Write Shorter Conditionals (Using Dictionaries) | Python Snippets #4

In the "Python Snippets" series we will cover a variety of topics around Python programming in a byte-sized format.

Each video covers a single topic that you might or might not be familiar with, and is meant to be a hands-on part of your road to being a better Python programmer.

This week: I'll teach you how to write shorter conditionals using dictionaries!

VIDEO CONTENT
---------------------------------------------------
0:00 - Intro
0:16 - Benefits and drawbacks
0:36 - Example case
0:50 - Using If-Else and Match-Case
1:10 - Dictionary Conditional
2:02 - Speed Comparison (1)
2:33 - Speed Comparison (2)

CREDITS
---------------------------------------------------
Introduction Audio: Dion P. van der Geest
Background Audio: Royalty Free Music from Bensound

Sign icon: Kiranshastry (from Flaticon)
Checkmark icon: Octopocto (from Flaticon)
Stop & Thumbs Down Icon: Freepik (from Flaticon)

Dictionary photo: Mega Pixel, Shutterstock

[appuser]

For this month example, declaring the dictionary as a global constant is the correct approach in my view. It is external and constant data that does not change. Perhaps this would all be better wrapped as a class, or passing a config object which has the dictionary declared as a member if it helps with encapsulation which ensures we do not redeclare the dictionary each time.

[cwatson]

I love the .get() method and how you can give a second arg as a fallback value. I'm curious though, how are you measuring the runtime? Those numbers seem a bit unrealistic. Maybe I'm missing something?

[dang]

You can also memoize the dict if you want to keep it in the function

[lawrencedoliveiro]

Note that what you have here is not a switch statement, but a switch _expression_ . You have something you can use inside an expression, which is not true of if-statements and match-statements.

Did you try it in the lazy-evaluation form?

{
key1 ∶ lambda ∶ case_key1_expr,
key2 ∶ lambda ∶ case_key2_expr,
key3 ∶ lambda ∶ case_key3_expr,
․․․
}․get(index, default_expr)]()

[ruudvanderham]

Thijmen,
There's an even faster (although not generic) solution:

def month(idx):
try:
return ("Not a month", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December")[(idx >= 1) and idx]
except Exception:
return "Not a month"

Without the check for negative and non integer values it is very elegant, IMHO:

def month(idx):
return ("Not a month", "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December")[idx]

[federicomuciaccia]

I think you can also use an Enum instead of a dict

[Simple_Simon_UK]

Why not use the built in functions of datetime:

import datetime

#provide month number
month_num = "3"
datetime_object = datetime.datetime.strptime(month_num, "%m")

month_name =
print("Short name: ", month_name)

full_month_name =
print("Full name: ", full_month_name)

[miguelvasquez]

Why to declare de dict outside is faster?

[abdelghafourfid]

make it even faster :
_get_month = dt.get
def month(idx):
return _get_month(idx, "not a month")