Python: Improve this by making the algorithm less complex

Question

I have a working block of code, but it seems there should be a more efficient algorithm, presumably with less loops or using a library/module.

This example version of the code takes a list of strings, reverse sort by len(), then build a new list:

gs = ["catan", "ticket to ride", "azul"]

mg = {}
for i in range(len(gs)):
        mg[i] = len(gs[i])

popularity = {k: v for k, v in sorted(mg.items(), key=lambda v: v[1], reverse=True)}
tg = []
for game in popularity.keys():
        tg.append(gs[game])

The production code does not set mg[i] to len() and the elements in the list are not necessarily strings, but the algorithm works otherwise.

For this, the same output is:

['ticket to ride', 'catan', 'azul']

How can I improve the efficiency of this algorithm?

Sort the list by the criteria largest number to smallest number. In this example the number is len(mg[i]). I'll add sample output — James Risner
– James Risner, Commented Jan 4, 2023 at 17:09

AKX · Accepted Answer · 2023-01-04 17:23:04Z

5

gs = ["catan", "ticket to ride", "azul"]
tg = sorted(gs, key=len, reverse=True)

has the same effect for a list of strings.

If you want to sort the list in place for more space efficiency,

gs.sort(key=len, reverse=True)

This also works for a custom "criteria function":

def my_criteria_function(value):
    return len(value) * value.count(" ")  # super secret sauce

...sort(key=my_criteria_function) # (or sorted(..., key=my_criteria_function)

edited Jan 4, 2023 at 17:23

answered Jan 4, 2023 at 17:10

AKX

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3 Comments

James Risner Over a year ago

len is the same code, production doesn't use len()

AKX Over a year ago

It doesn't matter; substitute that function for len then.

James Risner Over a year ago

Confirmed in production. Thanks.

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