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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")

How do I convert a column to Pandas Timestamps?

I have a column in my DataFrame with values like '2022-06-03T00:00:00.000Z' and I want to convert these (in place) to pd.Timestamp. I see many answers he on how to convert to np.datetime64 and on how do convert arbitrary columns of DataFrames, but can't figure out how to apply these to covering to pd.Timestamp.

Use from pd.to_datetime method
I think this solve your problem
Just you need to active utc argument in your method
import pandas as pd
lst = {'a':['Geeks', 'For'],'b':['2022-06-03T00:00:00.000Z','2024-03-03T00:00:00.000Z']}
df = pd.DataFrame(lst)
df['b']=pd.to_datetime(df['b'],utc=True)
type(df['b'][0])
pandas._libs.tslibs.timestamps.Timestamp

Convert time column (object type) to datetime format

I have a separate column with HH:MM: SS object type. I want to convert object type to DateTime format. Encountering the error below.
Dataset time column ["ACCIDENT_TIME]:

You can try the below one also:
import pandas as pd
df['col_name'] = pd.to_datetime(df['col_name'], errors='coerce')

if you want to change dtype of column you can do in below way:
df['col_name'] = df['col_name'].astype('datetime64[ns]')

Date Formatting Problem in pandas Dataframe

I have a Date column in my Dataframe, when I display the dates, The Dates format are merged, and are in random format.How to put them in right format? Like in dd/mm/yyyy

This is pseudo code since you did not gave us your code. It assumed that the column date of a dataframe df is correctly formatted as datetime.
You can use the vectorized datetime function strftime() with (see the docs):
df['date'].dt.strftime("%d/%m/%Y")
When you want to save the changes of the format, you need to assign it again to the date column, like this
df['date'] = df['date'].dt.strftime("%d/%m/%Y")

Faster solution for date formatting

I am trying to change the format of the date in a pandas dataframe.
If I check the date in the beginning, I have:
df['Date'][0]
Out[158]: '01/02/2008'
Then, I use:
df['Date'] = pd.to_datetime(df['Date']).dt.date
To change the format to
df['Date'][0]
Out[157]: datetime.date(2008, 1, 2)
However, this takes a veeeery long time, since my dataframe has millions of rows.
All I want to do is change the date format from MM-DD-YYYY to YYYY-MM-DD.
How can I do it in a faster way?

You should first collapse by Date using the groupby method to reduce the dimensionality of the problem.
Then you parse the dates into the new format and merge the results back into the original DataFrame.
This requires some time because of the merging, but it takes advantage from the fact that many dates are repeated a large number of times. You want to convert each date only once!
You can use the following code:
date_parser = lambda x: pd.datetime.strptime(str(x), '%m/%d/%Y')
df['date_index'] = df['Date']
dates = df.groupby(['date_index']).first()['Date'].apply(date_parser)
df = df.set_index([ 'date_index' ])
df['New Date'] = dates
df = df.reset_index()
df.head()
In my case, the execution time for a DataFrame with 3 million lines reduced from 30 seconds to about 1.5 seconds.

I'm not sure if this will help with the performance issue, as I haven't tested with a dataset of your size, but at least in theory, this should help. Pandas has a built in parameter you can use to specify that it should load a column as a date or datetime field. See the parse_dates parameter in the pandas documentation.
Simply pass in a list of columns that you want to be parsed as a date and pandas will convert the columns for you when creating the DataFrame. Then, you won't have to worry about looping back through the dataframe and attempting the conversion after.
import pandas as pd
df = pd.read_csv('test.csv', parse_dates=[0,2])
The above example would try to parse the 1st and 3rd (zero-based) columns as dates.
The type of each resulting column value will be a pandas timestamp and you can then use pandas to print this out however you'd like when working with the dataframe.

Following a lead at #pygo's comment, I found that my mistake was to try to read the data as
df['Date'] = pd.to_datetime(df['Date']).dt.date
This would be, as this answer explains:
This is because pandas falls back to dateutil.parser.parse for parsing the strings when it has a non-default format or when no format string is supplied (this is much more flexible, but also slower).
As you have shown above, you can improve the performance by supplying a format string to to_datetime. Or another option is to use infer_datetime_format=True
When using any of the date parsers from the answers above, we go into the for loop. Also, when specifying the format we want (instead of the format we have) in the pd.to_datetime, we also go into the for loop.
Hence, instead of doing
df['Date'] = pd.to_datetime(df['Date'],format='%Y-%m-%d')
or
df['Date'] = pd.to_datetime(df['Date']).dt.date
we should do
df['Date'] = pd.to_datetime(df['Date'],format='%m/%d/%Y').dt.date
By supplying the current format of the data, it is read really fast into datetime format. Then, using .dt.date, it is fast to change it to the new format without the parser.
Thank you to everyone who helped!