Is there a way to input a decimal into a expected integer string

Question

With my basic interest calculator I keep getting the error

 r = int(raw_input("What is the rate of growth?\n=> "))
 ValueError: invalid literal for int() with base 10: '.52'

the numbers I am using are principal of 20,000 Rate of 52% (.052 is what I put in) compoundings are Semiannually with 3 periods

    import math

    p = int(raw_input("what is the principal value?\n=> "))
    r = int(raw_input("What is the rate of growth?\n=> "))
    c = int(raw_input("How many compoundings per periods are taking place? 
    \n=> "))
    t = int(raw_input("How mant periods are taking place?\n=>  "))

    number = str(p*(1+r/c)**c*t)
    print(number)

edit: sorry for crappy variables it is what is on my homework ;)

Possible duplicate of [Python input processing when input is a float or string](https://stackoverflow.com/questions/35144721/python-input-processing-when-input-is-a-float-or-string) — ivan_pozdeev, Apr 18 '19 at 23:29

score 3 · Answer 1 · answered Apr 18 '19 at 23:29

No; that's a contradiction: an int does not have a decimal point. .052 is not an int; it's a float. Use the proper conversion:

rate = float(raw_input("What is the rate of growth?\n=> "))

Also, use meaningful variable names; single-letter variables tend to gang up and bite you later.

score 2 · Accepted Answer · answered Apr 18 '19 at 23:35

2

Why are you casting the input string as an int if you want a float?

Change

r = int(raw_input("What is the rate of growth?\n=> "))

to

r = float(raw_input("What is the rate of growth?\n=> "))

answered Apr 18 '19 at 23:35

Alec

8,529
8
37
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Is there a way to input a decimal into a expected integer string

2 Answers2