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Python input control loop

Python beginner here. Practicing user input control.
Trying to make user input loop to the beginning if anything but a whole number between 1 and 10 is used. Been trying for hours, tried using Try and Except commands but couldn't do it correctly. What am i doing wrong? Thank you.
Edit:
Thank you very much for your help everyone, however the problem is still not solved (but very close!) I'm trying to figure out how to loop back to the beginning if anything BUT a whole number is typed. Agent Biscuit (above) gave a great answer for floating numbers, but any word or letter that is typed still produces an error. I´m trying to understand how to loop when anything random (except whole numbers between 1 and 10) is typed. None of the above examples produced corrcct results. Thank you for your help
while True:
print("Enter a number between 1 and 10")
number = int(input())
if (number > 0) and (number < 10):
print("Thank you, the end.")
break
else number != (> 0 and < 10):
print("It has to be a whole number between 1 and 10.")
print("Please try again:")

I have identified some problems.
First, the input statement you are using would just raise an error if a float value is entered, because the int at the start requires all elements of the input to be a number, and . is not a number.
Second; your else statement. else is just left as else:, and takes no arguments or parameters afterwards.
Now, how to check if the number is not whole? Try this:
while True:
print("Enter a number between 1 and 10")
number = float(input())
if (number > 0) and (number < 10) and (round(number)==number):
print("Thank you, the end.")
break
else:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
This accepts a float value, but only accepts it if it is equal to a whole number, hence the (round(number)==number).
Hope that answers your question.

First of all, you can't use a condition in a else statement. Also, you need to use or operator instead of and if one of the conditions is acceptable.
So, your code needs to be like this
while True:
print("Enter a number between 1 and 10")
number = int(input())
if (number > 0) and (number < 10):
print("Thank you, the end.")
break
elif number < 0 or number >10:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")

Thanks to ack (above) for pointing me to a useful link. By studying another thread, I found the solution. It may not be perfect code, but it works 100%:
while True:
try:
print("Enter a number between 1 and 10")
number = float(input())
if (number > 0) and (number < 10) and (round(number)==number):
print("Thank you, the end.")
break
else:
print("\n")
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
print("\n")
continue
except ValueError:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
print("\n")

Inputting large numbers and check if it is prime (Python)

I am making a program using python wherein it checks large numbers if it is a prime number or not. (this is for RSA Encryption)
Here is my code:
p = int(input("Enter first prime number: "))
while(1):
if isPrime(p) is False:
print("%d is not a prime number!" % p)
p = int(input("Please enter first prime number again: "))
else:
print("%d is a prime number." % p)
break
q = int(input("Enter second prime number: "))
while(1):
if isPrime(q) is False:
print("%d is not a prime number!" % q)
q = int(input("Please enter second prime number again: "))
else:
print("%d is a prime number." % q)
break
p is the first large number and q is the second large number.
here is the function that checks if it is a prime number:
def isPrime(num):
if num > 1:
for i in range(2, num):
if(num % i) == 0:
return False
else:
return True
else:
return False
I tried running it using small numbers like 17 and 11 and it works but when I tried to input a 16-digit number, it doesn't work anymore.
Here is the sample run:
On the second image, when I entered a large number, it does not continue. I tried waiting for half an hour and see if it works but it still, it's just like that. Why is it so?

It's taking so long because you're looping over A LOT of numbers, every one up until your 16 digit one, which is a lot of loop iterations. Even if you're only doing constant work per iteration, getting through that many numbers will still take awhile. Loops in python are notoriously slow. Luckily, you can use sympy like so:
from sympy.ntheory import factorint
def isPrime(n):
return len(factorint(n)) == 1
and then things should be going much faster. For reference:
factorint(133453565475464563453)
took about half a second

creating a python program which ask the user for a number - even odd

I am new to python and I am taking a summer online class to learn python.
Unfortunately, our professor doesn't really do much. We had an assignment that wanted us to make a program (python) which asks the user for a number and it determines whether that number is even or odd. The program needs to keep asking the user for the input until the user hit zero. Well, I actually turned in a code that doesn't work and I got a 100% on my assignment. Needless to say our professor is lazy and really doesn't help much. For my own knowledge I want to know the correct way to do this!!! Here is what I have/had. I am so embarrassed because I know if probably very easy!
counter = 1
num = 1
while num != 0:
counter = counter + 1
num=int(input("Enter number:"))
while num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)

There are a couple of problems with your code:
You never use counter, even though you defined it.
You have an unnecessary nested while loop. If the number the user inputs is even, then you're code will continue to run the while loop forever.
You are incorrectly using the else clause that is available with while loops. See this post for more details.
Here is the corrected code:
while True:
# Get a number from the user.
number = int(input('enter a number: '))
# If the number is zero, then break from the while loop
# so the program can end.
if number == 0:
break
# Test if the number given is even. If so, let the
# user know the number was even.
if number % 2 == 0:
print('The number', number, 'is even')
# Otherwise, we know the number is odd. Let the user know this.
else:
print('The number', number, 'is odd')
Note that I opted above to use an infinite loop, test if the user input is zero inside of the loop, and then break, rather than testing for this condition in the loop head. In my opinion this is cleaner, but both are functionally equivalent.

You already have the part that continues until the user quits with a 0 entry. Inside that loop, all you need is a simple if:
while num != 0:
num=int(input("Enter number:"))
if num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)
I left out the counter increment; I'm not sure why that's in the program, since you never use it.

Use input() and If its only number specific input you can use int(input()) or use an If/else statement to check

Your code wasn't indented and you need to use if condition with else and not while
counter = 1
num = 1
while num != 0:
counter = counter + 1
num = int(input("Enter number:"))
if num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)
Sample Run
Enter number:1
Odd 1
Enter number:2
Even 2
Enter number:3
Odd 3
Enter number:4
Even 4
Enter number:5
Odd 5
Enter number:6
Even 6
Enter number:0
Even 0

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.

Collatz sequence ends at 4

I am writing a Collatz sequence program using the practice projects from chapter 3 of Automate the boring stuff with python.
The program outline is:
Write a function named collatz() that has one parameter named number.
If number is even, then collatz() should print number // 2 and return
this value. If number is odd, then collatz() should print and return
3 * number + 1.
Then write a program that lets the user type in an integer and that
keeps calling collatz() on that number until the function returns the
value 1.
My code runs however it stops on 4 rather than 1. For every number I have tried so far the output goes past 1 back to 4.
example output:
6,3,10,5,16,8,4,2,1,4
I am using python 3.4.2
def collatz(number):
if number % 2 == 0:
number = number //2
print(number)
return number
elif number % 2 == 1:
number = 3 * number + 1
print(number)
return number
print ("pick a number:")
while True:
try:
number = int(input())
while number != 1:
number = collatz(number)
collatz(number)
break
except ValueError:
print("Error: Please enter a valid integer")
print("Magic! You are down to 1.")

The problem is that you call collatz() once more after the loop finishes with 1. Just remove that line, and it works fine.
Also if you move the "pick a number" to the input function, you can avoid the new line after the question and are asked again every time, if you input an invalid value.
Additionally you should also check if the number is greater than or equal to 1, to avoid endless loops. The code to do all that would look like that:
while True:
try:
number = int(input("pick a number: "))
if number < 1:
print("Error: Please enter a integer greater than or equal to 1 ")
continue
while number != 1:
number = collatz(number)
# removed the additional call to collatz
break
except ValueError:
print("Error: Please enter a valid integer")
print("Magic! You are down to 1.")

def collatz(number):
number = number // 2 if number % 2 == 0 else 3 * number + 1
print(number)
return number
number = int(input("Pick a Number\n"))
while number != 1:
number = collatz(number)
print("Magic! You are down to 1.")