How to check variable against 2 possible values?

I have a variable s which contains a one letter string

s = 'a'

Depending on the value of that variable, I want to return different things. So far I am doing something along the lines of this:

if s == 'a' or s == 'b':
   return 1
elif s == 'c' or s == 'd':
   return 2
else: 
   return 3

Is there a better way to write this? A more Pythonic way? Or is this the most efficient?

Previously, I incorrectly had something like this:

if s == 'a' or 'b':
   ...

Obviously that doesn’t work and was pretty dumb of me.

I know of conditional assignment and have tried this:

return 1 if s == 'a' or s == 'b' ...

I guess my question is specifically to is there a way you can compare a variable to two values without having to type something == something or something == something

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 1

if s in ('a', 'b'):
    return 1
elif s in ('c', 'd'):
    return 2
else:
    return 3

Method 2

 d = {'a':1, 'b':1, 'c':2, 'd':2}
 return d.get(s, 3)

Method 3

If you only return fixed values, a dictionary is probably the best approach.

Method 4

if s in 'ab':
    return 1
elif s in 'cd':
    return 2
else:
    return 3

Method 5

return 1 if (x in 'ab') else 2 if (x in 'cd') else 3

Method 6

Maybe little more self documenting using if else:

d = {'a':1, 'b':1, 'c':2, 'd':2} ## good choice is to replace case with dict when possible
return d[s] if s in d else 3

Also it is possible to implement the popular first answer with if else:

  return (1 if s in ('a', 'b') else (2 if s in ('c','d') else 3))


All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

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