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Python Problems with Guessing Numbers

"""
I am new to coding in general and I have looked up how to fix this but when I input a letting than
a letting 2x then a number this happens
ges a number 1 tho 100 t
your ges is t
plz input a positive number below 100 that is not a word. exmp(42)
ges a number 1 tho 100 w
your ges is w
plz input a positive number below 100 that is not a word. exmp(42)
this is where I get confused
ges a number 1 tho 100 2
your ges is 2
2
your ges is t
plz input a positive number below 100 that is not a word. exmp(42)
ges a number 1 tho 100 2
your ges is 2
2
None
Traceback (most recent call last):
File "c:\Users\BertD\OneDrive\Desktop\random_number_gessing_game.py", line 20, in<module>
ges = int(ges)
ValueError: invalid literal for int() with base 10: 't'
I am concerned about how python is coming up with this?
It's a tuff puzzle to solve for me.
"""
import random
ges = input("ges a number 1 tho 100 ")
def valid_ges(num):
while True:
print("your ges is "+str(num))
if num.isdigit():
print(num)
break
elif num != num.isdigit():
print("plz input a positive number below 100 that is not a word. exmp(42)")
num = input("ges a number 1 tho 100 ")
elif num > 100:
print("try a number below 100")
num = input("pic a number 1 tho 100 ")
valid_ges(ges)
print(valid_ges(ges))
r = random.randint(1, 100)
ges = int(ges)
while ges != r:
if (ges <= 100):
print("your "+str(r - ges)+" away")
ges = abs(int(input("pic a number 1 tho 100 ")))
elif(ges == r):
break
print("yay you did it")
thank you for your time and have a nice day!

I don't think that giving you the solution will be beneficial for your learning process. Instead, I'll try to point out where your mistake is.
The main problem is in your function valid_guess. You successfully look if your input is a valid number. However, you never give back this information to the rest of your program. Instead, you do print(num) whenever the user input is valid, which only shows the valid guess on the screen. This value is however not stored anywhere.
You should have a good look at return in python. When you write a function that needs to process some information and then forward that information back to the rest of the program, you use return statements instead. Let me show this with an example:
def square(num):
print(num**2)
>>> x = 5
>>> square(x)
25
>>> xsquared = square(x)
25
>>> print(xsquared)
None
Do you see how the result is not stored in xsquared? It just shows the result to the user, then completely forgets about it!
If I however use this instead:
def square(num):
return(num**2)
>>> x = 5
>>> xsquared = square(x)
>>> print(xsquared)
25
The result is now successfully stored! Do note however that we didn't ask the program to show us the results this time. To do so, we need to add another print statement, as you can see in the last line!
Try to use this information to your problem now. Try to store the new guess in ges whenever you give a valid number.
As #pranav-hosangadi mentioned, you should have a close look at a python debugger and learn to work with it. This will help you understand your code much better! Good luck with learning, and feel free to comment if you have more questions!

This a python value error which occurs when you try to convert a string value that is not formatted as an integer.
Here, the line which causes the error is ges = int(ges).
Where when you first executed you are giving(ges) it a String t. It's not changing it's value while guessing again. Go through your code.

Collatz from automate the boring stuff

I know there are multiple posts on this question. But I could not post my code in any other way except by asking a question. Can someone please help me understand how I can stop n from being input into the collatz function each time the global scope executes.
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.
(Amazingly enough, this sequence actually works for any integer—sooner or later, using this sequence,
you’ll arrive at 1! Even mathematicians aren’t sure why.
Your program is exploring what’s called the Collatz sequence, sometimes called “the simplest impossible math problem.”)#
Remember to convert the return value from input() to an integer with the int() function; otherwise, it will be a string value.
desired output
3
10
5
16
8
4
2
1
Input Validation
Add try and except statements to the previous project to detect whether the user types in a noninteger string.
Normally, the int() function will raise a ValueError error if it is passed a noninteger string, as in int('puppy').
In the except clause, print a message to the user saying they must enter an integer.
def collatz(number):
if number%2==0:
number=number//2
print(number)
elif number%2==1:
number=3*number+1
print(number)
print('Enter number: ')
n=int(input())
while n!=1:
collatz(n)

You‘ve created an infinite loop, since your „n“ doesn‘t change within the loop and „n!=1“ is never met as long as the user doesn’t input “1” in the beginning.
Try this:
def collatz(number):
if number % 2 == 0:
number = number // 2
else:
number = 3 * number + 1
print(number)
return number
n = int(input("Enter number: "))
while n != 1:
n = collatz(n)

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

Creating a small program in python

I'm trying to make a program in Python 3.5 that asks the user to enter a number from 1 to 9. The the program will also guess a number from 1 to 9. If the user guesses the same number correctly, then the program will print Correct . Otherwise, the program will print Wrong. I wrote a program. Everything went fine with my code. However, when I guessed a number correctly, the program suddenly wrote Wrong instead of Correct. What am I doing wrong?
Here is my code:
print('Enter a number from 1 to 9')
x = int(input('Enter a number: '))
import random
random = print(random.randint(1,9))
if int(x) != random:
print ('Wrong')
else:
print ('Correct')

You are saving the result of a print() call (and masking random). print() returns None, so it will always be unequal to an integer, and therefore always "wrong."
import random
print('Enter a number from 1 to 9')
x = int(input('Enter a number: '))
r = random.randint(1,9)
if x != r:
print('Wrong')
else:
print('Correct')
Also note that I've moved the import statement to the top, avoided a second int() cast on something you've already done that to, and removed the space between the print reference and its arguments.

Here is the mistake,
random = print(random.randint(1,9))
You need to do something like this.
random = random.randint(1,9)
print(random)
Also, you have already converted the input to int so, you can do just this.
if x != random:

As pointed out your mistake is the line
random = print(random.randint(1,9))
But why?
functions (like print() take something, do something (with it) and give something back.
Example:
sum(3,4) takes 3 and 4, may add them and returns 7.
print("Hello World") on the other hand takes "Hello world", prints it on the screen but has nothing useful to give back, and therefore returns None (Pythons way to say "nothing").
You then assign None to the name random and test if it equals your number, which it (of course) doesn't.

python string and integer comparison

I want to print all the numbers until the given user input using while loop. Example:Enter:5 ==> 1 2 3 4 5 But the below program loops for ever.
user = str(input("Enter : "))
i = 1
while i < user:
print(i)
i = i + 1

ummm while i < int(user):?

Try this instead:
try:
user = int(raw_input('Enter: ')) # Cannot compare a string with an integer.
except ValueError:
print('Input should be an integer!')
i = 1
while True:
i += 1
if i > user:
break
print(i)
Note: In your code, even if we were to explicitly declare the input as an integer it would still not quite work the way you want it to. This is because in your code the while loop stops once i is equal to user (as the condition is while less than... and will thus not print out the final value, user. I therefore modified it so it breaks at the point where i is greater than user, meaning that the last printed value will be equal to user.
Example previous output where user = 5:
1
2
3
4
And with the new code:
1
2
3
4
5
It is however better to use a for loop here, if you are not set on using a while loop:
for i in range(1, user+1):
print(i)

input in Python 2.x will try to evaluate what the user enters, it is equivalent to
user = eval(raw_input(...))
In this case, you are explicitly converting whatever is supplied to a string (with str()). In Python 2.x, strings always compare larger than numbers, so i < user is always True.
It is wiser to use raw_input and convert to int. You can also simplify your code with a for loop:
user = int(raw_input("Enter : "))
for i in range(user):
print(i)

You are comparing an int to a str and that is why you are getting an infinite loop. You should compare the same type of variables
user = int(input("Enter: "))
should work