Develop Reference

Splitting Date in a Pandas Dataframe

Dataset
Hi, I Have a Index ['release_date'] in a format of month,date,year , I was trying to split this column by doing
test['date_added'].str.split(' ',expand=True) #code_1
but now it's creating a 4 columns and what really is happening is for some reason is it is simply for few rows it's shifting columns therefore creating a 4th column
code_1
This is the error I am facing
I tried splitting ['release_date'], I am expecting it to be splitted into 3 rows but for some reason few rows are being shifting to other column.
if someone wants to inspect that dataframe you can use google colab for it,
!gdown 1x-_Kq9qYrybB9-DxJHoeVlPabmAm6xbQ

you can use:
df['day'] = pd.DatetimeIndex(df['date_added']).day
df['Month'] = pd.DatetimeIndex(df['date_added']).month
df['year'] = pd.DatetimeIndex(df['date_added']).year

day, month, year = zip(*[(d.day, d.month, d.year) for d in df['date_added']])
df df.assign(day = day, month = month, year = year)

How to extract year and month from string in a dataframe

1. Question
I have a dataframe, and the Year-Month column contains the year and month which I want to extract.
For example, an element in this column is "2022-10". And I want to extract year=2022, month=10 from it.
My current solution is to use apply and lambda function:
df['xx_month'] = df['Year-Month'].apply(lambda x: int(x.split('-')[1]))
But it's super slow on a huge dataframe.
How to do it more efficiently?
2. Solutions
Thanks for your wisdom, I summarized each one's solution with the code:
(1) split by '-' and join #Vitalizzare
pandas.Series.str.split - split strings of a series, if expand=True then return a data frame with each part in a separate column;
pandas.DataFrame.set_axis - if axis='columns' then rename column names of a data frame;
pandas.DataFrame.join - if the indices are equal, then the frames stacked together horizontally are returned.
df = pd.DataFrame({'Year-Month':['2022-10','2022-11','2022-12']})
df = df.join(
df['Year-Month']
.str.split('-', expand=True)
.set_axis(['year','month'], axis='columns')
)
(2) convert the datatype from object (str) into datetime format #Neele22
import pandas as pd
df['Year-Month'] = pd.to_datetime(df['Year-Month'], format="%Y-%m")
(3) use regex or datetime to extract year and month #mozway
df['Year-Month'].str.extract(r'(?P<year>\d+)-(?P<month>\d+)').astype(int)
# If you want to assign the output to the same DataFrame while removing the original Year-Month:
df[['year', 'month']] = df.pop('Year-Month').str.extract(r'(\d+)-(\d+)').astype(int)
Or use datetime:
date = pd.to_datetime(df['Year-Month'])
df['year'] = date.dt.year
df['month'] = date.dt.month
3. Follow up question
But there will be a problem if I want to subtract 'Year-Month' with other datetime columns after converting the incomplete 'Year-Month' column from string to datetime.
For example, if I want to get the data which is no later than 2 months after the timestamp of each record.
import dateutil # dateutil is a better package than datetime package according to my experience
df[(df['timestamp'] - df['Year-Month'])>= dateutil.relativedelta.relativedelta(months=0) and (df['timestamp'] - df['Year-Month'])<= datetime.timedelta(months=2)]
This code will have type error for subtracting the converted Year-Month column with actual datetime column.
TypeError: Cannot subtract tz-naive and tz-aware datetime-like objects
The types for these two columns are:
Year-Month is datetime64[ns]
timestamp is datetime64[ns, UTC]
Then, I tried to specify utc=True when changing Year-Month to datetime type:
df[["Year-Month"]] = pd.to_datetime(df[["Year-Month"]],utc=True,format="%Y-%m")
But I got Value Error.
ValueError: to assemble mappings requires at least that [year, month,
day] be specified: [day,month,year] is missing
4. Take away
If the [day,month,year] is not complete for the elements in a column. (like in my case, I only have year and month), we can't change this column from string type into datetime type to do calculations. But to use the extracted day and month to do the calculations.
If you don't need to do calculations between the incomplete datetime column and other datetime columns like me, you can change the incomplete datetime string into datetime type, and extract [day,month,year] from it. It's easier than using regex, split and join.

df = pd.DataFrame({'Year-Month':['2022-10','2022-11','2022-12']})
df = df.join(
df['Year-Month']
.str.split('-', expand=True)
.set_axis(['year','month'], axis='columns')
)
pandas.Series.str.split - split strings of a series, if expand=True then return a data frame with each part in a separate column;
pandas.DataFrame.set_axis - if axis='columns' then rename column names of a data frame;
pandas.DataFrame.join - if the indices are equal, then the frames stacked together horizontally are returned.

You can use a regex for that.
Creating a new DataFrame:
df['Year-Month'].str.extract(r'(?P<year>\d+)-(?P<month>\d+)').astype(int)
If you want to assign the output to the same DataFrame while removing the original Year-Month:
df[['year', 'month']] = df.pop('Year-Month').str.extract(r'(\d+)-(\d+)').astype(int)
Example input:
Year-Month
0 2022-10
output:
year month
0 2022 10
alternative using datetime:
You can also use a datetime intermediate
date = pd.to_datetime(df['Year-Month'])
df['year'] = date.dt.year
df['month'] = date.dt.month
output:
Year-Month year month
0 2022-10 2022 10

You can also convert the datatype from object (str) into datetime format. This will make it easier to work with the dates.
import pandas as pd
df['Year-Month'] = pd.to_datetime(df['Year-Month'], format="%Y-%m")

Output and preserve groupby index structure without aggregate function

I have a dataframe that I want to groupby year and followed by months within each year. Due to the fact that the data are quite huge (recorded from 3 decades ago till now), I would like to have the output presented as shown below for subsequent calculation but without any aggregate function such ".mean()" behind.
However, I am unable to do so because groupby always require an .agg, else it will show this error: <pandas.core.groupby.generic.DataFrameGroupBy object at 0x0000022BF79A52E0>
On the other hand, I am a bit worried about importing as Series because I do not know how to set the parameters to get exactly the same format as below. Another reason is that I used the below lines to import the .csv into dataframe:
df=pd.read_csv(r'file directory', index_col = 'date')
df.index = pd.to_datetime(df.index)
For some weird reasons, if I define the date string format in pd.read_csv to import and subsequently, try to sort by other methods according to years and month, function or method that gets confused when the records have date starts off with 01(day)/01(month)/1990 and 01(day)/02(month)/1990. For example, it interprets the first number in Jan as day and the second number as month and sorts all chronologically but when it comes to Feb, when the day is should be 01, the method thought that 01 is the month and 02 is the day portion and move that Feb record to the Jan group.
Are there any ways to achieve the same format?
Methods shown in the post below does not seem to help me get the format I want: Pandas - Groupby dataframe store as dataframe without aggregating

IIUC:
You can use dayfirst parameter in to_datetime() and set that equal to True then create 'Year' and 'Month' column and make them index and sort index:
df=pd.read_csv(r'file directory')
df['date']=pd.to_datetime(df['date'],dayfirst=True)
df['Year']=df['date'].dt.year
df['Month']=df['date'].dt.month
df=df.set_index(['Year','Month']).sort_index()
OR in 3 steps via assign():
df=pd.read_csv(r'file directory')
df['date']=pd.to_datetime(df['date'],dayfirst=True)
df=(df.assign(Year=df['date'].dt.year,Month=df['date'].dt.month)
.set_index(['Year','Month']).sort_index())

you can iterate through the groups of the groupby results.
import pandas as pd
import numpy as np
rand = np.random.RandomState(1)
df = pd.DataFrame({'A': ['foo', 'bar'] * 3,
'B': rand.randn(6),
'C': rand.randint(0, 20, 6)})
groupby_obj = df.groupby(['A'])
for k, gdf in groupby_obj:
print('Groupby Key:', k)
print('Dataframe:\n', gdf, '\n')
you can apply all dataframe methods on gdf

Access only second row of column names in a dataframe

I want to read an excel file where the second line is a date in a string format and the first line is the weekday that corresponds to each date, and then change the second line from string to datetime. If I only read the second line as index, and completely skip the first line with the days, I do the following to convert it to a datetime:
Receipts_tbl.columns = pd.to_datetime(Receipts_tbl.columns)
How do I do that if I have a multiindexed dataframe, where the first line of the indices remains as weekdays, and I want the second to be converted to datetime?
Thanx

You didn't give an example of what your data source looks like, so I'm inferring.
If you use pd.read_excel with header=None, it will treat the first two rows as data and you can manipulate them to achieve your goal. Here's a minimum example, with an example "real" data row beneath:
df = pd.DataFrame([['Mon', 'Tues'], ['10-02-1995', '11-23-1997'],
[12, 32]])
# 0 1
#0 Mon Tues
#1 10-02-1995 11-23-1997
#2 12 32
Next, convert the first row to datetime as you said in your question.
df.loc[1] = pd.to_datetime(df.loc[1])
Create a multi-index from the first two rows, and set it as the dataframe's columns
df.columns = df.T.set_index([0,1]).index.set_names(['DOW', 'Date'])
Lastly, select from second row down, as the first two rows are now in the columns.
df = df.loc[2:].reset_index()
df
#DOW Mon Tues
#Date 812592000000000000 880243200000000000
#0 12 32
Note that DOW and Date are now a multilevel index for the columns, and the 'data' rows have been reindexed to start at 0.
Please let me know if I misunderstood your question.

Assuming you have this data in the clipboard
Day Date Data
Mo 2018-08-06 blah
Mo 2018-08-06 blah
Mo 2018-08-06 blah
Tu 2018-08-07 blah
Try
import pandas as pd
df = pd.read_clipboard().set_index(['Day', 'Date'])
to get a multiindexed example
Then change the Date to Datetime
df2 = df.reset_index()
df2.Date = pd.to_datetime(df2.Date, yearfirst=True)
Afterwards you can set the multiindex again, if you want.
Note, check out the documentation on to_datetime if your
datetime string is formatted differently. It assumes
month first, unless you set dayfirst or yearfirst to True.

Dropping rows from a Dataframe based on Date

How can I drop rows from Dataframe df if the dates associated with df['maturity_dt'] are less that today's date?
I am currently doing the following:
todays_date = datetime.date.today()
datenow = datetime.datetime.combine(todays_date, datetime.datetime.min.time()) #Converting to datetime
for (i,row) in df.iterrows():
if datetime.datetime.strptime(row['maturity_dt'], '%Y-%m-%d %H:%M:%S.%f') < datenow):
df.drop(df.index[i])
However, its taking too long and I was hoping to do something like: df = df[datetime.datetime.strptime(df['maturity_dt'], '%Y-%m-%d %H:%M:%S.%f') < datenow, but this results in the error TypeError: must be str, not Series
Thank You

Haven't tried it but maybe the pandas native functions will iterate faster. Something like:
df['dt']=pandas.Datetimeindex(df['maturity_dt'])
newdf=df.loc[df['dt']<=todays_date].copy()

Instead of parsing the date in each row, you could format your comparison date in the same format as these dates are stored and then you could just do a string comparison.
Also, if there is a way to drop multiple rows in a single call, you could use your loop just to gather the indices of those rows to be dropped, then use that call to drop them in bunches.