indexing data types with numbers that start with 0 in python

Question

I have a function that will convert 3 numbers into a date.

def conv(month, day, year):
    months = ('None','January','February','March','April','May','June','July','August','September','October','November','December')
    print(months[month],str(day)+',',year)

so if I run the function like this: conv(6,17,2016) the output will be:

June 17, 2016

so far so good, but what if I give 06 for the month instead of 6 how can I get the same output like above? is there any way to use 06 and somehow turning it into 6 for indexing or I have to only give 6 and there is no way for doing that?

Is there a reason you're not using the `datetime` module? It has the facility to deal with zero-padded values, see [here](https://docs.python.org/3/library/datetime.html#strftime-and-strptime-behavior) — roganjosh, Aug 30 '18 at 09:15
yeah I know I can use datetime but I just wondering if there is a way for that indexing — Ghost.007, Aug 30 '18 at 09:21

score 1 · Accepted Answer · answered Aug 30 '18 at 09:20

1

assuming you are aware that 06 will automatically raise a syntax error, that means you can only create a string such as "06".

with that said, simple use to convert to int

number = int(month)

answered Aug 30 '18 at 09:20

Wong Siwei

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yeah you are right I have to either give the 6 or give 06 as a string – Ghost.007 Aug 30 '18 at 09:38

Rahul K P · Answer 2 · 2018-08-30T09:29:58.663

1

I would suggest to use datetime,

In [31]: d = datetime.datetime.now().date()

In [32]: d.strftime('%B %d, %Y')
Out[32]: 'August 30, 2018'

Or use calender.

import calendar
def conv(month, day, year):
    print(calendar.month_name[int(month)],str(day)+',',year)

edited Aug 30 '18 at 09:29

answered Aug 30 '18 at 09:23

Rahul K P

15,740
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score 1 · Answer 3 · answered Aug 30 '18 at 09:28

1

This will be a little easier to digest:

import datetime

def conv(month, day, year):
    d = datetime.date(year, month, day)
    print(d.strftime("%B %d, %Y"))

answered Aug 30 '18 at 09:28

Dr G.

1,298
9
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Cut7er · Answer 4 · 2018-08-30T09:28:51.227

0

cast your 06 to int - it will then be a normal 6. So you will print by using print(months[int(month)],str(day)+',',year)

but as mentioned in the comments - just use datetime module instead :-)

edited Aug 30 '18 at 09:28

answered Aug 30 '18 at 09:18

Cut7er

1,209
9
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`def conv(int(month), day, year):` is invalid syntax – roganjosh Aug 30 '18 at 09:22
you sure? well, then it can be specified later in the code by doing `print(months[int(month)],str(day)+',',year)` – Cut7er Aug 30 '18 at 09:26
Yes I'm sure, I tested it :) – roganjosh Aug 30 '18 at 09:27

score 0 · Answer 5 · answered Aug 30 '18 at 09:40

You can convert tuple to datetime then use .strftime to format it:

>>> tup=(2018,8,30)
>>> from datetime import datetime
>>> datetime(*tup)
datetime.datetime(2018, 8, 30, 0, 0)
>>> print(datetime(*tup)) # print
2018-08-30 00:00:00
>>> datetime(*tup).strftime('%B %d, %y')
'August 30, 18'
>>>

indexing data types with numbers that start with 0 in python

5 Answers5