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.
Related
im trying to code a script that calculates how many minutes there are in x days and prints out a anwser, if the user input is a negative number, a string or a 0 it should print a error messege.
Here is how far i have come
# global var
user_input = input('Click here and type how many days you want to know the minutes of\nthen press enter...\n')
# def for calculating minutes in x days
def calc_minutes(user_input):
if user_input.isdigit() > 0:
user_input_int = int(user_input)
minutes_a_day = 1440
calc_anwser = minutes_a_day * user_input_int
return f'there are {calc_anwser} minutes in {user_input} days'
else:
if user_input == int():
return 'ERROR: type something else then a string'
if user_input == 0:
return 'type a number above 0'
if user_input < 0:
return ' type a positive number'
print(calc_minutes(user_input))
i cant figure it out, if i type a positive number it works, it calculates how many minutes there are in the days u typed and prints it out.
but if you type 0 as input it still runs the calculation instead of giving the error messege.
and if you type a string or negative number it says the code isnt right and displays the red error crash messege(not the error messege i typed to be displayed)
could anyone explain why it cant use the code ive written for the negative numbers, strings and 0?
user_input.isdigit() just tells you if the input only contains digits. It returns a boolean, not the value of the number. So you can't compare it to tell if the number is more than 0.
What's actually happening is that the boolean value True or False is converted to a number: True == 1 and False == 0. That's being compared with 0.
The best way to do this is to call int() to try to convert the input to an integer, and use try/except to detect the error if it can't be converted. If it converts successfully, check there to see if it's positive.
def calc_minutes(user_input):
try:
user_input_int = int(user_input)
if user_input_int <= 0:
return 'ERROR: type a positive number'
minutes_a_day = 1440
calc_anwser = minutes_a_day * user_input_int
return f'there are {calc_anwser} minutes in {user_input} days'
except ValueError:
return 'ERROR: type an integer'
In the first if operator you don't check whether the number is positive, but whether the string you got is a number.
Possible solution:
First, convert the input string to integer type
user_input = int(input('...'))
Then, check if it's non-negative:
if user_input > 0:
I am working on a program that requires the user to enter a number and will continue to loop until a positive number is given. When a positive number is given, it will alert the user and present them with the sum of the digits of their number. However, I thought I had written my code correctly, but it is giving me an incorrect answer. What have I done wrong and how can I fix this?
user_input = float(int(input("Please Enter Your Number:")))
s = 0
while user_input < 0:
float(int(input("Please Enter Another Number: ")))
if user_input > 0:
s += user_input%10
user_input //= 10
print("You've entered a positive number! The sum of the digits is: ", s)
Four things:
Not sure why you storing the input as float, int should suffice.
If you give a negative input, it will enter the while loop. However, in the while loop, you are not actually assigning the new input to user_input. Fix this by adding user_input =
The while loop guarantees user_input is >= 0, so if user_input > 0: is unnecessary.
Probably the most important, to calculate the sum of digits, you need to repeatedly divide and sum, not just do it once. So, add a while loop.
Final code:
user_input = int(input("Please Enter Your Number: "))
s = 0
while user_input < 0:
user_input = int(input("Please Enter Another Number: "))
while user_input:
s += user_input % 10
user_input //= 10
print("You've entered a positive number! The sum of the digits is: ", s)
The if statement is generally used to decide if something should be done once.
If you want to keep going until user_input becomes zero, you'll need a while.
Also, I'm not entirely certain why you're storing the number as a float, especially when you make that from an int anyway. It may as well just be an int.
Additionally, you're loop to re-enter the value if it was negative doesn't actually assign the new value to the variable.
And you probably also want to outdent the print statement lest it be done on every iteration of the loop you're about to add.
Of course, some may suggest a more Pythonic way of summing the digits of a positive number is a simple:
sum([int(ch) for ch in str(x)])
That works just as well, without having to worry about explicit loops.
Another way to solve this is using assert and a function:
def sum_num():
# try get user input
try:
user_in = input('Enter Number: ')
assert int(user_in) > 0
except AssertionError:
# we got invalid input
sum_num()
else:
s_d = sum([int(i) for i in user_in])
print('You\'ve entered a positive number! The sum of the digits is: ', s_d)
#run the function
sum_num()
So this will asked user input, if it is not greater than zero it will throw assertion error, which we catch and return the user to inputting the number by calling the function again. If all is well, we split the input into character and add them up. as list('12') gives ['1','2']. We convert to int and add them. :)
The awesome thing about this is you can add more too the asset to capture other issue as floats, character as invalid inputs. E.g.
Assuming literal_eval is important( from ast import literal_eval)
assert isinstance(literal_eval(user_in),int) and int(user_in)>0
Check if user_in is integer and it is greater than 0. So you won’t get issues when user inputs floats or characters.
I am having issues regarding python. My issue is that my code always prints the first if statements return even if the user inputs a valid integer. I genuinely do not know the issue. I have to allow the user to input a value that is restricted to 1-10 and I have to be able to output either, "Please input an integer between 1-10" or "That's not even a number silly!". Could someone please help me and do more than just write the code, but explain it too?
value = input("Please input an integer: ")
def numValue(value):
for value in range(-1000, 1000):
if (value < 1, value > 10):
return "Please input a number between 1-10."
elif (value >= 1, value <= 10):
return "Great choice!"
print (numValue(value))
First you need to tell python that you are expecting an integer. This is done by wrapping int(..) around input
You have an unnecessary for loop with range. It is not needed
The syntax to check if a number is between a range is start < num < end or (<=) if you want the range to be inclusive
You don't to add the inverse check in elif, using else will automatically fallback if original condition fails
value = int(input("Please input an integer: "))
def numValue(value):
if 1 <= value <= 10:
return "Great choice!"
else:
return "Please input a number between 1-10."
print (numValue(value))
Note: This is not my homework. Python is not taught in my college so i am doing it my myself from MITOCW.
So far i have covered while loop, input & print
Q) Write a program that asks the to input 10 integers, and then prints the largest odd number that was entered. If no odd number was entered it should print a message to the effect
How can i compare those 10 number without storing them in some list or something else? Coz i haven't covered that as if yet.
print "Enter 10 numbers: "
countingnumber=10
while countingnumber<=10:
number=raw_input():
if number % 2 == 0:
print "This is odd"
countingnumber=countingnumber+1
else:
print "This is even. Enter the odd number again"
i think the program will look something like this. But this has some unknown error & How can i compare all the numbers to get the largest odd number without storing those 10 numbers in the list.
print "Enter 10 numbers: "
countingNumber = 1
maxNumber = 0
while countingNumber<=10:
number=int(raw_input())
if (number % 2 == 0):
countingNumber = countingNumber+1
if (maxNumber < number):
maxNumber = number
else:
print "This is even. Enter the odd number again"
print "The max odd number is:", maxNumber
you can just define a maxnum variable and save the max in it! also you must use int(raw_input()) instead raw_input()
print "Enter 10 numbers: "
maxnum=0
for i in range(10):
number=int(raw_input())
if number%2 == 0:
print "This is odd"
if number>maxnum:
maxnum=number
else:
print "This is even. Enter the odd number again"
print "max odd is :{0}".format(maxnum)
DEMO:
Enter 10 numbers:
2
This is odd
4
This is odd
6
This is odd
8
This is odd
12
This is odd
14
This is odd
16
This is odd
100
This is odd
2
This is odd
4
This is odd
max odd is :100
Whenever I do input, I like to make sure I don't leave room for human error giving me bugs.
Because I put in extra checks I break code into a lot of separate function. This also gives code the quality of being non-coupled. ie) You can reuse it in other programs!!
def input_number():
while true:
input = raw_input("Enter Value: ")
if not input.isdigit():
print("Please enter numbers only!")
else:
return int(input)
Designing the input function in this fashion gives the code no opportunity to crash. We can now use it in a function to get odd numbers!
def input_odd_number():
while true:
input = input_number()
if input % 2 == 0:
print("Please enter odd numbers only!")
else:
return input
Now we can finally move onto the main code. We know we need ten numbers so lets make a for loop. We also now we need to hold onto the largest odd number, so lets make a variable to hold that value
def largest_odd(count = 10): // its always nice to make variables dynamic. The default is 10, but you can change it when calling!
max_odd = input_odd_number() // get the first odd number
for i in range(count - 1): // we already found the first one, so we need 1 less
new_odd = input_odd_number()
if new_odd > max_odd:
max_odd = new_odd
print("The largest odd value in '{}' inputs was: {}".format(count, max_odd)
In your solution are multiple flaws.
A syntax error: The colon in number=raw_input():.
raw_input returns a string and you have to cast it to an int.
Your while loop just runs one time, because you start with 10 and compare 10 <= 10. On the next iteration it will be 11 <= 10 and finishes.
Also you have mixed odd an even. even_number % 2 gives 0 and odd_number % 2 gives 1.
To get the biggest value you only need a additional variable to store it (See biggest_number in my solution). Just test if this variable is smaller then the entered.
You ask again if the number is odd, but you should take every number and test only against odd numbers.
A working solution is:
print "Enter 10 numbers"
count = 0
max_numbers = 10
biggest_number = None
while count < max_numbers:
number=int(raw_input("Enter number {0}/{1}: ".format(count + 1, max_numbers)))
if number % 2 == 1:
if biggest_number is None or number > biggest_number:
biggest_number = number
count += 1
if biggest_number is None:
print "You don't entered a odd number"
else:
print "The biggest odd number is {0}".format(biggest_number)
If you wonder what the format is doing after the string take a look in the docs. In short: It replaces {0} with the first statement in format, {1} with the second and so on.
here is the correct code for that:
print "Enter 10 numbers: "
countingnumber=1
MAX=-1
while countingnumber<=10:
number=int(raw_input())
if number%2==1:
if number>MAX:
MAX=number
if MAX==-1:
print "There Weren't Any Odd Numbers"
else:
print MAX
here are some notes about your errors:
1- you should cast the raw input into integer using int() function and the column after calling a function is not needed and therefor a syntax error
2- your while loop only iterates once because you initial counting number is 10 and after one iteration it would be bigger than 10 and the while body will be skipped.
3-an even number is a number that has no reminder when divided by 2 but you wrote it exactly opposite.
4- you don't need to print anything in the while loop, you should either print the biggest odd number or print "There Weren't Any Odd Numbers".
5- an additional variable is needed for saving the maximum odd number which is MAX.
Last note: in the code provided above you can combine the two ifs in the while loop in to one loop using 'and' but since you said you are a beginner I wrote it that way.
I was wondering how I would sum up the numbers they input for n even though it's an input and not int. I am trying to average out all the numbers they input.
n=print("Enter as many numbers you want, one at the time, enter stop to quit. ")
a=0
while n!="stop":
n=input("Start now ")
a+=1
In Python 2.x input() evaluates the input as python code, so in one sense it will return something that you can accept as an integer to start with, but may also cause an error if the user entered something invalid. raw_input() will take the input and return it as a string -> evaluate this as an int and add them together.
http://en.wikibooks.org/wiki/Python_Programming/Input_and_Output
You would be better off using a list to store the numbers. The below code is intended for Python v3.x so if you want to use it in Python v2.x, just replace input with raw_input.
print("Enter as many numbers you want, one at the time, enter stop to quit. ")
num = input("Enter number ").lower()
all_nums = list()
while num != "stop":
try:
all_nums.append(int(num))
except:
if num != "stop":
print("Please enter stop to quit")
num = input("Enter number ").lower()
print("Sum of all entered numbers is", sum(all_nums))
print("Avg of all entered numbers is", sum(all_nums)/len(all_nums))
sum & len are built-in methods on lists & do exactly as their name says. The str.lower() method converts a string to lower-case completely.
Here is one possibility. The point of the try-except block is to make this less breakable. The point of if n != "stop" is to not display the error message if the user entered "stop" (which cannot be cast as an int)
n=print("Enter as many numbers you want, one at the time, enter stop to quit. ")
a=0
while n!="stop":
n=input("Enter a number: ")
try:
a+=int(n)
except:
if n != "stop":
print("I can't make that an integer!")
print("Your sum is", a)