads. I've tried writing this as Python 3 functions with print("") but it still does not run.
Thanks to anyone who can help with this.
The while-cycle is never entered because the condition number != 24 never holds.
This is why there is no output.
Your loop never executes because you state that number = 24, and then right after, your while loop will only start if number != 24.
In addition, raw_input will yield a string, not an int so either ask for "24" or cast the raw_input to an int.
It also seems that you don't actually give the user a chance to guess the number at all; you only ask the user if s/he wants to keep playing.
You might want to do something like this:
number = 24
answer = ""
while answer != str(number):
answer = raw_input("Guess my lucky number, or type 'no' to quit.")
if answer == "no":
print "Okay, see you later"
break
elif answer != str(number):
print "wrong number"
if answer == str(number):
print "you got it right"
Here's the syntax issues:
answer = ""
while answer != "24":
answer = raw_input("Guess my lucky number! Do you want to keep guessing?")
if answer == "24":
# You can fill in the rest ...
Well, I don't want to straight out solve it for you, but take a look at your conditional in the while loop. Think about what happens, line-by-line, when you run it, particularly in the "number" variable.
The other answers are touching on the problem but there are more...
Yes you really should be checking the answer variable in your loop instead of the number variable, but also keep in mind that raw_input is going to give you a string. So you will not get an int 24, you will get a string "24". The idea here is to take the answer variable from raw_input and check that variable against both your number and the value "no"
number = "24"