Related
I have a hypothetical time series data frame, which is with some missing observations (assumption is that the data frame shall include all dates and corresponding values and for all the dates in the year). As we can see in the head and tail information, there are certain dates and corresponding values are missing (30th Jan & 29th Dec). There would be many more such in the data frame, sometimes missing observations for more than one consecutive date.
Is there a way that missing dates are detected and inserted into the data frame and corresponding values are filled with a rolling average with one week window (this would naturally increase the number of rows of the data frame)? Appreciate inputs.
df.head(3)
date value
0 2020-01-28 25
1 2020-01-29 32
2 2020-01-31 45
df.tail(3)
date value
3 2020-12-28 24
4 2020-12-30 35
5 2020-12-31 37
df.dtypes
date object
value int64
dtype: object
Create DaetimeIndex, then use DataFrame.asfreq with rolling and mean:
df['date'] = pd.to_datetime(df['date'])
df = df.set_index('date').asfreq('d').rolling('7D').mean()
If need all values by year use:
df['date'] = pd.to_datetime(df['date'])
idx = pd.date_range('2020-01-01','2020-12-31')
df = df.set_index('date').reindex(idx).rolling('7D').mean()
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?
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.
The df['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 etc. for different sampling intervals, and in case one has time series data with time stamp, we can go for granular sampling intervals such as 45Min for 45 min, 15Min for 15 min sampling etc.
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_pydatetime() #datetime method is deprecated
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: 100*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
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'])
SINGLE LINE: Adding a column with 'year-month'-paires:
('pd.to_datetime' first changes the column dtype to date-time before the operation)
df['yyyy-mm'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%Y-%m')
Accordingly for an extra 'year' or 'month' column:
df['yyyy'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%Y')
df['mm'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%m')
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
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]')
#KieranPC's solution is the correct approach for Pandas, but is not easily extendible for arbitrary attributes. For this, you can use getattr within a generator comprehension and combine using pd.concat:
# input data
list_of_dates = ['2012-12-31', '2012-12-29', '2012-12-30']
df = pd.DataFrame({'ArrivalDate': pd.to_datetime(list_of_dates)})
# define list of attributes required
L = ['year', 'month', 'day', 'dayofweek', 'dayofyear', 'weekofyear', 'quarter']
# define generator expression of series, one for each attribute
date_gen = (getattr(df['ArrivalDate'].dt, i).rename(i) for i in L)
# concatenate results and join to original dataframe
df = df.join(pd.concat(date_gen, axis=1))
print(df)
ArrivalDate year month day dayofweek dayofyear weekofyear quarter
0 2012-12-31 2012 12 31 0 366 1 4
1 2012-12-29 2012 12 29 5 364 52 4
2 2012-12-30 2012 12 30 6 365 52 4
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
There is two steps to extract year for all the dataframe without using method apply.
Step1
convert the column to datetime :
df['ArrivalDate']=pd.to_datetime(df['ArrivalDate'], format='%Y-%m-%d')
Step2
extract the year or the month using DatetimeIndex() method
pd.DatetimeIndex(df['ArrivalDate']).year
df['Month_Year'] = df['Date'].dt.to_period('M')
Result :
Date Month_Year
0 2020-01-01 2020-01
1 2020-01-02 2020-01
2 2020-01-03 2020-01
3 2020-01-04 2020-01
4 2020-01-05 2020-01
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!
Then I tried:
df['ArrivalDate'].apply(lambda(x):x[:-2])
I think here the proper input should be string.
df['ArrivalDate'].astype(str).apply(lambda(x):x[:-2])
I have the data set of customers with their policies, I am trying to find the number of months the customer is with us. (tenure)
df
cust_no poly_no start_date end_date
1 1 2016-06-01 2016-08-31
1 2 2017-05-01 2018-05-31
1 3 2016-11-01 2018-05-31
output should look like,
cust_no no_of_months
1 22
So basically, it should get rid of the months where there is no policy and count the overlapping period once not twice. I have to do this for every customers, so group by cust_no, how can i do this?
Thanks.
One way to do this is to create date ranges for each records, then use stack to get all the months. Next, take the unique values only to count a month only once:
s = df.apply(lambda x: pd.Series(pd.date_range(x.start_date, x.end_date, freq='M').values), axis=1)
ss = s.stack().unique()
ss.shape[0]
Output:
22
For multiple customers you can use groupby. Continuing with #ScottBoston's answer:
df_range = df.apply(lambda r: pd.Series(
pd.date_range(start=r.start_date, end=r.end_date, freq='M')
.values), axis=1)
df_range.groupby('cust_no').apply(lambda x: x.stack().unique().shape[0])
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?
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.
The df['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 etc. for different sampling intervals, and in case one has time series data with time stamp, we can go for granular sampling intervals such as 45Min for 45 min, 15Min for 15 min sampling etc.
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_pydatetime() #datetime method is deprecated
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: 100*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
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'])
SINGLE LINE: Adding a column with 'year-month'-paires:
('pd.to_datetime' first changes the column dtype to date-time before the operation)
df['yyyy-mm'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%Y-%m')
Accordingly for an extra 'year' or 'month' column:
df['yyyy'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%Y')
df['mm'] = pd.to_datetime(df['ArrivalDate']).dt.strftime('%m')
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
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]')
#KieranPC's solution is the correct approach for Pandas, but is not easily extendible for arbitrary attributes. For this, you can use getattr within a generator comprehension and combine using pd.concat:
# input data
list_of_dates = ['2012-12-31', '2012-12-29', '2012-12-30']
df = pd.DataFrame({'ArrivalDate': pd.to_datetime(list_of_dates)})
# define list of attributes required
L = ['year', 'month', 'day', 'dayofweek', 'dayofyear', 'weekofyear', 'quarter']
# define generator expression of series, one for each attribute
date_gen = (getattr(df['ArrivalDate'].dt, i).rename(i) for i in L)
# concatenate results and join to original dataframe
df = df.join(pd.concat(date_gen, axis=1))
print(df)
ArrivalDate year month day dayofweek dayofyear weekofyear quarter
0 2012-12-31 2012 12 31 0 366 1 4
1 2012-12-29 2012 12 29 5 364 52 4
2 2012-12-30 2012 12 30 6 365 52 4
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
There is two steps to extract year for all the dataframe without using method apply.
Step1
convert the column to datetime :
df['ArrivalDate']=pd.to_datetime(df['ArrivalDate'], format='%Y-%m-%d')
Step2
extract the year or the month using DatetimeIndex() method
pd.DatetimeIndex(df['ArrivalDate']).year
df['Month_Year'] = df['Date'].dt.to_period('M')
Result :
Date Month_Year
0 2020-01-01 2020-01
1 2020-01-02 2020-01
2 2020-01-03 2020-01
3 2020-01-04 2020-01
4 2020-01-05 2020-01
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!
Then I tried:
df['ArrivalDate'].apply(lambda(x):x[:-2])
I think here the proper input should be string.
df['ArrivalDate'].astype(str).apply(lambda(x):x[:-2])
We have some ready available sales data for certain periods, like 1week, 1month...1year:
time_pillars = pd.Series(['1W', '1M', '3M', '1Y'])
sales = pd.Series([4.75, 5.00, 5.10, 5.75])
data = {'time_pillar': time_pillars, 'sales': sales}
df = pd.DataFrame(data)
I would like to do two operations.
Firstly, create a new column of date type, df['date'], that corresponds to the actual date of 1week, 1month..1year from now.
Then, I'd like to create another column df['days_from_now'], taking how many days are on these pillars (1week would be 7days, 1month would be around 30days..1year around 365days).
The goal of this is then to use any day as input for a a simple linear_interpolation_method() to obtain sales data for any given day (eg, what are sales for 4Octobober2018? ---> We would interpolate between 3months and 1year).
Many thanks.
I'm not exactly sure what you mean regarding your interpolation, but here is a way to make your dataframe in pandas (starting from your original df you provided in your post):
from datetime import datetime
from dateutil.relativedelta import relativedelta
def create_dates(df):
df['date'] = [i.date() for i in
[d+delt for d,delt in zip([datetime.now()] * 4 ,
[relativedelta(weeks=1), relativedelta(months=1),
relativedelta(months=3), relativedelta(years=1)])]]
df['days_from_now'] = df['date'] - datetime.now().date()
return df
create_dates(df)
sales time_pillar date days_from_now
0 4.75 1W 2018-04-11 7 days
1 5.00 1M 2018-05-04 30 days
2 5.10 3M 2018-07-04 91 days
3 5.75 1Y 2019-04-04 365 days
I wrapped it in a function, so that you can call it on any given day and get your results for 1 week, 3 weeks, etc. from that exact day.
Note: if you want your days_from_now to simply be an integer of the number of days, use df['days_from_now'] = [i.days for i in df['date'] - datetime.now().date()] in the function, instead of df['days_from_now'] = df['date'] - datetime.now().date()
Explanation:
df['date'] = [i.date() for i in
[d+delt for d,delt in zip([datetime.now()] * 4 ,
[relativedelta(weeks=1), relativedelta(months=1),
relativedelta(months=3), relativedelta(years=1)])]]
Takes a list of the date today (datetime.now()) repeated 4 times, and adds a relativedelta (a time difference) of 1 week, 1 month, 3 months, and 1 year, respectively, extracts the date (i.date() for ...), finally creating a new column using the resulting list.
df['days_from_now'] = df['date'] - datetime.now().date()
is much more straightforward, it simply subtracts those new dates that you got above from the date today. The result is a timedelta object, which pandas conveniently formats as "n days".