This is the problem I have to solve : Write a program to sum a series of numbers entered by the user. The program should first prompt the user for how many numbers are to be summed. It should then input each of the numbers and print a total sum. This is what I have so far:
def excercise13():
print("Programming Excercise 13")
print("This program adds a series of numbers.")
while True:
try:
numberTimes = float(input("Enter how many numbers will be added: "))
except ValueError:
print("Invalid input.")
else:
break
numberTimes = int(numberTimes)
while True:
try:
for i in range(1,(numberTimes+1)):
("""I don't know what to put here""")
except ValueError:
print("Invalid input.")
else:
break
totalSum =
print("The sum of",nums,"is:",totalSum)
print()
excercise13()
I will go through the solution, based on your code, code block by code block.
def excercise13():
currentnumber = 0
Here we create the function excercise13() and set currentnumber to 0
print("Programming Excercise 13")
print("This program adds a series of numbers.")
while True:
try:
numberTimes = int(input("Enter how many numbers will be added: "))
except ValueError:
print("Invalid input.")
else:
break
You should use int instead of float. Can you imagine doing a process 3.5 times? This also reduces your previous repetition.
for x in range(numbertimes): #More pythonic way.
new_number = input ("Please enter a number to be added.")
currentnumber += new_number
The above code block makes the program ask for a new number numbertimes times. It then adds this number to currentnumber
totalSum = currentnumber
print("The sum of",nums,"is:",totalSum)
print()
This sets the totalSum to the final currentnumber
excercise13()
This starts your code.
Python has this functionality built in as the sum function.
def makesum():
try:
numbers = input('Enter the numbers to sum, comma seperated: ')
print 'The sum is {0}'.format(sum(numbers))
except:
print 'Input invalid. Try again.'
makesum()
makesum()
Related
I want to make an calculator for average but i'm facing some issues. I want the numbers entered by the users come as a print statement but it is just throwing the last entered value.Here is my code.
numlist = list()
while True:
inp = input(f"Enter a number, (Enter done to begin calculation): ")
if inp == 'done':
break
value = float(inp)
numlist.append(value)
average = sum(numlist) / len(numlist)
print(f"The Entered numbers are: ", inp)
print(f"average is = ", average)
Solution
we can print the list
numlist = list()
while True:
inp = input(f"Enter a number, (Enter done to begin calculation): ")
if inp == 'done':
break
value = float(inp)
numlist.append(value)
average = sum(numlist) / len(numlist)
print(f"The Entered numbers are: {numlist}")
print(f"average is = ", average)
As I understand, you want to print all the user inputs, however, as I see in your code, you are printing just a value but not the storing list, try to change your code to print(f"The Entered numbers are: ", numlist)
You are telling the user that their entered numbers are the last thing they typed in input, which will always be "done" because that is what breaks out of the loop. If you want to print the entered numbers; print numlist instead of inp.
I need some help regarding calculating averages and ranges. I am using built-in functions such as sum(), len(), etc. and cannot seem to calculate the average or range. I am using it to write a small piece of code for fun but cannot seem to get it to work. any help is much appreciated. Thank you!
x = 1
number_list = []
while x == 1:
input_number = input("PLease input an integer")
if str.isdigit(input_number) == True:
number_list.append(input_number)
else:
print("Please input a valid integer only.")
continueornot = input("Would you like to continue adding data? PLease input 'Yes' to continue, and anything else to quit.")
if continueornot == 'Yes':
x = 1
else:
print("Here is the maximum number:", max(number_list))
print("Here is the minimum number:", min(number_list))
print("Here is the count:", len(number_list))
print("Here is the average:" + sum(number_list) / len(number_list))
print("Here is the range:", range(number_list))
quit()
Change
if str.isdigit(input_number) == True:
number_list.append(input_number)
to
if input_number.isdigit():
number_list.append(int(input_number))
The error is because you're trying to do those operations on a list of strings.
You can also remove the check against True since that is implicitly checking the truthiness and since input_number is already a str, you can call the isdigit() method directly.
The problem is that you are appending strings to the list rather than integers and then you are applying arithmetic operations on it. So first you should convert the input number to int type.
Secondly range function will not give you the range of a list rather then it returns a sequence.
x = 1
number_list = []
while x == 1:
input_number = input("PLease input an integer")
if str.isdigit(input_number) == True:
input_number=int(input_number)
number_list.append(input_number)
else:
print("Please input a valid integer only.")
continueornot = input("Would you like to continue adding data? PLease input 'Yes' to continue, and anything else to quit.")
if continueornot == 'Yes':
x = 1
else:
print("Here is the maximum number:", max(number_list))
print("Here is the minimum number:", min(number_list))
print("Here is the count:", len(number_list))
print("Here is the average:" , sum(number_list) / len(number_list))
print("Here is the range:", max(number_list)-min(number_list))
I am completing questions from a python book when I came across this question.
Write a program which repeatedly reads numbers until the user enters "done". Once done is entered, print out total, count, and average of the numbers.
My issue here is that I do not know how to check if a user specifically entered the string 'done' while the computer is explicitly checking for numbers. Here is how I approached the problem instead.
#Avg, Sum, and count program
total = 0
count = 0
avg = 0
num = None
# Ask user to input number, if number is 0 print calculations
while (num != 0):
try:
num = float(input('(Enter \'0\' when complete.) Enter num: '))
except:
print('Error, invalid input.')
continue
count = count + 1
total = total + num
avg = total / count
print('Average: ' + str(avg) + '\nCount: ' + str(count) + '\nTotal: ' + str(total))
Instead of doing what it asked for, let the user enter 'done' to complete the program, I used an integer (0) to see if the user was done inputting numbers.
Keeping your Try-Except approach, you can simply check if the string that user inputs is done without converting to float, and break the while loop. Also, it's always better to specify the error you want to catch. ValueError in this case.
while True:
num = input('(Enter \'done\' when complete.) Enter num: ')
if num == 'done':
break
try:
num = float(num)
except ValueError:
print('Error, invalid input.')
continue
I think a better approach that would solve your problem would be as following :
input_str = input('(Enter \'0\' when complete.) Enter num: ')
if (input_str.isdigit()):
num = float(input_str)
else:
if (input_str == "done"):
done()
else:
error()
This way you control cases in which a digit was entered and the cases in which a string was entered (Not via a try/except scheme).
def main():
num = int(input('Please enter an odd number: '))
if False:
print('That was not a odd number, please try again.')
else:
print('Congrats, you know your numbers!')
def number():
if (num / 2) == 0:
return True,num
else:
return False,num
main()
I am trying to make it so that if the number entered is odd, it congratulates the user. If not then it should tell them to try again. I am trying to return the Boolean value to main and then when I try to use the code in the main function to prompt the user, it doesn't work.
Your functions are very odd and I'm not talking about numbers that aren't divisible by 2. Try this:
num = int(input('Please enter an odd number: '))
if num % 2 == 0:
print('Better luck next time??') # no really, you should go back to school (;
else:
print('Genius!!!')
Try this:
num_entry = int(input('Please enter an odd number: '))
def number():
return num_entry % 2 == 0
def main():
if number() == True:
print('Sorry, please try again.')
else:
print('Nice! You know your numbers!')
number()
main()
This should work!
Your code does look odd like Malik Brahimi mentioned. This may be because you are trying to write your python code like Java, which requires a main method. There is no such requirement in python.
If you would like to have your check for the "odd-ness" of the number wrapped in a defined function that you can call elsewhere, you should try writing it like this.
def odd_check(number):
if number % 2 == 0:
#This is the check the check formula, which calculates the remainder of dividing number by 2
print('That was not an odd number. Please try again.')
else:
print('Congrats, you know your numbers!')
num = int(input('Please enter an odd number: ')) #where variable is stored.
odd_check(num) #This is where you are calling the previously defined function. You are feeding it the variable that you stored.
If you would like a piece of code that will continue to ask your user to enter a number until they get it right, try something like this:
while True: #This ensures that the code runs until the user picks an odd number
number = int(input('Please enter an odd number: ')) #where variable is stored.
if number % 2 == 0: #the check formula, which calculates the remainder of dividing num by 2
print('That was not an odd number. Please try again.')
else:
print('Congrats, you know your numbers!')
break #This stops the "while" loop when the "else" condition is met.
I have this simple code:
var = 1
while var == 1 :
try:
num = int(raw_input("Enter a number :"))
except ValueError:
print "Thats not a number!"
continue
try:
num2 = int(raw_input("Enter another number :"))
except ValueError:
print "Thats not a number!"
continue
print "Sum of previous 2 inputs:="+str(num+num2)
print "Good bye!"
Now first continue statement does the job, but the second one, not. Because it goes back at the top of loop, but I need it to go back where second exception was caught, so it would ask to enter second number again, not first number.
Any ideas?
You can factor out entering a number to a function – this spares you writing the same code twice:
def input_int(prompt):
while True:
try:
return int(raw_input(prompt))
except ValueError:
print "That's not a valid integer!"
...
num = input_int("Please enter a number: ")
num2 = input_int("Please enter another number: ")