We Keep Coding

Pandas str split a column and add it to a dataframe

so I have a dataframe with this information and for the release date, I want to str,split so that it only display the year.
how can i str.split the release date so that it only shows the year and label it "release year"
enter image description here

df['release year'] = df['releaseDate'].apply(lambda x: x.split('-')[-1])

You can access the year and month attributes directly using the following:
import pandas as pd
df = pd.DataFrame({"data" : ["11-Jan-2020", "05-Feb-2020", "01-Mar-2020"]})
df["data"]=pd.to_datetime(df["data"], dayfirst=True)
df['year'] = df['data'].dt.year
df.head()
Your df will now look like:
data year
0 2020-01-11 2020
1 2020-02-05 2020
2 2020-03-01 2020

Split dataframe into two using Data as splitting point

I have a dataframe which has 100,000 rows and 24 columns; representing crime over a year period October 2019 - October 2020
I'm trying to split the my df into two one dataframe of all rows ranging from october 1st - 31st March and the second ranging from April 1st - October 31st;
Would anyone be able to kindly assist how using pandas?

Assuming the column is of datetime type. You can do like this :
import pandas as pd
split_data = pd.datetime(2020,03,31)
df_1 = df.loc[df['Date']<= split_date]
df_2 = df.loc[df['Date'] > split_date]
if the column containing date field is not datetime type. You should first convert it into datetime type.
df['Date'] = pd.to_datetime(df['Date'])

counting months between two days in dataframe

I have a dataframe with multiple columns, one of which is a date column. I'm interested in creating a new column which contains the number of months between the date column and a preset date. For example one of the dates in the 'start date' column is '2019-06-30 00:00:00' i would want to be able to calculate the number of months between that date and the end of 2021 so 2021-12-31 and place the answer into a new column and do this for the entire date column in the dataframe. I haven't been able to work out how i could go about this but i would like it in the end to look like this if the predetermined end date was 2021-12-31:
df =
|start date months
0|2019-06-30 30
1|2019-08-12 28
2|2020-01-24 23

You can do this using np.timedelta64:
end_date = pd.to_datetime('2021-12-31')
df['start date'] = pd.to_datetime(df['start date'])
df['month'] = ((end_date - df['start date'])/np.timedelta64(1, 'M')).astype(int)
print(df)
start date month
0 2019-06-30 30
1 2019-08-12 28
2 2020-01-24 23

Assume that start date column is of datetime type (not string)
and the reference date is defined as follows:
refDate = pd.to_datetime('2021-12-31')
or any other date of your choice.
Then you can compute the number of months as:
df['months'] = (refDate.to_period('M') - df['start date']\
.dt.to_period('M')).apply(lambda x: x.n)

Create a new column in a dataframe that shows Day of the Week from an already existing dd/mm/yy column? Python

I have a dataframe that contains a column with dates e.g. 24/07/15 etc
Is there a way to create a new column into the dataframe that displays all the days of the week corresponding to the already existing 'Date' column?
I want the output to appear as:
[Date][DayOfTheWeek]

This might work:
If you want day name:
In [1405]: df
Out[1405]:
dates
0 24/07/15
1 25/07/15
2 26/07/15
In [1406]: df['dates'] = pd.to_datetime(df['dates']) # You don't need to specify the format also.
In [1408]: df['dow'] = df['dates'].dt.day_name()
In [1409]: df
Out[1409]:
dates dow
0 2015-07-24 Friday
1 2015-07-25 Saturday
2 2015-07-26 Sunday
If you want day number:
In [1410]: df['dow'] = df['dates'].dt.day
In [1411]: df
Out[1411]:
dates dow
0 2015-07-24 24
1 2015-07-25 25
2 2015-07-26 26

I would try the apply function, so something like this:
def extractDayOfWeek(dateString):
...
df['DayOfWeek'] = df.apply(lambda x: extractDayOfWeek(x['Date'], axis=1)
The idea is that, you map over every row, extract the 'date' column, and then apply your own function to create a new row entry named 'Day'

Depending of the type of you column Date.
df['Date']=pd.to_datetime(df['Date'], format="d/%m/%y")
df['weekday'] = df['Date'].dt.dayofweek

Editing the date in pandas to show year only, by column

I am trying to understand how I can edit the dataframe in python using pandas so I can drop everything but the year.
Example: if the date is 2014-01-01, I want it to show 2014 and drop both the month and the date. All the dates are in a single column.
Thanks in advice!

You can convert the numpy.datetime64 date value to datetime using pd.to_datetime() and then you can extract year or month or day from it.
import numpy as np
date = np.datetime64('2014-01-01')
type(date)
Output:
numpy.datetime64
Convert this date to pandas datetime using pd.to_datetime.
date = pd.to_datetime(date)
type(date)
Output:
pandas._libs.tslibs.timestamps.Timestamp
Then you can extract the year using .year
date.year
Output:
2014
So, if you if you have a df:
df = pd.DataFrame({'date': [np.datetime64('2014-01-01'), np.datetime64('2015-01-01'), np.datetime64('2016-01-01')]})
df['date'] = pd.DatetimeIndex(df['date']).year
df
Output:
date
0 2014
1 2015
2 2016
Alternately, you can also do this
df = pd.DataFrame({'date': [np.datetime64('2014-01-01'), np.datetime64('2015-01-01'), np.datetime64('2016-01-01')]})
df['date'] = df['date'].apply(lambda x: x.strftime('%Y'))
df
Output:
date
0 2014
1 2015
2 2016
EDIT 1
Group by using year when the column has date values
df = pd.DataFrame({'date': [np.datetime64('2014-01-01'), np.datetime64('2015-01-01'), np.datetime64('2016-01-01')]})
df['date'] = pd.to_datetime(df['date'])
df.set_index('date', inplace=True)
df.groupby(df.index.year).size()
Output:
date
2014 1
2015 1
2016 1
You can still do the same even if you have removed the month and day from the date and only have year in your column
df = pd.DataFrame({'date': [np.datetime64('2014-01-01'), np.datetime64('2015-01-01'), np.datetime64('2016-01-01')]})
df['date'] = pd.DatetimeIndex(df['date']).year
df.groupby('date').size()
Output:
date
2014 1
2015 1
2016 1