Extracting just Month and Year from Pandas Datetime column (Python)

问题:

I have a Dataframe, df, with the following column:

df['ArrivalDate'] = ... 936   2012-12-31 938   2012-12-29 965   2012-12-31 966   2012-12-31 967   2012-12-31 968   2012-12-31 969   2012-12-31 970   2012-12-29 971   2012-12-31 972   2012-12-29 973   2012-12-29 ... 

The elements of the column are pandas.tslib.Timestamp.

I want to just include the year and month. I thought there would be simple way to do it, but I can't figure it out.

Here's what I've tried:

df['ArrivalDate'].resample('M', how = 'mean') 

I got the following error:

Only valid with DatetimeIndex or PeriodIndex  

Then I tried:

df['ArrivalDate'].apply(lambda(x):x[:-2]) 

I got the following error:

'Timestamp' object has no attribute '__getitem__'  

Any suggestions?

Edit: I sort of figured it out.

df.index = df['ArrivalDate'] 

Then, I can resample another column using the index.

But I'd still like a method for reconfiguring the entire column. Any ideas?

回答1:

You can directly access the year and month attributes, or request a datetime.datetime:

In [15]: t = pandas.tslib.Timestamp.now()  In [16]: t Out[16]: Timestamp('2014-08-05 14:49:39.643701', tz=None)  In [17]: t.to_datetime() Out[17]: datetime.datetime(2014, 8, 5, 14, 49, 39, 643701)  In [18]: t.day Out[18]: 5  In [19]: t.month Out[19]: 8  In [20]: t.year Out[20]: 2014 

One way to combine year and month is to make an integer encoding them, such as: 201408 for August, 2014. Along a whole column, you could do this as:

df['YearMonth'] = df['ArrivalDate'].map(lambda x: 1000*x.year + x.month) 

or many variants thereof.

I'm not a big fan of doing this, though, since it makes date alignment and arithmetic painful later and especially painful for others who come upon your code or data without this same convention. A better way is to choose a day-of-month convention, such as final non-US-holiday weekday, or first day, etc., and leave the data in a date/time format with the chosen date convention.

The calendar module is useful for obtaining the number value of certain days such as the final weekday. Then you could do something like:

import calendar import datetime df['AdjustedDateToEndOfMonth'] = df['ArrivalDate'].map(     lambda x: datetime.datetime(         x.year,         x.month,         max(calendar.monthcalendar(x.year, x.month)[-1][:5])     ) ) 

If you happen to be looking for a way to solve the simpler problem of just formatting the datetime column into some stringified representation, for that you can just make use of the strftime function from the datetime.datetime class, like this:

In [5]: df Out[5]:              date_time 0 2014-10-17 22:00:03  In [6]: df.date_time Out[6]:  0   2014-10-17 22:00:03 Name: date_time, dtype: datetime64[ns]  In [7]: df.date_time.map(lambda x: x.strftime('%Y-%m-%d')) Out[7]:  0    2014-10-17 Name: date_time, dtype: object 

回答2:

If you want new columns showing year and month separately you can do this:

df['year'] = pd.DatetimeIndex(df['ArrivalDate']).year df['month'] = pd.DatetimeIndex(df['ArrivalDate']).month 

or...

df['year'] = df['ArrivalDate'].dt.year df['month'] = df['ArrivalDate'].dt.month 

Then you can combine them or work with them just as they are.

回答3:

Best way found!!

the date_column has to be in date time format.

df['month_year'] = df.date_column.dt.to_period('M') 

You could also use D for Day, 2M for 2 months, 45Min for 45 min sampling etc.

回答4:

If you want the month year unique pair, using apply is pretty sleek.

    df['mnth_yr'] = df['date_column'].apply(lambda x: x.strftime('%B-%Y'))      

outputs month-year in one column.

don't forget to first change the format to date-time before, I generally forget :|

    df['date_column'] = pd.to_datetime(df['date_column']) 

回答5:

Thanks to jaknap32, I wanted to aggregate the results according to Year and Month, so this worked:

df_join['YearMonth'] = df_join['timestamp'].apply(lambda x:x.strftime('%Y%m')) 

Output was neat:

0    201108 1    201108 2    201108 

回答6:

df['year_month']=df.datetime_column.apply(lambda x: str(x)[:7]) 

This worked fine for me, didn't think pandas would interpret the resultant string date as date, but when i did the plot, it knew very well my agenda and the string year_month where ordered properly... gotta love pandas!

回答7:

You can first convert your date strings with pandas.to_datetime, which gives you access to all of the numpy datetime and timedelta facilities. For example:

df['ArrivalDate'] = pandas.to_datetime(df['ArrivalDate']) df['Month'] = df['ArrivalDate'].values.astype('datetime64[M]') 

回答8:

Extracting the Year say from ['2018-03-04']

df['Year'] = pd.DatetimeIndex(df['date']).year   

The df['Year'] creates a new column. While if you want to extract the month just use .month

文章来源: Extracting just Month and Year from Pandas Datetime column (Python)