Question

Sure I could write this myself, but before I go reinventing the wheel is there a function that already does this?

Answer 1

Given an instance x of datetime.date, (x.month-1)//3 will give you the quarter (0 for first quarter, 1 for second quarter, etc — add 1 if you need to count from 1 instead;-).

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Originally two answers, multiply upvoted and even originally accepted (both currently deleted), were buggy — not doing the -1 before the division, and dividing by 4 instead of 3. Since .month goes 1 to 12, it’s easy to check for yourself what formula is right:

for m in range(1, 13):
  print m//4 + 1,
print

gives 1 1 1 2 2 2 2 3 3 3 3 4 — two four-month quarters and a single-month one (eep).

for m in range(1, 13):
  print (m-1)//3 + 1,
print

gives 1 1 1 2 2 2 3 3 3 4 4 4 — now doesn’t this look vastly preferable to you?-)

This proves that the question is well warranted, I think;-).

I don’t think the datetime module should necessarily have every possible useful calendric function, but I do know I maintain a (well-tested;-) datetools module for the use of my (and others’) projects at work, which has many little functions to perform all of these calendric computations — some are complex, some simple, but there’s no reason to do the work over and over (even simple work) or risk bugs in such computations;-).