Trying to understand while loops but after typing in the input a number that is not in the list, it doesn't keep asking until it is one of the numbers in the list.
number = 12
while number not in [0,1,2,3,4]:
number = input("type a number: ")
print(number)
if number in [0,1,2,3,4]:
print('one of those numbers')
else:
break
print('not one of those numbers')
the above gives this if you type 4343:
type a number: 4343
4343
not one of those numbers
If you step through your code you can see that the "number" variable gets what you typed in, 4343. Then you check to see if number is in the list of 0-4. It's not, so execution proceeds down the "else" branch, which is a break. Break will "break" the execution of your loop, so the next thing printed is "not one of those numbers" and execution stops and you are not asked for another input.
If you were to type a number like "4" what would happen? Well, the same thing I think. That's because your input comes in as a string, not a number, and you are asking if the string "4" is in the list of numbers, and it isn't. You need to parse the input as an int, you can do that by wrapping your input call in the "int" function - int(input("Type a number: ")) and now "number" will be an int, not a string.
number = 12
while number not in [0,1,2,3,4]:
number = int(input("type a number: "))
print(number)
if number in [0,1,2,3,4]:
print('one of those numbers')
break
else:
print('not one of those numbers')
continue
As #inteoryx told that when you give an input it's a default string value. You have to convert it to integer and you used break statement if the number not in the list. break gets you out of the loop you should use continue statement instead. I added break in the if statement that will get you out of the loop if the number is in the list.
Related
Beginner question here (sorry). I am practicing conditional statements comparing integer inputs. Oddly enough, when I put in "5" as my first input, I am told it is larger than "10" (my second input). I must be missing something because I've stared much too long and can't figure it out.
Code:
print("Enter first number")
first = input()
print("Enter second number")
second = input()
if first > second:
print("Greatest is", first)
elif second > first:
print("Greatest is", second)
else:
print("Both are the same!")
Below are my inputs and the result:
Enter first number
5
Enter second number
10
Greatest is 5
Thank you for your help in advance.
Assuming this is Python, input() returns a string and "5" is lexicographically larger than "10".
You can use int(...) to convert a string into an integer:
first = int(input())
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.
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.
If I'm asking for a user input of numbers which continues as long as an empty string is not entered, if an empty string is entered then the program ends.
My current code is:
n=0
while n != "":
n = int(input("Enter a number: "))
But obviously this isn't exactly what I want. I could remove the int input and leave it as a regular input, but this will allow all types of inputs and i just want numbers.
Do i ago about this a different way?
calling int() on an empty string will cause a ValueError so you can encapsulate everything in a try block:
>>> while True:
try:
n = int(input('NUMBER: '))
except ValueError:
print('Not an integer.')
break
NUMBER: 5
NUMBER: 12
NUMBER: 64
NUMBER:
not a number.
this also has the added benefit of catching anything ELSE that isn't an int.
I would suggest using a try/except here instead.
Also, with using a try/except, you can instead change your loop to using a while True. Then you can use break once an invalid input is found.
Also, your solution is not outputting anything either, so you might want to set up a print statement after you get the input.
Here is an example of how you can put all that together and test that an integer is entered only:
while True:
try:
n = int(input("Enter a number: "))
print(n)
except ValueError:
print("You did not enter a number")
break
If you want to go a step further, and handle numbers with decimals as well, you can try to cast to float instead:
while True:
try:
n = float(input("Enter a number: "))
print(n)
except ValueError:
print("You did not enter a number")
break
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)