CODEWITHSUNDEEP
python
mark_mcnall
mark_mcnall

Reputation: 13

Python: Is there a more efficient way to convert int value of months?

I have a list of integers [1, 2...40, 41...] that represent months of a few years. i.e. 1 is January of year 1, 13 is January of year 2, and so on. When trying to convert into just the month integer (all Januarys to 1, etc.) I'm using month_int % 12 but it turns December into 0. I wrote the following small block to catch the Decembers but was wondering if there's a more efficient way?

for month_int in month_list:
    if not(bool(month_int % 12)):
        # Dec exception here
    else:
        # work as normal

Basically, I want to change lists like this:

[1, 2...12, 13, 14, 15...]
[1, 2...12, 1, 2, 3...]

Upvotes: 1

Views: 92

Answers (3)

user2390182
user2390182

Reputation: 73450

fodma1's answer has you covered, but you can do it even shorter:

month = month % 12 or 12

>>> l = [1, 2, 12, 13, 14, 15]
>>> [m % 12 or 12 for m in l]
[1, 2, 12, 1, 2, 3]

Upvotes: 1

Vltava
Vltava

Reputation: 39

Since there are more months that aren't December it would be slightly more efficient to have that condition first e.g.

for month_int in month_list:
if (bool(month_int % 12)):
    # normal
else:
    # December

Since it will only move to the else once every 12 iterations then.

Upvotes: 0

fodma1
fodma1

Reputation: 3535

You are indexing from 1 instead of zero. What you can do is subtract 1 and then add 1 to the remainder again:

month = (month - 1) % 12 + 1

Upvotes: 3

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