Use raw_input to have user input variable name and use its value

Question

I am creating a financial modeling tool that a user can use. I want the user to be able to be able to enter what method they want to use by typing the variable name. However, with the code I have, the raw_input just gives out a string of what they entered, not the variable name itself

loose = 5
accurate = 10
extreme = 15

method = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ") #the user would either type loose, accurate, or extreme

calculation = x*y*method #x and y is just a generic calculation and method should be equal to either 5, 10, or 15 based on whether the user chose loose, accurate, or extreme

I get an error that says method is not an int. What can I do to have it equal to the loose, accurate, or extreme values above?

MrAlexBailey · Accepted Answer · 2015-03-24 19:29:04Z

5

Change your variables into a dictionary and then use it to lookup the corresponding integer based on the user's input:

values = {'loose': 5,
          'accurate': 10,
          'extreme': 15}

method = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ') #the user would either type loose, accurate, or extreme

calculation = x*y*values[method] #x and y is just a generic calculation and method should be equal to either 5, 10, or 15

You can also use values.get(method, 0) in place of values[method], if the users enters something that is not in your dictionary then 0 will be used.

edited Mar 24, 2015 at 19:29

answered Mar 24, 2015 at 19:25

MrAlexBailey

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2 Comments

lsowen Over a year ago

This is a good way to do it. You will just need to make sure to handle the situation where someone puts in a method that doesn't match one of your choices.

MrAlexBailey Over a year ago

@Isowen Good point; I edited to show use of values.get(method, 0) which would handle the case of an input not being in the dictionary.

lsowen · Accepted Answer · 2015-03-24 19:30:15Z

1

~~The best~~One way would be something like:

loose = 5
accurate = 10
extreme = 15

method_selection = raw_input('Would you like the calculation to be loose, accurate, or extreme in its precision?: ')

if method_selection == "loose":
    method = loose
elif method_selection == "accurate":
    method = accurate
else:
    method = extreme

calculation = x*y*method

Adjust the order of the if statements to give you the proper "default" (ie someone puts in a method that doesn't match one of yours).

edited Mar 24, 2015 at 19:30

answered Mar 24, 2015 at 19:26

lsowen

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2 Comments

matt Over a year ago

The best way? I initially thought of this as well, but using a dictionary as per Jkdc's answer is much better. You can simply add keys to the dictionary instead of adding if/else code. Not to mention the simplicity and ability to add/remove keys at run-time.

lsowen Over a year ago

I agree, probably not the best way. Updating to reflect this.

Eli Korvigo · Accepted Answer · 2015-03-24 19:38:58Z

0

A direct solution is to use eval.

method = eval(raw_input(...))

Note, that it's not recommended to use eval, since it sets a user free to execute whatever he/she wants.

edited Mar 24, 2015 at 19:38

answered Mar 24, 2015 at 19:31

Eli Korvigo

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2 Comments

lsowen Over a year ago

While a potential method, it is usually best to avoid eval if at all possible.

Eli Korvigo Over a year ago

@ Yeah, I've added that to the answer along with an explanation. Yet, it is an option one should know.

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