I am getting an infinite loop. I am not sure on how to covert the result as the new number variable and put it back in the while loop.
#Collatz squence
import sys
def collatz():
try:
print('Enter a number')
number = int(input())
except:
ValueError
print('Please type an integer')
while number != 1:
if number %2 == 0:
result = number//2
print(result)
elif number %2 == 1:
result = 3*number + 1
print(result)
**result = number**
while number == 1:
print ('You have arrived at the number itself')
sys.exit()
collatz()
The following works:
#Collatz squence
import sys
def collatz():
try:
print('Enter a number')
number = int(input())
except ValueError:
print('Please type an integer')
sys.exit(1)
while number != 1:
if number %2 == 0:
result = number//2
print(result)
elif number %2 == 1:
result = 3*number + 1
print(result)
number = result # set the number to the result
while number == 1:
print ('You have arrived at the number itself')
sys.exit()
collatz()
Notice that I set the number to the result, in your code the number never changed, and so kept hitting the same block of code over and over. I also added a sys.exit call in the exception, we don't want to continue if someone entered in a bad value.
Related
Could you, please, help me to understand how should I use try/except block and count tries at the same time.
Here is the code without try/except block (it seems it's working fine):
import random
number = random.randint(1, 10)
tries = 3
name = input('Hi! What is your name?\n')
answer = input(f'{name}, let\'s play a game! Yes or No?\n')
if answer == 'Yes':
print(f'But be aware: you have only {tries} tries!\nReady?')
chat = input('')
print('Ok, guess a number from 1 to 10!')
while tries != 0:
choice = int(input('Your choice: '))
tries -= 1
if choice > number:
print('My number is less!')
elif choice < number:
print('My number is higher!')
else:
print('Wow! You won!')
break
print(f'You have {tries} tries left.')
if tries == 0 and choice != number:
print(f'Sorry, {name}, you lost... It was {number}. Try next time. Good luck!')
else:
print('No problem! Let\'s make it another time...')
This one is with try/except block.. Not sure where should I place 'choice' variable and where count 'tries', it keeps looping and looping:
import random
number = random.randint(1, 10)
tries = 3
name = input('Hi! What is your name?\n')
answer = input(f'{name}, let\'s play a game! Yes or No?\n')
if answer == 'Yes':
print(f'But be aware: you have only {tries} tries!\nReady?')
chat = input('')
print('Ok, guess a number from 1 to 10!')
while True:
try:
choice = int(input('Your choice: '))
if 0 < choice < 11:
while tries != 0:
tries -= 1
if choice > number:
print(f'My number is less!')
elif choice < number:
print(f'My number is higher!')
else:
print('Wow! You won!')
break
print(f'You have {tries} tries left.')
if tries == 0 and choice != number:
print(f'Sorry, {name}, you lost... It was {number}. Try next time. Good luck!')
else:
print(f'Hey {name}, I said, print a number from 1 to 10!')
except ValueError:
print('Please, enter a number!')
else:
print('No problem! Let\'s make it another time...')
Thanks!
I wrote an input function python program,
But when run that code , IDE show that, "this function need to pass argument"
Even though ,I didn't declare any argument enter image description here
please help me how to solve this problem , Thank you in advance
list_number = list()
def input():
while True:
try:
number = input("Enter your number in to list = ")
if number == "Quit":
break
number = int(number)
list_number.append(number)
print(list_number)
except ValueError as e:
print(e)
def diagram():
display = ""
for i in list_number:
for j in range(i):
display = display +"#"
print(display)
display = ""
input()
diagram()
Several errors are noticed at glance:
mixture of namespace
You declared list_number as a global variable, but you cannot set value to it
directly insides a function. Instead, you can let the function return a value,
or use global statement to temporary allow a function to set a value to
a global variable temperary.
Read more on offical document, or search keyword python namespace for
relative articles.
name collision on builtin keyword
Some special word are reserved by python and could not be used as variable or
function name, input is amoung them.
BTW: The title of your question and example code layout is confusion! Follow the
tour to learn how to ask a better question and improve layout, so that people
can help you out.
Example code: though the test part has some bug I don't solved...
# remove: move it to a main progress for future design
# list_number = list()
# rename: input is a reserved name of builtins, pick another word
def myinput(*pargs):
if pargs:
for arg in pargs:
try:
yield int(arg)
except ValueError:
pass
else:
count = 0
while True:
# move out of `try` statement as it won't raise any exceptions
# imply lowercase for easier string comparison
userinput = input("Enter your number in to list: ").lower()
if userinput in ['quit', 'q']:
# for interactive, give user a response
print("Quit input procedure. Preparing Diagram...")
break
try:
number = int(userinput)
except ValueError:
# raise a error and the output will print to output by default
# there is no need to `print` an error
# and, for improve, you can raise a more specific message
# and continue your program
msg = "The program wants a number as input, please try again.\n"
msg += "Type `Quit` to exit input procedure."
print(msg)
continue
except KeyboardInterrupt:
msg = "You pressed Interrupt Keystroke, program exit."
print(msg)
return 0
# print a message and pass the value intercepted
count += 1
print("%d: number %d is added to queue." % (count, number))
yield number
def diagram(numbers):
# there is no need to iter a list by index
# and I am **not** sure what you want from your origin code
# if what you wnat is:
# join number with "#" sign
# then just use the builtins str.join method
# valid: is_list_like
if is_list_like(numbers):
numstr = map(str, numbers)
ret = "#".join(numstr)
else:
ret = "Nothing to export."
return ret
def is_list_like(obj):
"""fork from pandas.api.types.is_list_like,
search c_is_list_like as keyword"""
return (
# equiv: `isinstance(obj, abc.Iterable)`
hasattr(obj, "__iter__") and not isinstance(obj, type)
# we do not count strings/unicode/bytes as list-like
and not isinstance(obj, (str, bytes))
)
def main(*pargs):
# get a generator of user input
# if passed in values, accept parameter as user input for test
msgout = ""
if pargs:
# bug: test input not filtered by int() function
list_number = list(myinput(pargs))
print("Run builtin test module.")
else:
list_number = list(myinput())
count = len(list_number)
# process your input by whatever means you need
if count == 1:
msgout += "Received %d number from user input.\n" % count
else:
msgout += "Received %d numbers from user input.\n" % count
msgout += "The diagram is:\n%s" % diagram(list_number)
print(msgout)
def test():
"""simulate user input"""
userinputs = [
['a', 1, 5, 4, 9, 'q'],
[999, 'Quit'],
['q'],
]
for userinput in userinputs:
main(*userinput)
# test bug:
# 1. charactor is printed as output, too
if __name__ == "__main__":
# remove test() if you don't need it
test()
main()
Well I would change your function name from input to something else because you cannot have any function named anything from base python named in your function, This is probably the reason for your error.
Like the others said, input() is a builtin function in Python. Try this following code:
list_number = list()
def input_func():
while True:
try:
number = input("Enter your number in to list = ")
if number == "Quit":
break
number = int(number)
list_number.append(number)
print(list_number)
except ValueError as e:
print(e)
def diagram():
display = ""
for i in list_number:
for j in range(i):
display = display + "#"
print(display)
display = ""
input_func()
diagram()
Also, nice to note that try should be used more precisely only where the exception is expected to be thrown. You could rewrite input_func with that in mind, such as:
def input_func():
while True:
number = input("Enter your number in to list = ")
if number == "Quit":
break
try:
number = int(number)
except ValueError as e:
print(e)
else:
list_number.append(number)
print(list_number)
My task is:
Write a program that repeatedly prompts a user for integer numbers until the user enters 'done'. Once 'done' is entered, print out the largest and smallest of the numbers. If the user enters anything other than a valid number catch it with a try/except and put out an appropriate message and ignore the number. Enter 7, 2, bob, 10, and 4 and match the output below.
I want to ignore any invalid integer and print 'Invalid Output' message after calculating the maximum and minimum number.But it always prints the invalid message right after user input. How am i supposed to solve this?
Thanks in advance.
Code:
largest = None
smallest = None
while True:
num = input("Enter a number: ")
if num == "done":
break
try:
n = int(num)
except:
print('Invalid input')
if largest is None:
largest=n
elif n>largest:
largest=n
elif smallest is None:
smallest=n
elif n<smallest:
smallest=n
print("Maximum", largest)
print('Minimum', smallest)
You could store that invalid output as a boolean variable
largest = None
smallest = None
is_invalid=False
while True:
num = input("Enter a number: ")
if num == "done":
break
try:
n = int(num)
except:
is_invalid=True
if largest is None:
largest=n
elif n>largest:
largest=n
elif smallest is None:
smallest=n
elif n<smallest:
smallest=n
print("Maximum", largest)
print('Minimum', smallest)
if is_invalid:
print('Invalid Input')
This can be one solution
largest = None
smallest = None
errors = False
while True:
num = input('Please type a number : ')
if num == 'done':
break
try:
number = int(num)
#your logical operations and assignments here
except ValueError:
errors = True
continue
if errors:
print('Invalid input')
else:
print('Your Results')
Hope this helps :)
If you are familiar with lists,you can use list to solve your problem effectively,
here is a modified version of your code which uses list,
num1=[]
while True:
num = input("Enter a number: ")
num1.append(num)
if num == "done":
break
for i in num1:
try:
i = int(i)
except:
print('Invalid input:',i)
num1.remove(i)
print("Maximum", max(num1))
print('Minimum', min(num1))
output:
Enter a number: 34
Enter a number: 57
Enter a number: 89
Enter a number: ds
Enter a number: 34
Enter a number: do
Enter a number: done
Invalid input: ds
Invalid input: do
Maximum 89
Minimum 34
hope this helps,let me know if anything is incorrect.
Try removing the (int) in try block. Because you want only integer input in try block so if does not satisfy the condition of integer input it will execute the except block.
Your code should look like this in try block:
try:
n = num
I'm trying to validate multiple user inputs, can be of diff. data types.
Is there any better way to do User Input validation in Python3 ? or Improve this block of code.
def validate_number(message):
i = 0
while i < 4:
try:
userInput = int(input(message))
except ValueError:
i += 1
if i == 4:
print('Max try reached !!!')
return False
else:
print("Not an integer! Try again.")
continue
else:
return userInput
break
#------------------------------------------
a = validate_number('Enter 1st No: ')
if a:
b = validate_number('Enter 2nd No: ')
if a and b:
sum = a + b
print('Result is : %s' %(sum))
print('Result is : {} '.format(sum))
print(f'Result is : {sum}')
this is my suggestion:
def validate_number(message):
for i in range(4):
try:
return int(input(message))
except ValueError:
print("Not an integer! Try again.")
print("Max try reached !!!")
a simple for loop in order to count the number of tries.
as it is the function will return None if no valid input is given. you may have to tweak that according to your needs.
I'm not sure what I'm doing wrong here. I'm trying to limit user input to 1-6 (dice game). The logic is working but when a ValueError() is raised it doesn't prompt the user again.
try:
while True:
choice = input('Enter number to hold - type D to roll: ')
print(choice)
if choice == 'D':
return choice_list
elif len(choice) > 1 or choice not in '12345':
raise ValueError()
else:
choice_list[int(choice) - 1] = 'X'
printer(roll_list, choice_list)
except ValueError:
print ("Invalid input")
Because you're exiting your loop at Exception. You should write a code like this:
while True:
try:
choice = input('Enter number to hold - type D to roll: ')
print(choice)
if choice == 'D':
return choice_list
elif len(choice) > 1 or choice not in '12345':
raise ValueError()
else:
choice_list[int(choice) - 1] = 'X'
printer(roll_list, choice_list)
except ValueError:
print ("Invalid input")
Use while loop before try code which will keep looping try and catch block like this
while True:
try:
# Your Code
except ValueError:
# Your Code