CODEWITHSUNDEEP
pythonif-statementrefactoring
Andy
Andy

Reputation: 89

Python: Solve this messy if else structure better

I've been stuck on this for a while. I haven't actually written this code yet, but if I were to it would look something like this;

i = 0  # this is an int I get from somewhere else
s = "" # this is the string that I ultimately want to use for something
if i < 1:
    s = "string1"
elif i < 5:
    s = "string2"
elif i < 17:
    s = "string3"

and so on. Basically I want to assign a different string to assign different strings to different intervals and depending on which interval i falls in, I assign the corresponding string to s.

The example above is really messy and looks like a pain to expand/change. I was thinking this could be somehow solved with a dict, but I can't quite seem to figure it out.

Upvotes: 0

Views: 110

Answers (3)

mcsoini
mcsoini

Reputation: 6642

If you define a dictionary with the limit values in the right order you can just pull the next-higher element using a simple generator comprehension:

map_max_string = {1: "string1", 5: "string2", 17: "string3"}
s = next(val for key, val in map_max_string.items() if i < key)

If you don't want to rely on the assumption that the dict is ordered, you can of course still sort it by the key:

s = next(val for key, val in sorted(map_max_string.items()) 
         if i < key)

Upvotes: 1

user_na
user_na

Reputation: 2273

Since python 3.7 dicts are ordered. So you could define your thresholds as key of a dict and link them to a string:

D = {1: 'string1',
     5: 'string2',
     17: 'string3'}

x = 10
s = ""
for i in D.keys():
    if x < i:
        s = D[i]
        break

This way you have the definitions close by and save some if.

Upvotes: 3

Robert Kwiatkowski
Robert Kwiatkowski

Reputation: 461

Without providing any details it's hard to give you any advice. However if you want "pythonic oneliner" here you go:

s = "string1" if i < 1 else ("string2" if i < 5 else "string3")

Upvotes: 0

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