I am trying to write a program that reads numbers input by the user until the user types done. If the user types a non-number other than "done," I want to return an error message like "please enter a number number. When the user types "done", I want to calculate the total of the numbers, the number count and the average. I have tried to create a while loop with try and except to catch the non-numeric error other than done. That is part of the trick, a string entry is an error unless the string is "done." Here is the beginning of my code without any attempt to create a file that can be totaled, counted and maxed.
bank = 0
number = 0
while True:
try:
number = int(raw_input("Enter an integer ( such as 49 or 3 or 16) \n"))
bank = bank + number
print 'You entered--- ', number, 'Your running total is ', bank
except:
if number == 'done':
print 'Done'
else:
if number == 'done':
print 'Done'
else:
print 'Your entry was non-numberic. Please enter a number.'
bank = number + bank
When I run this and enter "done" I get the "else:" response and a new input line. I do not get the "Done" print from if number == "done"
Answer written in python 3
The exception used is ValueError because the compiler catches this error as a result of the conversion done in line 7 so i just added the continue in line 19 to make it skip the error and go back to the start.
bank = 0
count = 0
while True:
try:
number = input('enter an integer:\n')
if number != 'done':
bank += int(number)
print('you entered -- ', number, 'your total is ', bank)
count += 1
elif number == 'done':
print('Done')
print('you entered %d numbers' % count)
print('Your total is %s' % bank)
average = bank/count
print('Your average is %.02f' % average)
break
except ValueError:
print('oops!! that was not an integer')
continue
Related
I am trying to create password Generate in ruby. At the moment all is working just got stuck at the final piece of generating the password.
I asked user if he/she would like the password to include numbers, lowercase or uppercase.
If YES, user will enter 1 and 0 for NO.
I used the code below to generate password if everything is 1. Meaning user want to include numbers, lowercase and uppercase.
if numbers == 1 && lowercase == 1 && uppercase == 1
passGen = [(0..9).to_a + ('A'..'Z').to_a + ('a'..'z').to_a].flatten.sample(10)
end
p passGen
This works 90% of the time. 10% of the time the generated password will not include say any numbers. But everything else present. I am not sure if this is because of the size or length of Array from which the password is sampled.
Anyway lets go to the main problem below
Here is the problem, I am struggling to write the code to generate password if one or more of input is 0. That's if user don't want to include numbers. Or no numbers and uppercase etc . As I can't predict what user may want or not want. I need help on this please.
Thank you.
You will need to make your input array more dynamic:
passGen = []
passGen += (0..9).to_a if numbers == 1
passGen += ('A'..'Z').to_a if uppercase == 1
passGen += ('a'..'z').to_a if lowercase == 1
passGen.sample(10).join
Now, to tackle your other issue with missing characters - this is caused as you are simply taking 10 random characters from an array. So it can just take, for example, all digits.
To tackle this you need to get one character from each generator first and then generate the remaining characters randomly and shuffle the result:
def generators(numbers:, lowercase:, uppercase:)
[
(0..9 if numbers),
('A'..'Z' if uppercase),
('a'..'z' if lowercase)
].compact.map(&:to_a)
end
def generate_password(generators:, length:, min_per_generator: 1)
chars = generators.flat_map {|g| Array.new(min_per_generator) { g.sample }}
chars += Array.new(length - chars.length) { generators.sample.sample }
chars.shuffle.join
end
gens = generators(numbers: numbers == 1, uppercase == 1, lowercase: lowercase == 1)
Array.new(10) { generate_password(generators: gens, length: 10) }
The code doesn't know it needs to include a digit/letter from every group. The sample takes random signs and since you a basically sampling 0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz there is a possibility that all the signs will not be digits.
The easiest way to fix it is to check if a sign from every group is in the "password" and then replace a random sign with a sign from group that is not present.
If I were to program this I would do it like that
def random_from_range(range)
range.to_a.sample.to_s
end
def passGen(numbers, lowercase, uppercase)
result = ''
possibleSigns = []
if numbers == 1
range = (0..9)
result += random_from_range(range)
possibleSigns += range.to_a
end
if lowercase == 1
range = ('A'..'Z')
result += random_from_range(range)
possibleSigns += range.to_a
end
if uppercase == 1
range = ('a'..'z')
result += random_from_range(range)
possibleSigns += range.to_a
end
desired_lenth = 10
while result.length < desired_lenth
result += possibleSigns.sample.to_s
end
result
end
puts passGen(1,1,1)
By saying (0..9).to_a + ('A'..'Z').to_a + ('a'..'z').to_a, you're creating an Array of 10 + 26 + 26 = 62 elements, and then you pick only 10 elements out of it.
At your place I'd wrap password generation around an until block:
def generate_password_with_digits_and_caps
[(0..9).to_a + ('A'..'Z').to_a + ('a'..'z').to_a].flatten.sample(10).join
end
passGen = ''
until passGen.match(/[A-Z]/) && passGen.match(/[a-z]/) && passGen.match(/\d/)
passGen = generate_password_with_digits_and_caps
end
This could also work (closer to your snipppet):
if numbers == 1 && lowercase == 1 && uppercase == 1
passGen = ''
until passGen.match(/[A-Z]/) && passGen.match(/[a-z]/) && passGen.match(/\d/)
passGen = [(0..9).to_a + ('A'..'Z').to_a + ('a'..'z').to_a].flatten.sample(10).join
end
end
Start with something simple and stupid:
passGen = (('0'..'9').to_a.sample(1)+ ('A'..'Z').to_a.sample(1)+('a'..'z').to_a.sample(8).shuffle).join
Technically speaking, this already fulfills your requirement. From the viewpoint of aesthetics and security, the disadvantage here is that the number of upper case characters is always 8. A more elegant solution would be to find three non-zero integers which add up to 10, and can be used as the arguments for the sample call. Also, if no numbers are requested, you simply pass 0 as argument to sample.
Since this exceeds the scope of your question, and I don't even know whether you want to go so far, I don't elaborate on this here further.
I'm an absolute beginner to Python and am tasked with creating a program that does a few things:
Inputs employee names into a list.
Inputs that employee's salary after inputting their name.
Totals the salaries in a list, (2 lists: names[] and salaries[]).
Finds the average salary after totaling.
Prints employees that earn within $5,000 of the average salary (Where I'm stuck).
Please see my code below:
# function to total the salaries entered into the "newSalary" variable and "salaries[]".
def totalSalaries(salaries):
total = 0
for i in salaries:
total += i
return total
# Finds the average salary after adding and dividing salaries in "salaries[]".
def averageSalaries(salaries):
l = len(salaries)
t = totalSalaries(salaries)
ave = t / l
return ave
# Start main
def main():
# Empty names list for "name" variable.
names = []
# Empty salaries list for "salary" and "newSalary" variables.
salaries = []
# Starts the loop to input names and salaries.
done = False
while not done:
name = input("Please enter the employee name or * to finish: ")
salary = float(input("Please enter the salary in thousands for " + name + ": "))
# Try/except to catch exceptions if a float isn't entered.
# The float entered then gets converted to thousands if it is a float.
try:
s = float(salary)
# Message to user if a float isn't entered.
except:
print("Please enter a valid float number.")
done = False
newSalary = salary * 1000
# Break in the loop, use * to finish inputting Names and Salaries.
if name == "*":
done = True
# Appends the names into name[] and salaries into salaries[] if * isn't entered.
# Restarts loop afterwards if * is not entered.
else:
names.append(name)
salaries.append(newSalary)
# STUCK HERE. Need to output Names + their salaries if it's $5,000 +- the total average salary.
for i in range(len(salaries)):
if newSalary is 5000 > ave < 5000:
print(name + ", " + str(newSalary))
# Quick prints just to check my numbers after finishing with *.
print(totalSalaries(salaries))
print(averageSalaries(salaries))
main()
Any info is greatly appreciated. I hope the rest of the functions and logic in this program makes sense.
Your haven't written your iterator correctly. With an array, you can just use for element in array: and the loop will iterate over array by putting each element in element. So your for loop becomes for salary in salaries.
Also, You need to split your condition in two and use additions and substractions. Your code should check if the salary is higher or equal to the average minus 5000 and if it is lower or equal to the average plus 5000. If you want to formalize this mathematically it would be:
Salary >= Average - 5000
and
Salary <= Average + 5000
So the condition at line becomes if salary >= (average - 5000) and salary <= (average + 5000)
Finally, you do not call averageSalaries before entering the loop, so the average salary haven't been computed yet. You should call the function and put the result in a variable before your for loop.
I'm creating a method where the user puts in a "PIN" number to access different methods.. Yes it's an ATM..
I'm trying to figure out how to make it so that the user gets a specific amount of tries until the program exit's..
After a bunch of research I haven't very well found anything useful.. The only thing I've found is to use .to_s.split(//) in order to add the number of the try into an empty array. Which doesn't make sense to me because why would you make an integer into a string..?
So my question is, how do you make it so that users only have a certain amount of tries, 3, until they get kicked out of the program?
Main source:
#!/usr/bin/env ruby
################
#ATM Rewrite
#
#Creator Lost Bam Not completed yet.
#
#11/19/15
##################
require_relative 'checking.rb'
require_relative 'savings.rb'
require_relative 'exit.rb'
require_relative 'loan.rb'
require_relative 'transfer.rb'
require_relative 'redirect.rb'
class ATM
attr_accessor :name, :checking_account, :savings_account, :pin_number, :transfer, :loan
def initialize( name, checking_account, savings_account )
#name = name
#checking_account = checking_account
#savings_account = savings_account
#pin_number = pin_number
end
end
##############
def pin
x = []
puts "Enter PIN Number:"
input = gets.chomp.to_i
if input == 1234
menu
else
x += 1.to_s.split(//) #<= This is what I found to convert integer to Array
puts "Invalid PIN, try again:"
input = gets.chomp
if x == 3
bad_pin
end
end
end
############
def menu #add #{name} on line 41
puts <<-END.gsub(/^\s*>/, ' ')
>
>Welcome thank you for choosing Bank of Bam.
>You may choose from the list below of what you would like to do
>For checking inquiries press '1'
>For savings account information press '2'
>To transfer funds press '3'
>To either apply for a loan, or get information on a loan press '4'
>To exit the system press '5'
>
END
input = gets.chomp
case input.to_i
when 1
checking_information
when 2
savings_information
when 3
transfer_funds
when 4
loan_info
when 5
exit_screen
else
puts "Invalid option please try again"
menu
end
end
def bad_pin
abort('Invalid PIN exiting sytem..')
exit
end
pin
Tried something new:
def pin
x = 3
puts "Enter PIN(#{x} attempts left):"
pin_num = gets.chomp
case pin_num.to_i
when 1234
menu
else
puts "Invalid PIN"
x -=1
return pin
if x == 0
bad_pin
end
end
end
It doesn't increment the number down it just keeps saying 3 tries left:
Enter PIN(3 attempts left):
4567
Invalid PIN
Enter PIN(3 attempts left):
45345
Invalid PIN
Enter PIN(3 attempts left):
6456
Invalid PIN
Enter PIN(3 attempts left):
4564
Invalid PIN
Your problem is that every time you recall the method the value of x resets again. You need to have a loop inside the pin method that'll keep track of attempts.
def pin
x = 3
while (x > 0) do
puts "Enter PIN(#{x} attempts left):"
pin_num = gets.chomp
case pin_num.to_i
when 1234
menu
else
puts "Invalid PIN"
x -=1
puts "no tries left" if x == 0
break if x == 0
end
end
end
Stay in the method. Recalling the method starts you back at three attempts.
import sgenrand
# program greeting
print("The purpose of this exercise is to enter a number of coin values")
print("that add up to a displayed target value.\n")
print("Enter coins values as 1-penny, 5-nickel, 10-dime,and 25-quarter.")
print("Hit return after the last entered coin value.")
print("--------------------")
#print("Enter coins that add up to 81 cents, one per line.")
total = 0
#prompt the user to start entering coin values that add up to 81
while True:
final_coin= sgenrand.randint(1,99)
print ("Enter coins that add up to", final_coin, "cents, on per line")
user_input = int(input("Enter first coin: "))
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
total = total + user_input
while total != final_coin:
user_input = int(input("Enter next coin:"))
total = total + user_input
if user_input == input(" "):
break
if total > final_coin:
print("Sorry - total amount exceeds", (final_coin))
if total < final_coin:
print("Sorry - you only entered",(total))
if total== final_coin:
print("correct")
goagain= input("Try again (y/n)?:")
if goagain == "y":
if goagain == "n":
print("Thanks for playing ... goodbye!" )
I've been trying to create this loop, so it can repeat the whole program at the end when the user accepts/ if he accepts to do it again at the end.
I know you have to have a while statement around your whole program, but with my while true statement at the top, it only repeats the first part of my program and not the whole thing.
Well one thing you should be careful of is that you do not set the
total = 0
at the start of each loop. So when the user plays the game again. He will continue using his previous total. You should move total = 0 to the start of the loop
while True:
total = 0
Additionally, you need to deindent you first
while True
statement as it does not align properly with the rest of your code.
Finally, you need to allow the user to exit the while loop after he selects No for trying again.
if goagain == "n":
print("Thanks for playing ... goodbye!" )
break
This can be done by applying a break statement.
I am working on a homework assignment that a user will input a grade percentage and it will output a letter grade. My issue is that I would like to restrict the user to only the number keys and a decimal point. If the user inputs anything else they will be prompted with an error message and will have a chance to input again. Here is my code that will work without the decimal, but I need the int to be float. Please help! Any feedback will be greatly appreciated!!
def percentLoop()
while True:
a = input('Enter a percent: ')
try:
number = int(a)
if (0< number <= 100):
return number
else:
print ('Enter a percent between 0 and 100.')
except:
print ('Please enter a percent between 0 and 100.')
Thanks for looking at what I have.
I haven't done python for ages, but is it just a matter of changing number = int(a) to number= float(a) ?