How do I avoid walls of If statements?

Question

I heard several times that coding like this is suboptimal:

if weapon == "sword":
 print("Knight") 
elif weapon == "katana":
 print("Samurai") 
elif weapon == "axe":
 print("Viking")

How do I write such code optimally?

You may now think about accepting an answer or comment one to get details ;) to reward those who spent time for you ;) — azro, Mar 20 '20 at 08:44

score 5 · Accepted Answer · edited Apr 12 '20 at 14:52

5

You can store these associations in a dictionary

weapons_roles = {
    "sword": "Knight",
    "katana": "Samurai",
    "axe": "Viking"
}

Print something, whenever the key is not in the dict
```
print(weapons_roles.get(weapon, "No role"))
```

Print a role only if the weapon is known

if weapon in weapons_roles:
    print(weapons_roles[weapon])

edited Apr 12 '20 at 14:52

marc_s

answered Mar 19 '20 at 12:05

azro

1

This is the best answer since it handles the cases where the weapon might not exist in the dict – Chris Doyle Mar 19 '20 at 12:08
Slight nit: this prints something whether there is a match or not, while the original code only printed something if there was a match (i.e., there was no `else` clause). That can be handled by using a `try` statement to catch (and ignore) a `KeyError` raised by `weapons_roles[weapon]`. – chepner Mar 19 '20 at 12:13
@chepner or just with an if – azro Mar 19 '20 at 12:28
I might continue to use `get` with the `if` statement, to avoid repeating the lookup. `role = weapon_roles.get(weapon); if role is not None` or `if (role := weapon_roles.get(weapon)) is not None:`. This is even a slighter nit, though. – chepner Mar 19 '20 at 12:50

score 1 · Answer 2 · answered Mar 19 '20 at 12:05

1

Try this below :

def example_function(weapon):
    weapon_dict = {'sword': 'Knight', 'katana': 'Samurai', 'axe': 'Viking'}
    return weapon_dict[weapon]

answered Mar 19 '20 at 12:05

Abhishek Kulkarni

2 Answers2