Better coding: simple way to cover all writing possibilities?

Question

I am writing a battleship game to practice. When starting a game, I want to ask a question related to the difficulty of the game, and it goes as follows:

Let's start a new game. Do you want to play right away? The number of battleships will be randomized, as well as their size (y/n)

When coding the answer I want to identify the yes/no question with all writing possibilities (yes, YES, yES, Yes, etc.) but my code gets like:

easy_mode = raw_input("Let's start a new game. Do you want to play right away? The number of battleships will be randomized, as well as their size (y/n)");
if easy_mode == ("yes" or "Yes" or "YES" or "yEs" or "yeS" or "yES" or "YEs" or "YeS"):
(...)

It is indeed ugly and I was wondering whether there was an efficient method to do this.

Not only is it ugly, **it doesn't work**. `('yes' or ...)` evaluates to `'yes'`, so you're really just testing `if easy_mode == 'yes':`. — jonrsharpe, Sep 22 '15 at 09:26
Why not a regular expression? Why English only when you can ask in every language? Too complicated - you've got bigger problems than this to worry about. — duffymo, Sep 22 '15 at 09:26
@duffymo *"Why not a regular expression?"* - because then you have [two problems](http://regex.info/blog/2006-09-15/247)! — jonrsharpe, Sep 22 '15 at 09:29
Your comment made me laugh @jonrsharpe. You are perfectly correct. — duffymo, Sep 22 '15 at 09:30

hjpotter92 · Accepted Answer · 2015-09-22T09:30:54.963

4

Since you're allowing your players to have single character answers as well, put that check in your code too:

if easy_mode.lower() in {'yes','y'}:
    (...)

edited Sep 22 '15 at 09:30

answered Sep 22 '15 at 09:28

hjpotter92

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If you make it a set `{'yes', 'y'}` it's slightly more efficient – jonrsharpe Sep 22 '15 at 09:29

score 2 · Answer 2 · answered Sep 22 '15 at 09:25

2

Try converting it to lowercase and then do checking.

easy_mode = raw_input("Let's start a new game. Do you want to play right away? The number of battleships will be randomized, as well as their size (y/n)");
if easy_mode.lower() == 'yes':

answered Sep 22 '15 at 09:25

Avinash Raj

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score 0 · Answer 3 · edited Sep 22 '15 at 10:06

0

You can just convert easy_mode string to lower-case and after that you will only need to compare it with full-lowercase version of string, like this:

if easy_mode.lower() == 'yes':
    pass

edited Sep 22 '15 at 10:06

GwynBleidD

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answered Sep 22 '15 at 09:26

tobyodavies

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score 0 · Answer 4 · answered Sep 22 '15 at 09:26

0

Yes, you have. You can compare after coverting to the input to lower case or uppercase first.

if easy_mode.lower() == "yes"

Or,

if easy_mode.upper() == "YES"

answered Sep 22 '15 at 09:26

Ahsanul Haque

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Better coding: simple way to cover all writing possibilities?

4 Answers4