This question already has answers here:
Limiting user input to a range in Python
(5 answers)
Closed 3 years ago.
I am trying to create a loop where the user is given choices 1-8 and if they do not choose 1-8, it loops them back around re-enter number 1-8. I am attempting to use a while loop with two conditions. What am I missing?
fm_select = int(input("Enter a number 1-8"))
while fm_select <= 8 and fm_select >= 1:
Your ranges are wrong. You want the while loop to fail when they are correct, since you're trying to break out of the loop. So, you want your loop to check against every number that isn't between one and eight. Instead, do
fm_select = 0
while (fm_select < 1 or fm_select > 8):
fm_select = int(input("Enter a number between one and eight: "))
"As long as their input is less than one or higher than eight, keep asking"
something like this should work
while(True):
fm_select = int(input("Enter a number 1-8"))
if 0 < fm_select < 8:
break
print("try again")
print("you have entered %d" %(fm_select) )
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print("Welcome to the number program")
number=input("Please give me a number \n")
number=int(number)
total_number=0
entries=0
while number>0:
total_number=total_number+number
print(total_number)
number=input("Please give me another number! \n")
number=int(number)
entries= int(entries)+1
if number < 0 :
print("Sorry, this value needs to be positive. Please enter a
different number.")
if number == -999:
print(total_number)
print(entries)
print(total_number/entries)
I'm in a beginners programming class, and the book is not very helpful at times. I'm trying to write a basic program that takes positive numbers, totals them, and averages them out at the end. Also rejects negative numbers, and asks if -999 is entered I print the average of all entries, amount of entries, and the value tally. Any advice or tips I can learn from to improve it would be helpful. Thanks!
The program runs ok, it just doesn't write out some things I wanted
From what you wrote and the comments in your code I am guessing that you want the program to continue running and asking for input if you enter a non-positive number. In that case I would rewrite it as:
print("Welcome to the number program")
total_number = 0
entries = 0
while True:
number = input("Please give me a number \n")
number = int(number)
if number == -999:
break
if number <= 0:
print("Sorry, this value needs to be positive. Please enter a different number.")
continue
total_number = total_number + number
entries += 1
print(total_number)
print(total_number)
print(entries)
print(total_number / entries)
Also, you can increment numbers with entries += 1
In most cases you should NOT create variables first, however this case you should. Create number = 0, tally = 0 and total_number = 0 first
Accept your first number inside your while loop and handle all of the logic in there as well.
Your while loop should continue to loop until the final condition is met which seems to be number == -999
Should tally be incremented if you enter a negative number? I assume not. What about a 0? Wrap the increment for tally and the addition to total_number in an if number > -1: condition. Use an if else to check for number == -999, and an else for handling invalid entries.
Finally, move your print statements outside of your while loop. It also doesn't need a condition around it because now, if you've exited your while loop, that condition has been satisfied.
Final note here, and this is just a nice to know/have and purely syntactic sugar, MOST languages support abbreviated incrementing. Theres a better word for it, but the gist is simply this.
total_number += number
# is exactly the same as
total_number = total_number + number
# but way nicer to read and write :)
print("Welcome to the number program")
number = 0
total_number = 0
entries = 0
while number != -999:
number = input("Please enter a number! \n")
number = int(number)
if number >= 0
total_number += number
entries += 1
print("Current sum: " + total_number)
elif number == -999:
break
else
print("Sorry, this value needs to be positive.")
print("Sum of entries: "+str(total_number))
print("Number of entries: " + str(entries))
print("Average entry: " +str(total_number/entries))
I have rewritten your code. But I am not sure what the goal was. Anyways, if the value were ever to be under 0 the loop would have been exited and a new value would have never been accepted from an input.
Also some things I have written more elegant.
print("Welcome to the number program")
number=int(input("Please give me a number \n"))
total_number=0
entries=0
while number > 0:
total_number += number
print(total_number)
number = int(input("Please give me another number! \n"))
entries += 1
if number == -999:
print(total_number)
print(entries)
print(total_number/entries)
break
elif number < 0:
number = input("Sorry, this value needs to be positive. Please enter a different number!")
This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 12 months ago.
I'm new to python and I am trying to figure out how do I enter an error message if the user inputs a number less than zero.
Here is my code to give you a better understanding and I thank you all in advance for any advice.
# Python Program printing a square in star patterns.
length = int(input("Enter the side of the square : "))
for k in range(length):
for s in range(length):
if(k == 0 or k == length - 1 or s == 0 or s == length - 1):
print('*', end = ' ')
else:
print('*', end = ' ')
print()
Here is a simple and straight forward way to use a while loop to achieve this. Simple setting an integer object of check to 0. Then the while loop will evaluate check's value, then take user input, if the user input is greater than 0, let's set out check object to 1, so when we get back to the top of the loop, it will end.
Otherwise, if the if check fails, it will print a quick try again message and execute at the top of the loop again.
check = 0
while check == 0:
length = int(input("Enter the side of the square : "))
if length > 0:
check = 1
else:
print("Please try again.")
You can use this code after the input statement and before for loop.
if length < 0 :
Print("invalid length")
Break
This question already has answers here:
Check if a number is odd or even in Python [duplicate]
(6 answers)
Closed 1 year ago.
number = int(input("Type your number to check even or odd :"))
for number in range (1,100):
if(number%2) == 0:
print("This is even number")
elif number > 100:
print("Enter the valid number from 1 to 100")
else:
print("This is ODD number")
i am a beginner in python language , I have written code to read the number as EVEN or ODD in for loop condition between (1,100). correct me if making any mistakes in my code .
Why are you using for loop, just check the condition like if number > 100;the number is invalid.Check this example
nos=int(input())
if(nos>100):
print("Enter the valid number from 1 to 100 ")
else:
if(nos % 2 ==0):
print("Number is Even")
else:
print("Number is Odd")
There are mistakes in your code.
1.Indentation error at the 2nd line.(Remove whitespace before for loop.)
2.The name of the input variable and the iterator name in for loop is same. So your intended logic would run on the numbers from 1 ,2, 3 ..... 99. It never runs on the user entered value. So change the name of any variable. Both cant be 'number'.
3.Although you change the name of the variable, you initialised for loop with 100 iterations so you see output 100 times.
so if you want to check the numbers between given range which are even or odd you can try this..
num = int(input(" Please Enter the Maximum Number : "))
for number in range(1, num+1):
if(number % 2 == 0):
print("{0} is Even".format(number))
print("{0} is Odd".format(number))
This question already has answers here:
How can I read inputs as numbers?
(10 answers)
Closed 4 years ago.
Well, I am a beginner and my variable (guess) input doesn't work with the if statement:
When I put on the input numbers from 0 to 9, I want it prints out Right number!, else it print out the other message:
guess = input("Choose a number between 0-9: ")
if guess <=9 and guess >=0:
print("Right number!")
else:
print("The number you entered is out of range try again!")
The input() function returns a string, not a number (e. g. "127", not 127). You have to convert it to a number, e. g. with the help of int() function.
So instead of
guess = input("Choose a number between 0-9: ")
use
guess = int(input("Choose a number between 0-9: "))
to obtain an integer in guest variable, not a string.
Alternatively you may reach it in 2 statements - the first may be your original one, and the second will be a converting one:
guess = input("Choose a number between 0-9: ")
guess = int(guess)
Note:
Instead of
if guess <=9 and guess >=0:
you may write
if 0 <= guess <= 9: # don't try it in other programming languages
or
if guess in range(10): # 0 inclusive to 10 exclusive
This question already has an answer here:
Python Checking 4 digits
(1 answer)
Closed 1 year ago.
I want to write a program that only accepts a 4-digit input from the user.
The problem is that I want the program to accept a number like 0007 but not a number like 7 (because it´s not a 4 digit number).
How can I solve this? This is the code that I´ve wrote so far:
while True:
try:
number = int(input("type in a number with four digits: "))
except ValueError:
print("sorry, i did not understand that! ")
if number > 9999:
print("The number is to big")
elif number < 0:
print("No negative numbers please!")
else:
break
print("Good! The number you wrote was", number)
But if I input 7 to it it will just say Good! The number you wrote was 7
Before casting the user's input into an integer, you can check to see if their input has 4 digits in it by using the len function:
len("1234") # returns 4
However, when using the int function, Python turns "0007" into simple 7. To fix this, you could store their number in a list where each list element is a digit.
If it's just a matter of formatting for print purposes, modify your print statement:
print("Good! The number you wrote was {:04d}", number)
If you actually want to store the leading zeros, treat the number like a string. This is probably not the most elegant solution but it should point you in the right direction:
while True:
try:
number = int(input("Type in a number with four digits: "))
except ValueError:
print("sorry, i did not understand that! ")
if number > 9999:
print("The number is to big")
elif number < 0:
print("No negative numbers please!")
else:
break
# determine number of leading zeros
length = len(str(number))
zeros = 0
if length == 1:
zeros = 3
elif length == 2:
zeros = 2
elif length == 3:
zeros = 1
# add leading zeros to final number
final_number = ""
for i in range(zeros):
final_number += '0'
# add user-provided number to end of string
final_number += str(number)
print("Good! The number you wrote was", final_number)
pin = input("Please enter a 4 digit code!")
if pin.isdigit() and len(pin) == 4:
print("You successfully logged in!")
else:
print("Access denied! Please enter a 4 digit number!")