13

I have a dataset of daily data. I need to get only the data of the first day of each month in the data set (The data is from 1972 to 2013). So for example I would need Index 20, Date 2013-12-02 value of 0.1555 to be extracted. The problem I have is that the first day for each month is different, so I cannot use a step such as relativedelta(months=1), how would I go about of extracting these values from my dataset?

Is there a similar command as I have found in another post for R?

R - XTS: Get the first dates and values for each month from a daily time series with missing rows

17 2013-12-05 0.1621
18 2013-12-04 0.1698
19 2013-12-03 0.1516
20 2013-12-02 0.1555
21 2013-11-29 0.1480
22 2013-11-27 0.1487
23 2013-11-26 0.1648
William Miller
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tadalendas
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4 Answers4

15

I would groupby the month and then get the zeroth (nth) row of each group.

First set as index (I think this is necessary):

In [11]: df1 = df.set_index('date')

In [12]: df1
Out[12]:
             n     val
date
2013-12-05  17  0.1621
2013-12-04  18  0.1698
2013-12-03  19  0.1516
2013-12-02  20  0.1555
2013-11-29  21  0.1480
2013-11-27  22  0.1487
2013-11-26  23  0.1648

Next sort, so that the first element is the first date of that month (Note: this doesn't appear to be necessary for nth, but I think that's actually a bug!):

In [13]: df1.sort_index(inplace=True)

In [14]: df1.groupby(pd.TimeGrouper('M')).nth(0)
Out[14]:
             n     val
date
2013-11-26  23  0.1648
2013-12-02  20  0.1555

another option is to resample and take the first entry:

In [15]: df1.resample('M', 'first')
Out[15]:
             n     val
date
2013-11-30  23  0.1648
2013-12-31  20  0.1555

Thinking about this, you can do this much simpler by extracting the month and then grouping by that:

In [21]: pd.DatetimeIndex(df.date).to_period('M')
Out[21]:
<class 'pandas.tseries.period.PeriodIndex'>
[2013-12, ..., 2013-11]
Length: 7, Freq: M

In [22]: df.groupby(pd.DatetimeIndex(df.date).to_period('M')).nth(0)
Out[22]:
    n       date     val
0  17 2013-12-05  0.1621
4  21 2013-11-29  0.1480

This time the sortedness of df.date is (correctly) relevant, if you know it's in descending date order you can use nth(-1):

In [23]: df.groupby(pd.DatetimeIndex(df.date).to_period('M')).nth(-1)
Out[23]:
    n       date     val
3  20 2013-12-02  0.1555
6  23 2013-11-26  0.1648

If this isn't guaranteed then sort by the date column first: df.sort('date').

Andy Hayden
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3

One way is to add a column for the year, month and day:

df['year'] = df.SomeDatetimeColumn.map(lambda x: x.year)
df['month'] = df.SomeDatetimeColumn.map(lambda x: x.month)
df['day'] = df.SomeDatetimeColumn.map(lambda x: x.day)

Then group by the year and month, order by day, and take only the first entry (which will be the minimum day entry).

df.groupby(
    ['year', 'month']
).apply(lambda x: x.sort('day', ascending=True)).head(1)

The use of the lambda expressions makes this less than ideal for large data sets. You may not wish to grow the size of the data by keeping separately stored year, month, and day values. However, for these kinds of ad hoc date alignment problems, sooner or later having these values separated is very helpful.

Another approach is to group directly by a function of the datetime column:

dfrm.groupby(
    by=dfrm.dt.map(lambda x: (x.year, x.month))
).apply(lambda x: x.sort('dt', ascending=True).head(1))

Normally these problems arise because of a dysfunctional database or data storage schema that exists one level prior to the Python/pandas layer.

For example, in this situation, it should be commonplace to rely on the existence of a calendar database table or a calendar data set which contains (or makes it easy to query for) the earliest active date in a month relative to the given data set (such as, the first trading day, the first week day, the first business day, the first holiday, or whatever).

If a companion database table exists with this data, it should be easy to combine it with the dataset you already have loaded (say, by joining on the date column you already have) and then it's just a matter of applying a logical filter on the calendar data columns.

This becomes especially important once you need to use date lags: for example, lining up a company's 1-month-ago market capitalization with the company's current-month stock return, to calculate a total return realized over that 1-month period.

This can be done by lagging the columns in pandas with shift, or trying to do a complicated self-join that is likely very bug prone and creates the problem of perpetuating the particular date convention to every place downstream that uses data from that code.

Much better to simply demand (or do it yourself) that the data must have properly normalized date features in its raw format (database, flat files, whatever) and to stop what you are doing, fix that date problem first, and only then get back to carrying out some analysis with the date data.

ely
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3
import pandas as pd
dates = pd.date_range('2014-02-05', '2014-03-15', freq='D') 
df = pd.DataFrame({'vals': range(len(dates))}, index=dates)
g = df.groupby(lambda x: x.strftime('%Y-%m'), axis=0)
g.apply(lambda x: x.index.min())
#Or depending on whether you want the index or the vals
g.apply(lambda x: x.ix[x.index.min()])
Isaac Laughlin
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1

The above didn't work for me because I needed more than one row per month where the number of rows every month could change. This is what I did:

dates_month = pd.bdate_range(df['date'].min(), df['date'].max(), freq='1M')
df_mth = df[df['date'].isin(dates_month)]
citynorman
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