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How to cleanly ask user for input and allow several types?

So this is a prompt for user input, and it works just fine. I was printing some names and associated (1 based) numbers to the console for the user to choose one. I am also giving the option to quit by entering q.
The condition for the number to be valid is a) it is a number and b) it is smaller or equal than the number of names and greater than 0.
while True:
number = str(input("Enter number, or q to quit. \n"))
if number == "q":
sys.exit()
try:
number = int(number)
except:
continue
if number <= len(list_of_names) and number > 0:
name = list_of_names[number-1]
break
There is no problem with this code, except I find it hard to read, and not very beautiful. Since I am new to python I would like to ask you guys, how would you code this prompt more cleanly? To be more specific: How do I ask the user for input that can be either a string, or an integer?

A bit simpler:
while True:
choice = str(input("Enter number, or q to quit. \n"))
if choice.lower() == "q":
sys.exit()
elif choice.isdigit() and (0 < int(choice) <= len(list_of_names)):
name = list_of_names[int(choice)-1]
break

Just downcase it.
number = str(input("Enter number, or q to quit. \n"))
number = number.lower()
That will make the q lower case so it doesn't matter if they press it with shift if they press something else just make a if statement that sets a while loop true.

Python IF ELSE statement not working [duplicate]

This question already has an answer here:
Simple Python IF statement does not seem to be working
(1 answer)
Closed 6 years ago.
I have my first program I am trying to make using python. The IF ELSE statement is not working. The output remains "Incorrect" even if the correct number is inputted by the user. I'm curious if it's that the random number and the user input are different data types. In saying that I have tried converting both to int with no avail.
Code below:
#START
from random import randrange
#Display Welcome
print("--------------------")
print("Number guessing game")
print("--------------------")
#Initilize variables
randNum = 0
userNum = 0
#Computer select a random number
randNum = randrange(10)
#Ask user to enter a number
print("The computer has chosen a number between 0 and 9, you have to guess the number!")
print("Please type in a number between 0 and 9, then press enter")
userNum = input('Number: ')
#Check if the user entered the correct number
if userNum == randNum:
print("You have selected the correct number")
else:
print("Incorrect")

On Python 3 input returns a string, you have to convert to an int:
userNum = int(input('Number: '))
Note that this will raise a ValueError if the input is not a number.

If you are using Python 3, change the following line:
userNum = input('Number: ')
to
userNum = int(input('Number: '))
For an explanation, refer to PEP 3111 which describes what changed in Python 3 and why.

I need to figure out how to make my program repeat. (Python coding class)

I am a beginner student in a python coding class. I have the majority of the done and the program itself works, however I need to figure out a way to make the program ask if wants a subtraction or an adding problem, and if the user would like another question. I asked my teacher for assistance and he hasn't gotten back to me, so I'm simply trying to figure out and understand what exactly I need to do.
import random
x = int(input("Please enter an integer: "))
if x < 0:
x = 0
print('Negative changed to zero')
elif x == 0:
print('Zero')
elif x == 1:
print('Single')
else:
print('More')
maximum = 10 ** x;
maximum += 1
firstnum = random.randrange(1,maximum) # return an int from 1 to 100
secondnum = random.randrange(1, maximum)
compsum = firstnum + secondnum # adds the 2 random numbers together
# print (compsum) # print for troubleshooting
print("What is the sum of", firstnum, " +", secondnum, "?") # presents problem to user
added = int(input("Your answer is: ")) # gets user input
if added == compsum: # compares user input to real answer
print("You are correct!!!")
else:
print ("Sorry, you are incorrect")

You'll want to do something like this:
def foo():
print("Doing good work...")
while True:
foo()
if input("Want to do more good work? [y/n] ").strip().lower() == 'n':
break
I've seen this construct (i.e., using a break) used more often than using a sentinel in Python, but either will work. The sentinel version looks like this:
do_good_work = True
while do_good_work:
foo()
do_good_work = input("Want to do more good work? [y/n] ").strip().lower() != 'n'
You'll want to do more error checking than me in your code, too.

Asking users for input is straightforward, you just need to use the python built-in input() function. You then compare the stored answer to some possible outcomes. In your case this would work fine:
print('Would you like to test your adding or subtracting skills?')
user_choice = input('Answer A for adding or S for subtracting: ')
if user_choice.upper() == 'A':
# ask adding question
elif user_choice.upper() == 'S':
# ask substracting question
else:
print('Sorry I did not understand your choice')
For repeating the code While loops are your choice, they will repeatedly execute a statement in them while the starting condition is true.
while True: # Condition is always satisfied code will run forever
# put your program logic here
if input('Would you like another test? [Y/N]').upper() == 'N':
break # Break statement exits the loop
The result of using input() function is always a string. We use a .upper() method on it which converts it to UPPERCASE. If you write it like this, it doesn't matter whether someone will answer N or n the loop will still terminate.

If you want the possibility to have another question asked use a while loop and ask the user for an input. If you want the user to input whether (s)he want an addition or substraction you already used the tools to ask for such an input. Just ask the user for a string.

"Try and Except" to differentiate between strings and integers? [duplicate]

This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 7 years ago.
I'm a beginner in the Python language. Is there a "try and except" function in python to check if the input is a LETTER or multiple LETTERS. If it isn't, ask for an input again? (I made one in which you have to enter an integer number)
def validation(i):
try:
result = int(i)
return(result)
except ValueError:
print("Please enter a number")
def start():
x = input("Enter Number: ")
z = validation(x)
if z != None:
#Rest of function code
print("Success")
else:
start()
start()
When the above code is executed, and an integer number is entered, you get this:
Enter Number: 1
Success
If and invalid value however, such as a letter or floating point number is entered, you get this:
Enter Number: Hello
Please enter a number
Enter Number: 4.6
Please enter a number
Enter Number:
As you can see it will keep looping until a valid NUMBER value is entered. So is it possible to use the "try and except" function to keep looping until a letter is entered? To make it clearer, I'll explain in vague structured English, not pseudo code, but just to help make it clearer:
print ("Hello this will calculate your lucky number")
# Note this isn't the whole program, its just the validation section.
input (lucky number)
# English on what I want the code to do:
x = input (luckynumber)
So what I want is that if the variable "x" IS NOT a letter, or multiple letters, it should repeat this input (x) until the user enters a valid letter or multiple letters. In other words, if a letter(s) isn't entered, the program will not continue until the input is a letter(s). I hope this makes it clearer.

You can just call the same function again, in the try/except clause - to do that, you'll have to adjust your logic a bit:
def validate_integer():
x = input('Please enter a number: ')
try:
int(x)
except ValueError:
print('Sorry, {} is not a valid number'.format(x))
return validate_integer()
return x
def start():
x = validate_integer()
if x:
print('Success!')

Don't use recursion in Python when simple iteration will do.
def validate(i):
try:
result = int(i)
return result
except ValueError:
pass
def start():
z = None
while z is None:
x = input("Please enter a number: ")
z = validate(x)
print("Success")
start()

Python does not accept my input as a real number

# Math Quizzes
import random
import math
import operator
def questions():
# Gets the name of the user
name= ("Alz")## input("What is your name")
for i in range(10):
#Generates the questions
number1 = random.randint(0,100)
number2 = random.randint(1,10)
#Creates a Dictionary containg the Opernads
Operands ={'+':operator.add,
'-':operator.sub,
'*':operator.mul,
'/':operator.truediv}
#Creast a list containing a dictionary with the Operands
Ops= random.choice(list(Operands.keys()))
# Makes the Answer variable avialabe to the whole program
global answer
# Gets the answer
answer= Operands.get(Ops)(number1,number2)
# Makes the Sum variable avialbe to the whole program
global Sum
# Ask the user the question
Sum = ('What is {} {} {} {}?'.format(number1,Ops,number2,name))
print (Sum)
global UserAnswer
UserAnswer= input()
if UserAnswer == input():
UserAnswer= float(input())
elif UserAnswer != float() :
print("Please enter a correct input")
def score(Sum,answer):
score = 0
for i in range(10):
correct= answer
if UserAnswer == correct:
score +=1
print("You got it right")
else:
return("You got it wrong")
print ("You got",score,"out of 10")
questions()
score(Sum,answer)
When I enter a float number into the console the console prints out this:
What is 95 * 10 Alz?
950
Please enter a correct input
I'm just curious on how I would make the console not print out the message and the proper number.

this is a way to make sure you get something that can be interpreted as a float from the user:
while True:
try:
user_input = float(input('number? '))
break
except ValueError:
print('that was not a float; try again...')
print(user_input)
the idea is to try to cast the string entered by the user to a float and ask again as long as that fails. if it checks out, break from the (infinite) loop.

You could structure the conditional if statement such that it cause number types more than just float
if UserAnswer == input():
UserAnswer= float(input())
elif UserAnswer != float() :
print("Please enter a correct input")

Trace through your code to understand why it doesn't work:
UserAnswer= input()
This line offers no prompt to the user. Then it will read characters from standard input until it reaches the end of a line. The characters read are assigned to the variable UserAnswer (as type str).
if UserAnswer == input():
Again offer no prompt to the user before reading input. The new input is compared to the value in UserAnswer (which was just entered on the previous line). If this new input is equal to the previous input then execute the next block.
UserAnswer= float(input())
For a third time in a row, read input without presenting a prompt. Try to parse this third input as a floating point number. An exception will be raised if this new input can not be parsed. If it is parsed it is assigned to UserAnswer.
elif UserAnswer != float() :
This expression is evaluated only when the second input does not equal the first. If this is confusing, then that is because the code is equally confusing (and probably not what you want). The first input (which is a string) is compared to a newly created float object with the default value returned by the float() function.
Since a string is never equal to a float this not-equals test will always be true.
print("Please enter a correct input")
and thus this message is printed.
Change this entire section of code to something like this (but this is only a representative example, you may, in fact, want some different behavior):
while True:
try:
raw_UserAnswer = input("Please enter an answer:")
UserAnswer = float(raw_UserAnswer)
break
except ValueError:
print("Please enter a correct input")