I am a beginner at python and I'm struggling with one of my (simple) college assignments. I have been given the following instructions:
A bank is offering a savings account where a yearly fee is charged. Write
a program that lets the user enter
An initial investment.
The yearly interest rate in percent.
The yearly fee.
the program should then calculate the time it takes
for the investment to double. The interest is added on once per year.
An example run of the program:
Enter the investment: 1000
Enter the interest rate: 10
Enter the fee: 10
The investment doubles after 7 years.
I have formulated the following code but am receiving an error message with regards to t. I would really appreciate if I could get some help, thanks!:
t=0
p=float(input("Enter the investment:"))
a=float(input("Enter the interest rate:"))
m=float(input("Enter the fee:"))
i=(float(a/100))
f=p
while f<=(2*p):
f=(float(f*((1+i)**t)-m)
t=t+1
print("The investment doubles after",t,"years")
I tried to write this in a way that was very easy to follow and understand. I edited it with comments to explain what is happening line by line. I would recommend using more descriptive variables. t/p/a/m/f may make a lot of sense to you, but going back to this program 6 months from now, you may have issues trying to understand what you were trying to accomplish. NOTE You should use input instead of raw_input in my example if using Python 3+. I use 2.7 so I use raw_input.
#first we define our main function
def main():
#investment is a variable equal to user input. it is converted to a float so that the user may enter numbers with decimal places for cents
investment = float(raw_input("Starting Investment: "))
#interest is the variable for interest rate. it is entered as a percentage so 5.5 would equate to 5.5%
interest = float(raw_input("Interest Rate as %, ex: 5.5 "))
#annual_fee is a variable that will hold the value for the annual fee.
annual_fee = float(raw_input("Annual Fee: "))
#years is a variable that we will use with a while loop, adding 1 to each year (but we wait until within the loop to do this)
years = 1
#we use a while loop as opposed to a for loop because we do not know how many times we will have to iterate through this loop to achieve a result. while true is always true, so this segment is going to run without conditions
while True:
#this is a variable that holds the value of our total money per year, this is equal to the initial investment + investment * interest percentage - our annual fee per year
#I actually had to try a few different things to get this to work, a regular expression may have been more suited to achieve an interest % that would be easier to work with. do some research on regular expressions in python as you will sooner or later need it.
total_per_year = investment + (years * (investment * (interest / 100))) - (annual_fee * years)
#now we start adding 1 to our years variable, since this is a while loop, this will recalculate the value of total_per_year variable
years += 1
#the conditional statement for when our total_per_year becomes equal to double our initial investment
if total_per_year >= 2 * investment:
#print years value (at time condition is met, so it will be 5 if it takes 5 years) and the string ' Years to Double Investment'
print years,' Years to Double Investment'
#prints 'You will have $' string and then the value our variable total_per_year
print 'You will have $', total_per_year
#this will break our while loop so that it does not run endlessly
break
#here is error handling for if the fee is larger than investment + interest
if (years * annual_fee) >= (years * (investment * (interest / 100))):
print('Annual Fee Exceeds Interest - Losing Money')
break
#initial call of our main function/begins loop
main()
Related
I have a problem I need to calculate future tuition depending on the input that the user puts. So for example the tuition is 5,000 per year and increases by 7% every year. If a user inputs 6 the program should print the total cost of tuition for years six, seven, eight and nine. So far I have this.
year = 1
n = int(input())
tuition = 5000
for i in range (n,n + 3):
tuition = tuition * 1.07
year = year + 1
print (tuition)
Regardless of what the user inputs, your for loop will run through 3 iterations.
It looks like you're trying to add a 7% increase (compound) every year for 3 years.
You don't need a loop for that.
e.g.,
tuition = 5_000
years = 3
increase = 1.07
print(tuition * increase ** years)
Output:
6125.215
A few hints
The problem with your program is that you are looping using a for loop but you are not using the value i within your for loop. You should use it. Additionally think about how many times your print() statement is executed. Only once, but you need it for every year so think about moving it into the for loop as well.
One other thing: input() takes a string as a parameter so you can provide some description of what a user is supposed to enter. You should use this as well to make your program usable.
n = int(input("Please enter a start year:"))
I came across a MIT open source Python coding practice as
Suppose you want to be able to afford the down payment in three years. How much should you save each month to achieve this? In this problem, you are going to write a program to answer that question. To simplify things, assume:
1. Your semi-annual raise is .07 (7%)
2. Your investments have an annual return of 0.04 (4%)
3. The down payment is 0.25 (25%) of the cost of the house.
4. The cost of the house that you are saving for is $1M.
You are now going to try to find the best rate of savings to achieve a down payment on a $1M house in 36 months. Since hitting this exactly is a challenge, we simply want your savings to be within $100 of the required down payment. Write a program to calculate the best savings rate, as a function of your starting salary. You should use [bisection search] to help you do this efficiently. You should keep track of the number of steps it takes your bisections search to finish. Limit floats to two decimals of accuracy (i.e., we may want to save at 7.04% - or 0.0704 in decimal - but we are not going to worry about the delta between 7.041% and 7.039%). This means we can search for an integer between 0 and 10000 (using integer division), and then convert it to a decimal percentage (using float division) to use when we are calculating the current_savings after 36 months. Using this range gives us only a finite number of numbers that we are searching over, as opposed to the infinite number of decimals between 0 and 1. This range will help prevent infinite loops. The reason we use 0 to 10000 is to account for two additional decimal places in the range 0% to 100%. Your code should print out a decimal (e.g. 0.0704 for 7.04%).
Keep in mind that it may not be possible to save on a down payment in a year and a half for some salaries. In this case your function should notify the user that it is not possible to save for the down payment in 36 months with a print statement.
EXAMPLE OUTPUTEnter the starting salary: 150000Best savings rate: 0.4411Steps in bisection search: 12
The following is one of the solutions that I found.
# user input
annual_salary = float(input('Enter your annual salary: '))
# static variables and initializers
semi_annual_raise = 0.07
r = 0.04
portion_down_payment = 0.25
total_cost = 1000000
steps = 0
current_savings = 0
low = 0
high = 10000
guess_rate = (high + low)//2
# Use a while loop since we check UNTIL something happens.
while abs(current_savings - total_cost*portion_down_payment) >= 100:
# Reset current_savings at the beginning of the loop
current_savings = 0
# Create a new variable for use within the for loop.
for_annual_salary = annual_salary
# convert guess_rate into a float
rate = guess_rate/10000
# Since we have a finite number of months, use a for loop to calculate
# amount saved in that time.
for month in range(36):
# With indexing starting a zero, we need to calculate at the beginning
# of the loop.
if month % 6 == 0 and month > 0:
for_annual_salary += for_annual_salary*semi_annual_raise
# Set monthly_salary inside loop where annual_salary is modified
monthly_salary = for_annual_salary/12
# Calculate current savings
current_savings += monthly_salary*rate+current_savings*r/12
# The statement that makes this a bisection search
if current_savings < total_cost*portion_down_payment:
low = guess_rate
else:
high = guess_rate
guess_rate = (high + low)//2
steps += 1
# The max amount of guesses needed is log base 2 of 10000 which is slightly
# above 13. Once it gets to the 14th guess it breaks out of the while loop.
if steps > 13:
break
# output
if steps > 13:
print('It is not possible to pay the down payment in three years.')
else:
print('Best savings rate:', rate)
print('Steps in bisection search:', steps)
Why is it necessary to reset variable value current_savings and to create a for_annual_salary before the FOR loop? The current_savings has already been defined as 0 at the beginning, and why does it creates a brand new variable for_annual_salary instead of using annual_salary in FOR loop?
If you don't assign annual_salary to a variable and your loop changed the value of it then you can access its previous value which user entered.
eg:
annual_salary = input("Enter salary")
annual_salary = 10000 + 1000
print(annual_salary ) #would give you 11000, And you got some bug in your code and you want to debug what user entered the salary.
You print("annual_salary") #And you get updated salary not what user entered.
If you created a variable like below:
annual_salary = input("Enter salary")
new_annual_salary = 10000 + 1000 # some operations
print(new_annual_salary ) #Every operation was performed on new variable and if code goes wrong somewhere,you still can find new and old value of annual_salary.
print(annual_salary)
Same for current_savings
And you need to set those variables to initially 0 because when you performing operation you need to provide some number/initial value.
c = 0
for i in range(3):
c+= i
print(c) # You will get 3 not 0
d #If you don't initialize your value and performing operations you'll get undefined error
d+= 1
print(d)
NameError: name 'd' is not defined
I am trying to create a program that asks the user their principal, interest rate, and total amount of years. I want my program to show them the amount of total return they would expect for each year. I want it to start at year 1. When I run my script, it only shows one year's worth total interest. Here is what I have so far.
#Declare the necessary variables.
princ = 0
interest = 0.0
totYears = 0
year = 1
#Get the amont of principal invested.
print("Enter the principal amount.")
princ = int(input())
#Get the interest rate being applied.
print("Enter the interest rate.")
interest = float(input())
#Get the total amount of years principal is invested.
print ("Enter the total number of years you're investing this amonut.")
totYears = int(input())
for years in range(1, totYears):
total=year*interest*princ
years += 1
print (total)
Thank you any help is appreciated.
There are problems here:
for years in range(1, totYears):
total=year*interest*princ
years += 1
print (total)
You change years within the loop. Don't. The for statement takes care of that for you. Your interference makes years change by 2 each time through the loop.
Every time through the loop, you throw away the previous year's interest and compute a new one. Your print statement is outside the loop, so you print only the final value of total.
Your loop index is years, but you've computed on the variable year, which is always 1. A programming technique I picked up many years ago is never to use a plural variable name.
Perhaps you need this:
for years in range(1, totYears):
total = years * interest * princ
print (total)
def main():
#Get amount of principal, apr, and # of years from user
princ = eval(input("Please enter the amount of principal in dollars "))
apr = eval(input("Please enter the annual interest rate percentage "))
years = eval(input("Please enter the number of years to maturity "))
#Convert apr to a decimal
decapr = apr / 100
#Use definite loop to calculate future value
for i in range(years):
princ = princ * (1 + decapr)
print('{0:5d} {0:5d}'.format(years, princ))
I'm trying to print the years and the principal value in a table, but when I print all that comes out is two columns of 10.
So you have several problems. The first problem is a display issue.
Your output statement print('{0:5d} {0:5d}'.format(years, princ)) has several issues.
printing years instead of i, so it's always the same value instead of incrementing
the 0 in the format statement{0:5d} means the 0'th element out of the following values, so you're actually printing years twice, the second one should be 1 instead of 0
you're using d to print what should be a floating point value, d is for printing integers, you should be using something along the lines of {1:.2f} which means "print this number with 2 decimal places
Once you've corrected those you'll still see incorrect answers because of a more subtle problem. You're performing division with integer values rather than floating point numbers, this means that any decimal remainders are truncated, so apr / 100 will evaluate to 0 for any reasonable apr.
You can fix this problem by correcting your input. (As a side note, running eval on user input is usually an incredibly dangerous idea, since it will execute any code that is entered.) Instead of eval, use float and int to specify what types of values the input should be converted to.
The following is corrected code which implements the above fixes.
#Get amount of principal, apr, and # of years from user
princ = float(input("Please enter the amount of principal in dollars "))
apr = float(input("Please enter the annual interest rate percentage "))
years = int(input("Please enter the number of years to maturity "))
#Convert apr to a decimal
decapr = apr / 100
#Use definite loop to calculate future value
for i in range(years):
princ = princ * (1 + decapr)
print('{0} {1:.2f}'.format(i, princ))
Having some trouble grasping why this "quick math" formula I was taught in high school does not seem to work correctly.
The premise is to take your hourly salary, double it and add three Zeros, the result will roughly equate to your yearly salary if you work full time 50 weeks out of the year.
# Preface
print '---> Want to know your yearly salary? <---'.upper()
# Question
money = raw_input("How much money do you earn per hour?")
# Math Work
mult = money * 2
result = mult + str(000)
# Answer
print "you make roughly $%r per year, Working full-time for 50 weeks out of the year" % result
Result:
my result looks something like this: "you make roughly $10100 per year, working full-time for 50 weeks out of the year"
I must be making a mistake in my expression...Simply put, I just do not know
You got all the types wrong.
raw_input acquires a string, so money is acquired as such. Thus, when you do mult=money*2 you are not doubling a number, but a string; writing money*2 thus has the effect of creating a string that is the concatenation of two copies of the string you provided. If you enter 10, mult will be '1010'.
Also, in str(000) 000 is an integer, so it's actually a plain 0; str(000) thus results in '0', which is concatenated to your doubled-string. 1010 concatenated with '0' => 10100.
What you actually want is
# Question
money = int(raw_input("How much money do you earn per hour?"))
# Math Work
mult = money * 2
result = str(mult) + "000"
By the way, adding zeroes and the like is fine for humans, but since we are dealing with a computer you can just multiply by 2000:
result = 2000*int(raw_input("How much money do you earn per hour?"))
You're trying to do math with a string. Convert it into an integer first:
money = int(raw_input("How much money do you earn per hour?"))
and multiply instead of trying to add a string to the end
result = money * 2000
Though if you really wanted to, you could convert the integer back to a string to add 3 zeros to the end:
mult = money * 2
strmult = str(mult)
result = strmult + '000'
The raw_input() function returns a string.
When you multiply money by a number, instead of multiplying the integer value, you are multiplying the string value. This results in the variable's new value being a multiple of the string, or the string repeated multiple times. I would suggest using the money=int(money) function on money to turn it into an integer, or better yet money=float(money) to get a floating-point number.
try this
money=int(input('how much you make an hour'))
final_yearly=money*2000
print(final_yearly)
You do realize the following would give you the desired answer, right?
#Math Work
mult = money * 2000
First, money is a string, when you read user input. So when the user inputs 10, you get '10'.
So when you do money*2, you don't get the expected 20. Rather, you get '10'*2, which is '10' concatenated twice, i/e/ '1010'.
Next, 000 is an int that evaluates to 0, the str of which is '0'. What you wanted to add is '000'
I would go about your task this way:
# Preface
print '---> Want to know your yearly salary? <---'.upper()
# Question
money = int(raw_input("How much money do you earn per hour?"))
# Math Work
mult = money * 2
result = str(mult) + "000"
Alternatively, you could do this as well:
# Preface
print '---> Want to know your yearly salary? <---'.upper()
# Question
money = int(raw_input("How much money do you earn per hour?"))
# Math Work
result = money*2000 # because adding three 0s is the same as multiplying by 1000
# Preface
print '---> Want to know your yearly salary? <---'.upper()
# Question
money = raw_input("How much money do you earn per hour?")
# Math Work
result = str(int(money)*2) + '000'
# Answer
print "you make roughly $%r per year, Working full-time for 50 weeks out of the year" % result