How to change the date from YYYY-MM-DD to YYDD-MM-YY in python

Question

I have a MySQL database, and there is a column in which I have stored the date. Now the requirement is I need to replace the dd with YY. and I want to do this using python(Pandas)

for eg date that i have is 2032-11-16, and I want 2016-11-32.

uptill now my approach is shown below

df = pd.read_csv('p1006_freight_rates.csv')
df['year'] = zip(*df['date'].apply(lambda x: (x[:4], x[5:])))

the above approach is absolutely wrong, I have also tried the string replace but that too had some issue.

and also please tell me how to save the update in mysql.

Answer 1

datetime module could help you with that:

datetime.datetime.strptime(date_string, format1).strftime(format2)

datetime.datetime.strptime("2013-1-25", '%Y-%m-%d').strftime('%m/%d/%y')

prints "01/25/13" .

If you can't live with the leading zero, try this:

dt = datetime.datetime.strptime("2013-1-25", '%Y-%m-%d')
print '{0}/{1}/{2:02}'.format(dt.month, dt.day, dt.year % 100)

This prints "1/25/13" .

EDIT: This may not work on every platform:

datetime.datetime.strptime("2013-1-25", '%Y-%m-%d').strftime('%-m/%d/%y')

Answer 2

A simple and straight forward method would be like this:

xx = [
    '2032-11-16',
    '2011-22-33'
]
df = pd.DataFrame(xx,columns=['date'])
df['new_date'] = df.date.map(lambda x:x[:2]+x[-2:]+x[4:8]+x[2:4])
df

Answer 3

Well, if you have a date string that you KNOW is in the format YYDD-MM-YY, then just use this:

new_str = old_str;
new_str[2:4] = old_str[8:10]
new_str[8:10] = old_str[2:4]