Parsing date string in Python

Question

How can I rewrite following clause:

if u'января' in date_category['title']:
    month = 1
elif u'февраля' in date_category['title']:
    month = 2
elif u'марта' in date_category['title']:
    month = 3
elif u'апреля' in date_category['title']:
    month = 4
elif u'мая' in date_category['title']:
    month = 5
elif u'июня' in date_category['title']:
    month = 6
elif u'июля' in date_category['title']:
    month = 7
elif u'августа' in date_category['title']:
    month = 8
elif u'сентября' in date_category['title']:
    month = 9
elif u'октября' in date_category['title']:
    month = 10
elif u'ноября' in date_category['title']:
    month = 11
elif u'декабря' in date_category['title']:
    month = 12

It just looks ugly.

@Marcin Of course. It is a parser, I have a string with russian dates. I strptime() won't work for these. date-util package too won't work for it. I'm parsing day with simple regexp and now I need to parse month somehow. The next string just takes parsed month and day and creates datetime object. — evgeniuz
– evgeniuz, Commented Apr 8, 2012 at 11:04
Howcome you are checking if the month is in the date_category['title'], is it possible for you to just check the index of the month in a month list? — jamylak
– jamylak, Commented Apr 8, 2012 at 11:07
@Marcin In russian %B for january would be 'январь', and 'January, 23' would be '23 января', so this just won't work. — evgeniuz
– evgeniuz, Commented Apr 8, 2012 at 11:08
@jamylak date_category['title'] contains not only month, but some other information. — evgeniuz
– evgeniuz, Commented Apr 8, 2012 at 11:11
@Shark I'm fairly certain you don't understand how python date parsing works. %B refers only to the month, and will deal with russian dates if you set your locale correctly. — Marcin
– Marcin, Commented Apr 8, 2012 at 11:11

Marcin · Accepted Answer · 2012-04-08 11:22:15Z

7

To solve this, parse your date information using the python datetime module. It has support for locales, and will sort this out. If genitive forms are really the issue, then just map those forms to the dative, then map back on output.

To answer your actual question -

Consider that you are pairing up data with integers. If you can transform your data into the format (month, integer), you can drive your code with data.

for (n, month) in enumerate(u'января, feb, mar, apri, may, jun, jul, aug, sep, oct, nov, декабря'.split(', '), 1):# the parameter 1 specifies to start counting from 1. h/t @san4ez
    if month in date_category['title']: return n

edited Apr 8, 2012 at 11:22

answered Apr 8, 2012 at 11:03

Marcin

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7 Comments

Adam Matan Over a year ago

Consider adding some sort of type check, e.g. return d[title] if d in title else None

San4ez Over a year ago

enumerate(iterable, 1) starts enumerating from 1. So no need to increment n when return value

Marcin Over a year ago

@AdamMatan (a) That is not in OP's code (b) This is a loop. It only returns if something is found. If you fall off the end of the function, you get None anyway.

Adam Matan Over a year ago

@Marcin Perhaps, but IMO explicit is better than implicit. A matter of taste.

Adam Matan Over a year ago

I meant creating the dictionary separately, and querying it. It's a different approach. @Marcin

|

Community · Accepted Answer · 2017-05-23 10:29:09Z

4

Get system support for Russian internationalization: OS locale support for use in Python

Use the locale module to set the locale: How to get unicode month name in Python?

Use strptime: Handling international dates in python

Demonstration: Locale troubles

Alternatively, just do:

months = ['Jan', 'Feb', 'Mar', ...]
monthToNumber = dict((name,i+1) for i,name in enumerate(months))

monthToNumber[date_category['title']]

edited May 23, 2017 at 10:29

CommunityBot

11 silver badge

answered Apr 8, 2012 at 11:06

ninjagecko

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1 Comment

Marcin Over a year ago

But he can't! Python doesn't know about Russia!

jamylak · Accepted Answer · 2012-04-08 11:15:56Z

2

months = (u'января', u'февраля', u'марта', u'апреля', u'мая', u'июня', u'июля', u'августа', u'сентября', u'октября', u'ноября', u'декабря')
month = next(i for i,name in enumerate(months,1) if name in date_category['title'])

edited Apr 8, 2012 at 11:15

answered Apr 8, 2012 at 11:05

jamylak

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Comments

Abhijit · Accepted Answer · 2012-04-08 12:39:54Z

1

You may also use the index function in list to get a mapping of months to months number

months = (u'января', u'февраля', u'марта', u'апреля', u'мая', u'июня', u'июля', u'августа', u'сентября', u'октября', u'ноября', u'декабря')
month = months.index(date_category['title'])+1

edited Apr 8, 2012 at 12:39

answered Apr 8, 2012 at 11:19

Abhijit

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5 Comments

jamylak Over a year ago

The problem is that the date_category['title'] also contains other information so it can't be used as an index.

Abhijit Over a year ago

@jamylak: Never came to my mind. In that case if OP mentions about the format, it can most possibly be reduced to the month name.

Abhijit Over a year ago

Any one who is down-voting, please leave a comment.

Joel Cornett Over a year ago

@Abhijit: couple of errors in your answer. namely month = moths and the fact that you index index. It should be months.index(round brackets).

Joel Cornett Over a year ago

@jamylak: I didn't realize that date_category['title'] contained more than the names of the months. I guess I'm going to have to revise my answer as well.

Joel Cornett · Accepted Answer · 2012-04-08 12:53:48Z

1

Create a dict of the months, with corresponding indices:

months = [u'января', u'февраля', u'марта', u'апреля', u'мая', u'июня', u'июля', u'августа', u'сентября', u'октября', u'ноября', u'декабря']

Then you can simply do:

month = max(enumerate((date_category['title'].find(month) for month in months), start = 1), key = lambda x: x[1])[0]

edited Apr 8, 2012 at 12:53

answered Apr 8, 2012 at 12:31

Joel Cornett

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1 Comment

evgeniuz Over a year ago

I think you haven't read another answers and comments. I cannot use date_category['title'] as index, because month name is just little part of it.

Thiem Nguyen · Accepted Answer · 2012-04-08 11:02:08Z

0

Simply create a hash array which maps your string to corresponding month number, then you can simply iterate through it, check where the key is in date_category['title'], and in all cases, set month to the corresponding value. Good luck.

answered Apr 8, 2012 at 11:02

Thiem Nguyen

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1 Comment

Marcin Over a year ago

Why would you need a hashtable?

Karl Knechtel · Accepted Answer · 2012-04-08 20:31:10Z

0

Here's something a bit more perverse, because I enjoy proving the futility of saying "there should be only one way to do it":

pattern = re.compile(u'(января)|(февраля)|(марта)|(апреля)|(мая)|(июня)|(июля)|(августа)|(сентября)|(октября)|(ноября)|(декабря)')

month = map(bool, x.search(date_category['title']).groups()).index(True) + 1

edited Apr 8, 2012 at 20:31

answered Apr 8, 2012 at 12:17

Karl Knechtel

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