My program breaks if a person inputs a letter instead of a interger the base of the problem is around here
while quesses<6 :
guesses = guesses+1
guess = input("Guess a number 1 to 100)
if int(guess) ==r:
print("correct you took" + str(guesses) + "tries")
break
the problem is if a letter is used it breaks what could I do to make it evaluate that it is a string the redo then input without increasing the value of guesses
Use some function to do the number parsing.
Parse the user input inside some try-except block.
If that succeeded => increase counter and return result and counter.
If not => call your number parsing function again.
Some simple number guessing game would be:
from random import choice
def get_number(current):
result = None
try:
result = int(
input('>>> [Round #{}] Guess a number between 1 and 100: '.format(
current + 1
))
)
except ValueError:
print('!!> [ERROR] Please try again')
if result is not None:
current += 1
return result, current
return get_number(current=current)
guesses = 0
unknown = choice(range(0, 100+1))
# print('>>> [debug] unknown is:', unknown)
while guesses < 6:
guess, guesses = get_number(guesses)
if guess == unknown:
print('>>> you win!', 'took', guesses, 'guesses')
break
print('>>> game over')
Related
To begin, a definition:
A polydivisible number is an integer number where the first n digits of the number (from left to right) is perfectly divisible by n. For example, the integer 141 is polydivisible since:
1 % 1 == 0
14 % 2 == 0
141 % 3 == 0
I'm working on a recursive polydivisible checker, which, given a number, will check to see if that number is polydivisible, and if not, recursively check every other number after until it reaches a number that is polydivisible.
Unfortunately, my code doesn't work the way I want it to. Interestingly, when I input a number that is already polydivisible, it does its job and outputs that polydivisible number. The problem occurs when I input a non-polydivisible number, such as 13. The next polydivisible number should be 14, yet the program fails to output it. Instead, it gets stuck in an infinite loop until the memory runs out.
Here's the code I have:
def next_polydiv(num):
number = str(num)
if num >= 0:
i = 1
print(i)
while i <= len(number):
if int(number[:i]) % i == 0:
i += 1
print(i)
else:
i = 1
print(i)
num += 1
print(num)
else:
return num
else:
print("Number must be non-negative")
return None
I'm assuming the problem occurs in the else statement inside the while loop, where, if the number fails to be polydivisible, the program resets i to 0, and adds 1 to the original number so it can start checking the new number. However, like I explained, it doesn't work the way I want it to.
Any idea what might be wrong with the code, and how to make sure it stops and outputs the correct polydivisible number when it reaches one (like 14)?
(Also note that this checker is only supposed to accept non-negative numbers, hence the initial if conditional)
The mistake is that you are no updating number after incrementing num.
Here is working code:
def next_polydiv(num):
number = str(num)
if num >= 0:
i = 1
print(i)
while i <= len(number):
if int(number[:i]) % i == 0:
i += 1
print(i)
else:
i = 1
print(i)
num += 1
print(num)
number = str(num) # added line
else:
return num
else:
print("Number must be non-negative")
return None
I have a similar answer to #PranavaGande, the reason is I did not find any way to iterate an Int. Probably because there isn't one...Duh !!!
def is_polydivisible(n):
str_n = str(n)
list_2 = []
for i in range(len(str_n)):
list_2.append(len(str_n[:i+1]))
print(list_2)
list_1 = []
new_n = 0
for i in range(len(str_n)):
new_n = int(str_n[:i+1])
list_1.append(new_n)
print(list_1)
products_of_lists = []
for n1, n2 in zip(list_1, list_2):
products_of_lists.append(n1 % n2)
print(products_of_lists)
for val in products_of_lists:
if val != 0:
return False
return True
Now, I apologise for this many lines of code as it has to be smaller. However every integer has to be split individually and then divided by its index [Starting from 1 not 0]. Therefore I found it easier to list both of them and divide them index wise.
There is much shorter code than mine, however I hope this serves the purpose to find if the number is Polydivisible or Not. Of-Course you can tweak the code to find the values, where the quotient goes into decimals, returning a remainder which is Non-zero.
while True:
numbers = input('> ')
if numbers == 'done':
break
total = 0
for number in numbers:
if numbers == int:
total = total + numbers
print(total)
I've had a hard time understanding exactley what u want to do with this code, please use a proper code block next time, with proper indentation. My guess is that u want to get a input number like 345 and add 3+4+5 as a output. If the input is not a int it should break the loop. Ive come up with 2 diffrent solutions, depending on what you need.
This code will simply take the input and check if it is "done", if it is not "done" it will try to add. This is a easy to understand solution but it will produce a error if the input is any diffrent string than "done".
while True:
numbers = input(">")
if numbers == "done":
break
else:
total = 0
for number in numbers:
total += int(number)
print(total)
This approach will test for the "done" string again, but afterwards will also check if the input can be converted into a int. if not the error is captured and it will return "invalid input". If u want the programm to terminate at any string u can just put break in the except section.
while True:
numbers = input(">")
if numbers == "done":
break
else:
try:
testing = int(numbers)
total = 0
for number in numbers:
total += int(number)
except:
total = "invalid input"
print(total)
im a Beginner myself and if a experiencend person can show me a better way to do this i would be very interested
while True:
numbers = input('> ')
if numbers == 'done':
break
total = 0
for number in numbers:
if numbers == int:
total = total + numbers
print(total)
Assuming this as your code:
Here's a solution to your problem-->
numbers=[]
while True:
a=input('>')
if a=='done':
break
else:
numbers.append(a)
total=0
for number in numbers:
total = total + int(number)
print(total)
By default everything gets accepted as string so we convert it to integer to find total.
Also we use list to store all the values we accept.
Another Solution is-->
numbers=[]
while True:
a=input('>')
if a=='done':
break
else:
numbers.append(a)
p=map(int,numbers)
print(sum(p))
Hope you understand the solution :-)
total = 0
average = 0
count = 0
while True:
numbers = input('> ')
if numbers == 'done': break
try:
total = int(numbers) + total
count = count + 1
except:
print('nope')
try:
average = total / count
except:
print('error')
print(total)
print(average)
print(count)
Try this:
total = 0
number_of_inputs = 0
while True:
number_string = input('Enter a number: ')
try:
total += float(number_string)
number_of_inputs += 1
except ValueError:
break # we weren't given a number, so exit the loop
# Now that we're outside of the loop, print out the total:
print('The total is:', total)
if number_of_inputs > 0:
average = total / number_of_inputs
print('The average is:', average)
else:
print('The average cannot be calculated, as no inputs were given.')
Do you see what's happening? The while loop keeps ask for and adding integers to total until a non-integer (like "done") is given. Once it gets that non-integer, the int() function will fail, and the exception it throws will get caught, and the code will immediately break out of the while loop.
And once out of the loop, the total and the average are printed out.
A few things you should be aware of:
If the user gave no inputs (which is possible here), the total will correctly print out as 0, but if you try to calculate the average, you will error, due to diving by number_of_inputs (which is also 0). That is why I check that number_of_inputs is greater than zero before I even attempt to calculate the average.
Originally I used int() to convert the string to a number, but I changed it to use float() instead. I figure that since you want to calculate an average, averages are not necessarily integers (even if all the inputs are), so there's no point in enforcing integer input. That is why I changed the int() to float(), but whether or not you want to use it is up to you.
ValueError isn't a function; it's an Exception. At this point you probably don't know what Exceptions are, so just know that they are special cases that can happen, and they're often used for catching errors, such as bad input values.
In the code I posted above, the loop is always expecting numerical input. But as soon as we have input that can't be converted to a number, the program then says, "Hey, I have an exception to what we're expecting! The exception is that there's an error in the value!" Then the program, instead of continuing to the next line of code (which is number_of_inputs += 1) will then execute the block of code under the except ValueError: section. And in the code above, all it does is call break, which exits the loop.
Once out of the loop, the code prints out the total and the average.
If it weren't for the try: and except ValueError: lines in the code, then the program would abruptly end (with a lengthy error message) once someone gave a non-numerical input. That happens because the call to float() wouldn't know how to convert a value like "done" to a number, so it does nothing more than just quitting.
However, by using try: and except ValueError:, we are anticipating that someone might give non-numerical input. When that happens (which it will, when the user is finished giving inputs) -- instead of quitting -- we want an alternate action to take. And we specify that alternate action to be a simple break out of the loop -- which will allow the program to continue with whatever is after the loop.
I hope this makes sense. If it doesn't, it will make more sense once you start learning about Exceptions in Python.
I'm a new learner of python, when I try to write a Collatz function I find that pycharm shows me one line is unreachable. I wonder why the function can't run the code
def Collatz(numBer):
if numBer%2 == 0:
return numBer//2
else:
return 3*numBer+1
print(numBer) #this code is unreachale
print('Please input the number:')
numBer = int(input())
while numBer != 1:
Collatz(numBer)
print(Collatz(numBer)) #because the former code are unreachable,so I write this to print the results
numBer = Collatz(numBer)
All code within the same scope below a return statement is unreachable because the function will finish its execution there. In your case you are returning the result so there is no need to rerun the function again to print it. Just take it in a variable and use it:
def Collatz(numBer):
if numBer%2 == 0:
return numBer//2
else:
return 3*numBer+1
print('Please input the number:')
numBer = int(input())
while numBer != 1:
numBer = Collatz(numBer)
print(numBer)
Hello welcome to Stack Overflow!
The reason why the print is "unreachable" is because of the return before the print. return ends a control flow so any code after the return is disregarded. Basically, the control flow goes like this (based on your function):
"Is the numBer divisible by 2?"
"If yes, then give me the integer division of that number and 2"
"Otherwise, give me the 3*number + 1"
If you wanted to print the number before you return it, it would be best to store it first into a variable and then return that variable, like so:
def Collatz(numBer):
if Collatz % 2 == 0:
value = numBer // 2
else:
value = 3 * numBer + 1
print(value)
return value
Apologies if similar questions have been asked but I wasn't able to find anything to fix my issue. I've written a simple piece of code for the Collatz Sequence in Python which seems to work fine for even numbers but gets stuck in an infinite loop when an odd number is enter.
I've not been able to figure out why this is or a way of breaking out of this loop so any help would be greatly appreciate.
print ('Enter a positive integer')
number = (int(input()))
def collatz(number):
while number !=1:
if number % 2 == 0:
number = number/2
print (number)
collatz(number)
elif number % 2 == 1:
number = 3*number+1
print (number)
collatz(number)
collatz(number)
Your function lacks any return statements, so by default it returns None. You might possibly wish to define the function so it returns how many steps away from 1 the input number is. You might even choose to cache such results.
You seem to want to make a recursive call, yet you also use a while loop. Pick one or the other.
When recursing, you don't have to reassign a variable, you could choose to put the expression into the call, like this:
if number % 2 == 0:
collatz(number / 2)
elif ...
This brings us the crux of the matter. In the course of recursing, you have created many stack frames, each having its own private variable named number and containing distinct values. You are confusing yourself by changing number in the current stack frame, and copying it to the next level frame when you make a recursive call. In the even case this works out for your termination clause, but not in the odd case. You would have been better off with just a while loop and no recursion at all.
You may find that http://pythontutor.com/ helps you understand what is happening.
A power-of-two input will terminate, but you'll see it takes pretty long to pop those extra frames from the stack.
I have simplified the code required to find how many steps it takes for a number to get to zero following the Collatz Conjecture Theory.
def collatz():
steps = 0
sample = int(input('Enter number: '))
y = sample
while sample != 1:
if sample % 2 == 0:
sample = sample // 2
steps += 1
else:
sample = (sample*3)+1
steps += 1
print('\n')
print('Took '+ str(steps)+' steps to get '+ str(y)+' down to 1.')
collatz()
Hope this helps!
Hereafter is my code snippet and it worked perfectly
#!/usr/bin/python
def collatz(i):
if i % 2 == 0:
n = i // 2
print n
if n != 1:
collatz(n)
elif i % 2 == 1:
n = 3 * i + 1
print n
if n != 1:
collatz(n)
try:
i = int(raw_input("Enter number:\n"))
collatz(i)
except ValueError:
print "Error: You Must enter integer"
Here is my interpretation of the assignment, this handles negative numbers and repeated non-integer inputs use cases as well. Without nesting your code in a while True loop, the code will fail on repeated non-integer use-cases.
def collatz(number):
if number % 2 == 0:
print(number // 2)
return(number // 2)
elif number % 2 == 1:
result = 3 * number + 1
print(result)
return(result)
# Program starts here.
while True:
try:
# Ask for input
n = input('Please enter a number: ')
# If number is negative or 0, asks for positive and starts over.
if int(n) < 1:
print('Please enter a positive INTEGER!')
continue
#If number is applicable, goes through collatz function.
while n != 1:
n = collatz(int(n))
# If input is a non-integer, asks for a valid integer and starts over.
except ValueError:
print('Please enter a valid INTEGER!')
# General catch all for any other error.
else:
continue
New to programming. How to count and print iterations(attempts) I had on guessing the random number?` Let's say, I guessed the number from 3-rd attempt.
import random
from time import sleep
str = ("Guess the number between 1 to 100")
print(str.center(80))
sleep(2)
number = random.randint(0, 100)
user_input = []
while user_input != number:
while True:
try:
user_input = int(input("\nEnter a number: "))
if user_input > 100:
print("You exceeded the input parameter, but anyways,")
elif user_input < 0:
print("You exceeded the input parameter, but anyways,")
break
except ValueError:
print("Not valid")
if number > user_input:
print("The number is greater that that you entered ")
elif number < user_input:
print("The number is smaller than that you entered ")
else:
print("Congratulation. You made it!")
There are two questions being asked. First, how do you count the number of iterations? A simple way to do that is by creating a counter variable that increments (increases by 1) every time the while loop runs. Second, how do you print that number? Python has a number of ways to construct strings. One easy way is to simply add two strings together (i.e. concatenate them).
Here's an example:
counter = 0
while your_condition_here:
counter += 1 # Same as counter = counter + 1
### Your code here ###
print('Number of iterations: ' + str(counter))
The value printed will be the number of times the while loop ran. However, you will have to explicitly convert anything that isn't already a string into a string for the concatenation to work.
You can also use formatted strings to construct your print message, which frees you from having to do the conversion to string explicitly, and may help with readability as well. Here is an example:
print('The while loop ran {} times'.format(counter))
Calling the format function on a string allows you replace each instance of {} within the string with an argument.
Edit: Changed to reassignment operator