Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 5 years ago.
Improve this question
I'm just starting to learn Python and have a question regarding an exercise that came up in the textbook I'm reading.
I found a solution that does function, but I'm wondering if there is a simpler/recommended solution?
Problem:
numXs = int(input('How many times should I print the letter X? '))
toPrint = ' '
#concatenate X to print numXs times print(toPrint)
My solution:
numXs = int(input('How many times should I print the letter X? '))
toPrint = ''
while (toPrint == ''):
if numXs == 0:
numXs = int(input('Enter an integer != 0 '))
else:
toPrint = abs(numXs) * 'X'
print(toPrint)
I'd suggest you get all your data checking and correction handled up front, so your actual algorithm can be much simpler:
numXs = 0
while numXs <= 0:
numXs = int(input('How many times should I print the letter X? '))
if numXs <= 0:
print('Enter an integer > 0')
print('X' * numXs)
A simple solution would be
try:
print("x" * int(input("how many times?")))
except:
print("you entered an invalid number")
If you want a zero or negative check
try:
num = int(input("how many times?"))
if num > 0:
print("x" * num)
else:
print("need a positive integer")
except:
print("not a number")
If you want this to happen forever, just wrap it in a while loop
while True:
#code above
You are not handling wrong conversions (input of "Eight") - so you could shorten it to:
print(abs(int(input("How many?")))*"X") #loosing the ability to "requery" on 0
Same functionality:
numXs = abs(int(input('How many times should I print the letter X? ')))
while not numXs: # 0 is considered FALSE
numXs = abs(int(input('Enter an integer != 0 ')))
print(numXs * 'X')
Safer:
def getInt(msg, errorMsg):
'''Function asks with _msg_ for an input, if convertible to int returs
the number as int. Else keeps repeating to ask with _errorMsg_ until
an int can be returned'''
i = input(msg)
while not i.isdigit():
i = input(errorMsg)
return int(i)
I am using isdigit() to decide if the input is a number we can use. You could also do try: and except: around the int(yourInput) so you can catch inputs that are not conversible to integers:
def getIntWithTryExcept(msg, errorMsg):
'''Function asks with _msg_ for an input, if convertible to int returs
the number as int. Else keeps repeating to ask with _errorMsg_ until
an int can be returned'''
i = input(msg) # use 1st input message
num = 0
try:
num = int(i) # try convert, if not possible except: will happen
except:
while not num: # repeat with 2nd message till ok
i = input(errorMsg)
try:
num = int(i)
except: # do nothing, while is still True so it repeats
pass
return int(i) # return the number
Related
number = 0
number_list = []
while number != -1:
number = int(input('Enter a number'))
number_list.append(number)
else:
print(sum(number_list)/ len(number_list))
EDIT: Have found a simpler way to get the average of the list but if for example I enter '2' '3' '4' my program calculates the average to be 2 not 3. Unsure of where it's going wrong! Sorry for the confusion
Trying out your code, I did a bit of simplification and also utilized an if statement to break out of the while loop in order to give a timely average. Following is the snippet of code for your evaluation.
number_list = []
def average(mylist):
return sum(mylist)/len(mylist)
while True:
number = int(input('Enter a number: '))
if number == -1:
break
number_list.append(number)
print(average(number_list));
Some points to note.
Instead of associating the else statement with the while loop, I revised the while loop utilizing the Boolean constant "True" and then tested for the value of "-1" in order to break out of the loop.
In the average function, I renamed the list variable to "mylist" so as to not confuse anyone who might analyze the code as list is a word that has significance in Python.
Finally, the return of the average was added to the end of the function. If a return statement is not included in a function, a value of "None" will be returned by a function, which is most likely why you received the error.
Following was a test run from the terminal.
#Dev:~/Python_Programs/Average$ python3 Average.py
Enter a number: 10
Enter a number: 22
Enter a number: 40
Enter a number: -1
24.0
Give that a try and see if it meets the spirit of your project.
converts the resulting list to Type: None
No, it doesn't. You get a ValueError with int() when it cannot parse what is passed.
You can try-except that. And you can just use while True.
Also, your average function doesn't output anything, but if it did, you need to call it with a parameter, not only print the function object...
ex.
from statistics import fmean
def average(data):
return fmean(data)
number_list = []
while True:
x = input('Enter a number')
try:
val = int(x)
if val == -1:
break
number_list.append(val)
except:
break
print(average(number_list))
edit
my program calculates the average to be 2 not 3
Your calculation includes the -1 appended to the list , so you are running 8 / 4 == 2
You don't need to save all the numbers themselves, just save the sum and count.
You should check if the input is a number before trying to convert it to int
total_sum = 0
count = 0
while True:
number = input("Enter a number: ")
if number == '-1':
break
elif not number.isnumeric() and not (number[0] == "-" and number[1:].isnumeric()):
print("Please enter numbers only")
continue
total_sum += int(number)
count += 1
print(total_sum / count)
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 3 years ago.
Improve this question
my code not stop in line 9-10 when inserted words, I would like it to continue only when numbers are entered
from cs50 import get_int
numbers = []
while True:
number = get_int("number: ")
if not number: # here #
break # here #
if number not in numbers:
numbers.append(number)
print()
for number in numbers:
print(number)
need stopping if input not number
You can parse the input string to be a number or not in a while True loop, if it is not a number, break the loop, else keep asking.
numbers = []
while True:
#Ask for input
s = input("number: ")
number = 0
#Try to parse the string as a number, if you cannot, break the loop
try:
number = int(s)
except:
break
#If you can parse the string as a number, add it to the list
numbers.append(number)
print(numbers)
Sample outputs will be
number: 123
number: 456
number: 789
number: abc
[123, 456, 789]
In Python 3, you can use the following to ensure number is an integer (whole number):
isinstance(number, int)
If either an integer or a float (number with decimals) is allowed, you can use the following:
isinstance(number, (int, float))
I highly recommend reading more about this subject in this answer.
You don't need to define a new function to do that.
numbers = []
while True:
number = input("number: ")
try:
number = int(number)
if number not in numbers:
numbers.append(number)
except:
break
print (numbers)
try:
if number == int(number):
if number not in numbers:
numbers.append(number)
except:
#invalid input will throw you into this block
break
I have this assignment to create a program that asks user to input any positive integer repeatedly or type anything else to end and generate the sum, count and average of the numbers. My teacher wants all the code in these this structure with these three def’s only
This is the code I have, any suggestions on how to get it to work?
def calcAverage(total,count):
sum = 0
count = 0
average = sum / count
def inputNumber(message):
while True:
try:
userInput = int(input(message))
count = count + 1
sum = sum + entry
if userInput < 0:
raise ValueError
except ValueError:
main()
else:
return userInput
break
entry = inputNumber('Type any positive integer, anything else to quit')
def main():
print('Sum')
print(sum)
print('Average')
print(average)
print('Total Numbers')
print(count)
The question is not well explained + we don't really get what the boundaries are. Moreover, you should clearly state what is not working. Now, to give you some hint, this is how I would do it:
input = None
L = list()
while True:
try:
input = int(input('Type any positive integer, anything else to quit: '))
if input < 0:
break
else:
L.append(input)
except:
break
S = sum(L)
I think you don't need to use exceptions here. Condition statements would make it clearer.
I would put the valid inputs in a list until the user make a invalid input. When it happens, just get out of your while loop using a break statement and return the result.
Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 6 years ago.
Improve this question
Instructions: Program needs to ask the user for a number. For example "5". The program outputs the number 15, as 1+2+3+4+5=15.
I am a novice and am stuck at the beginning:
n = (input("Insert a number: "))
while n != 0:
Please guide me what to do further
You can do it like this:
num = int(input("Choose a number: "))
total = sum(range(num + 1))
If you HAVE to do it using a while loop, you can do it this way:
total = 0
counter = 0
max = int(input("Choose a number: "))
while counter <= max:
total += counter
counter += 1
print(total)
n = int(input("Insert a number: "))
nums = range(1,n+1)
print sum(nums)
if you want to do the same thing with while loop:
n = int(input("Insert a number: "))
sum =0
while n>0:
sum+=n
n-=1
print sum
Maybe you can use something like this code:
try:
nr_in = int(input("Enter some number: "))
nr_out = 0
tmp = 0
while tmp < nr_in:
tmp += 1
nr_out += tmp
print(nr_out)
except:
print("This is not a number!")
This isn't the shortest and most pythonic way, but I think it might be easier to understand for you.
Hope this helps!
Do you use python 3 or 2? In python2, raw_input is recommended.
Basically all you need is to convert the string to numeric value ...
inp = input("number: ")
try:
n = int(inp)
except ValueError:
print("Please give me a number")
sys.exit(1)
Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 9 years ago.
Improve this question
I have an assignment. It was suppose to do the following:
--> Take an integer input(say 100)
--> Add the digits till the sum is a single digit number(1)
My program till now is:
goodvalue1=False
goodvalue2=False
while (goodvalue1==False):
try:
num=input("Please enter a number: ")
except ValueError:
print ("Wrong input. Try again.")
else:
goodvalue1=True
if (goodvalue1==True):
ListOfDigits=list(map(int,str(num)))
sum=10
while(sum>9):
Sum=sum(ListOfDigits)
if (Sum>9):
ListOfDigits=list(map(int,str(Sum)))
Sum=sum(ListOfDigits)
Those booleans are not needed. You can factor the code down to:
while True:
try:
num = int(input("Please enter a number: ")) # Note how I've added int()
break # Breaks out of the loop. No need for a boolean.
except ValueError:
print("Wrong input. Try again.")
I don't see why you called list(map(int, str(num))); but I think you were intending to put int() around your input. So I added one in above. Now it can catch an error :).
Now, to get one digit, you can use another while loop here:
while num > 9:
num = sum(map(int, str(num)))
Pretty much this creates [1, 0, 0] which sum() then calls on. This repeats until it is no longer a two digit number (or three, four, etc)
So altogether:
while True:
try:
num = int(input("Please enter a number: ")) # Note how I've added int()
break # Breaks out of the loop. No need for a boolean.
except ValueError:
print("Wrong input. Try again.")
while num > 9: # While it is a two digit number
num = sum(map(int, str(num)))
Just note that for conditional statements, it's never pythonic to do a == True or b == False.
From the PEP:
Don't compare boolean values to True or False using ==.
Yes: if greeting:
No: if greeting == True:
Worse: if greeting is True:
My take on this:
inp = None
while inp is None:
try:
inp = int(input('Enter number here: '))
except ValueError:
print('Invalid Input, try again')
summed = sum(map(int, str(inp)))
while summed > 9:
summed = sum(map(int, str(summed)))
print('The result is {}'.format(summed))
For an explanation #Haidro did a good job: https://stackoverflow.com/a/17787707/969534
You're very close. Here's what you need to change:
Sum=sum(ListOfDigits)
while(Sum>9):
Sum=sum(ListOfDigits)
if (Sum>9):
ListOfDigits=list(map(int,str(Sum)))
Sum=sum(ListOfDigits)
In this code, you have a while loop that executes when sum is bigger than 9. So why use another variable Sum (also, it makes for really difficult-to-read code)? Do this instead:
while(sum>9):
sum=sum(ListOfDigits)
ListOfDigits=list(map(int,str(sum)))
This is only to show you what went wrong with your code. I wouldn't recommend using it (look below for what I would do). First, you mix variable-naming conventions, which is a very bad idea, especially when you work in a team (even otherwise, can you imagine looking at your code a month or six months from now?).
Second, you don't ever use goodvalue2; what's it there for?
Third, if goodvalue1 is only ever going to be a bool, then why check if (goodvalue1==True)? if goodvalue1 is clearer and more pythonic.
Please, for the love of all that is good, use some spaces in your code. Eyes get very strained after looking at expressions like ListOfDigits=list(map(int,str(num))) for a while. Try ListOfDigits = list(map(int, str(num))) instead.
Personally, I would do this:
num = None
while num is None:
try:
num = int(raw_input("Enter a number: "))
except ValueError:
num = None
num = sum(int(i) for i in str(num))
while num > 9:
num = sum(int(i) for i in str(num)) # this uses a list comprehension. Look it up, they're very useful and powerful!
RECURSION !
Calculate the sum of the digits. Check if the sum has one digit or multiple digit. If one digit, that is your answer, else, call the function on the sum again.
def oneDigitSum(n):
if n < 10:
return n
else:
return oneDigitSum(sum([int(i) for i in str(n)]))
# [f(elem) for elem in li] = [f(a), f(b), .... ] where li = [a, b, ... ]
# sum returns the total of the numbers in list
while True: # continue this loop for eternity, until it gets break
try:
num=int(input("Please enter a number: "))
print(oneDigitSum(num))
break # printed the sum, now I can break the loop peace fully
except ValueError:
print ("Wrong input. Try again.")
continue # oops, looks like wrong input, lets continue the loop