Smart Loop List Creation in Python for Django Choice fields

Question

So. The following isn't very 'smart' ;)

MONTHS = (
    ('Jan', 'Jan'),
    ('Feb', 'Feb'),
    ('Mar', 'Mar'),
    ('Apr', 'Apr'),
    ('May', 'May'),
    ('Jun', 'Jun'),
    ('Jul', 'Jul'),
    ('Aug', 'Aug'),
    ('Sep', 'Sep'),
    ('Oct', 'Oct'),
    ('Nov', 'Nov'),
    ('Dec', 'Dec'),
)

YEARS = (
    ('1995', '1995'),
    ('1996', '1996'),
    ('1997', '1997'),
    ('1998', '1998'),
    ('1999', '1999'),
    ('2000', '2000'),
    ('2001', '2001'),
    ('2002', '2002'),
    ('2003', '2003'),
    ('2004', '2004'),
    ('2005', '2005'),
    ('2006', '2006'),
    ('2007', '2007'),
    ('2008', '2008'),
    ('2009', '2009'),
    ('2010', '2010'),
)

I'm newer to python, and would love to produce stuff like this 'pythonically'.

Such as,

a list of year tuples from 1995 to current year.
a list of the abbreviated months of the year

Thanks Stackers'

Perhaps the question is how to produce lists like this, with 'for loops' I suppose. — Daryl, Sep 28 '10 at 05:21
Btw. why are you using tuples consisting of two identical values? — ecik, Sep 28 '10 at 09:51
@ecik: That's how django's choices works in forms/models. See http://www.djangoproject.com/documentation/models/choices/. The first item in the tuple is the actual value, and the second is what gets displayed. — Sam Dolan, Sep 28 '10 at 18:20

ars · Accepted Answer · 2010-09-28T18:48:13.750

16

In [17]: from datetime import datetime

In [18]: tuple((str(n), str(n)) for n in range(1995, datetime.now().year + 1))
Out[18]:
(('1995', '1995'),
 ('1996', '1996'),
 ('1997', '1997'),
 ('1998', '1998'),
 ('1999', '1999'),
 ('2000', '2000'),
 ('2001', '2001'),
 ('2002', '2002'),
 ('2003', '2003'),
 ('2004', '2004'),
 ('2005', '2005'),
 ('2006', '2006'),
 ('2007', '2007'),
 ('2008', '2008'),
 ('2009', '2009'),
 ('2010', '2010'))

In [19]: import calendar

In [20]: tuple((m, m) for m in calendar.month_abbr[1:])
Out[20]:
(('Jan', 'Jan'),
 ('Feb', 'Feb'),
 ('Mar', 'Mar'),
 ('Apr', 'Apr'),
 ('May', 'May'),
 ('Jun', 'Jun'),
 ('Jul', 'Jul'),
 ('Aug', 'Aug'),
 ('Sep', 'Sep'),
 ('Oct', 'Oct'),
 ('Nov', 'Nov'),
 ('Dec', 'Dec'))

edited Sep 28 '10 at 18:48

answered Sep 28 '10 at 05:29

ars

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Calendar module also provides an attribute called month_abbr which is probably more suitable here. – ecik Sep 28 '10 at 09:55
To get the current year you should do `import datetime; datetime.datetime.now().year + 1` as the second arg of the range. – Sam Dolan Sep 28 '10 at 18:19

Annika Backstrom · Answer 2 · 2010-09-29T12:53:45.493

8

Try using zip() to make a list of two-tuples.

MONTHS = ('Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec')
somemonth = models.TextField(max_length=3, choices=zip(MONTHS,MONTHS))

choices will be set to [('Jan', 'Jan'), ('Feb', 'Feb'), ...].

In response to the comments on this answer, the "tuple" list comprehension version would be:

tuple((m, m) for m in MONTHS)

Versus the zip version:

tuple(zip(MONTHS, MONTHS))

But strictly speaking, Django doesn't need a tuple of choices, so:

zip(MONTHS, MONTHS)

edited Sep 29 '10 at 12:53

answered Sep 28 '10 at 18:53

Annika Backstrom

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Hey I like the 'zip()' method... but using the tuple() seems much less work! – Daryl Sep 29 '10 at 08:54
@Daryl I've updated the answer to compare the list comprehension to `zip()`. – Annika Backstrom Sep 29 '10 at 12:56

John Kugelman · Answer 3 · 2010-09-28T05:29:52.743

See module time.

>>> import time

For the months we can use strptime to turn a month number 1-12 into a struct_time, and then use strftime to pull the month name out.

>>> [time.strftime('%b', time.strptime(str(i), '%m')) for i in range(1, 13)]
['Jan', 'Feb', 'Mar', 'Apr', 'May', 'Jun', 'Jul', 'Aug', 'Sep', 'Oct', 'Nov', 'Dec']

The years are more straightforward, we just need to know the current year (plus one, due to the way range works).

>>> [str(i) for i in range(1995, time.localtime().tm_year + 1)]
['1995', '1996', '1997', '1998', '1999', '2000', '2001', '2002', '2003', '2004', '2005', '2006', '2007', '2008', '2009', '2010']

I added str(i) to have the years returned as strings since that's the way you wrote them. If integers are okay then you can drop the whole list comprehension and simply use range(...).

Dr Manhattan · Answer 4 · 2015-10-25T16:32:58.490

1

import datetime
from django.db import models
from django.utils.dates import MONTHS


class MyCoolModel(models.Model):
    YEARS = YEARS = [(i,)*2 for i in range(1995, datetime.today().year + 1)]

    month = models.PositiveSmallIntegerField(choices=MONTHS.items())
    year = models.PositiveIntegerField(choices=YEARS)

edited Oct 25 '15 at 16:32

answered Oct 25 '15 at 16:23

Dr Manhattan

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score 0 · Answer 5 · answered Sep 28 '10 at 15:51

Although some of the other answers might be "smarter", I dislike the idea of using a loop to define a tuple. Perhaps the following is a more readable compromise :

MONTHS = (
    ('Jan',) * 2,
    ('Feb',) * 2,
    ('Mar',) * 2,
    ('Apr',) * 2,
    ('May',) * 2,
    ('Jun',) * 2,
    ('Jul',) * 2,
    ('Aug',) * 2,
    ('Sep',) * 2,
    ('Oct',) * 2,
    ('Nov',) * 2,
    ('Dec',) * 2,
)

YEARS = (
    ('1995',) * 2,
    ('1996',) * 2,
    ('1997',) * 2,
    ('1998',) * 2,
    ('1999',) * 2,
    ('2000',) * 2,
    ('2001',) * 2,
    ('2002',) * 2,
    ('2003',) * 2,
    ('2004',) * 2,
    ('2005',) * 2,
    ('2006',) * 2,
    ('2007',) * 2,
    ('2008',) * 2,
    ('2009',) * 2,
    ('2010',) * 2,
)

Personally, for constants, I rather be more verbose and stick to tuples as oppose to defining a list with list comprehensions. It's more for personal taste and readability than anything else. — Belmin Fernandez, Sep 28 '10 at 18:43

Smart Loop List Creation in Python for Django Choice fields

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