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How to split a date index into separate day , month ,year column in pandas

I have dataset df1:
df1
I did a column and index transpose previously:
df1 = df.T
The dataset df previously looked like this:
df
I have already use the .to_datetime function to convert my dates:
df1.index = pd.to_datetime(df1.index).strftime('%Y-%m')
How could I split my date index and add them to new 'year' and 'month' columns on the right of the table?
I tried:
df1['month'] = df.index.month
df1['year'] = df.index.year
However, it is returning me the following error:
AttributeError: 'Index' object has no attribute 'day'
This is actually a follow up to another question raised before here
I wasn't able to add comment over there as I am a new account holder to stack overflow.
Thank you everyone, I am a new learner so please bear with me.

Try this
df.index = pd.to_datetime(df.index)
df['day'] = df.index.day
df['month'] = df.index.month
df['year'] = df.index.year

If your dates are index then your code should have worked. However, if the dates are in date column then try:
df['day'] = df.date.dt.day
df['month'] = df.date.dt.month
df['year'] = df.date.dt.year

import calendar as cal
import locale
df.Dates = pd.to_datetime(df.Dates)
df['Year'] = df.Dates.dt.year
df['Month'] = df.Dates.dt.month_name()
df['Day'] = df.Dates.dt.day

Try this:
df['time'] = pd.to_datetime(df['time'])
df['Which Day'] = df['time'].dt.day_name()
df['Year'] = df['time'].dt.year
df['Month'] = df['time'].dt.month_name())

Converting dates to datetime64 results in day and month places getting swapped

I am pulling a time series from a csv file which has dates in "mm/dd/yyyy" format
df = pd.read_csv(lib_file.csv)
df['Date'] = df['Date'].apply(lambda x:datetime.strptime(x,'%m/%d/%Y').strftime('%d/%m/%Y'))
below is the output
I convert dtypes for ['Date'] from object to datetime64
df['Date'] = pd.to_datetime(df['Date'])
but that changes my dates as well
how do I fix it?

Try this:
df['Date'] = pd.to_datetime(df['Date'], infer_datetime_format=True)
This will infer your dates based on the first non-NaN element which is being correctly parsed in your case and will not infer the format for each and every row of the dataframe.

just using the below code helped
df = pd.read_csv(lib_file.csv)
df['Date'] = pd.to_datetime(df['Date])

How to split csv into 2 dataframe with the condition

My idea is seperate both of the "String" then convert both dataframe into same datetime format. I try the code
data['date'] = pd.to_datetime(data['date'])
data['date'] = data['date'].dt.strftime('%Y-%m-%d')
but there are some error on the output. The 13/02/2020 will become 2020-02-13 that is what i want. But the 12/02/2020 will become 2020-12-02.
My dataframe have 2 type of date format. Which is YYYY-MM-DD and DD/MM/YYYY.
dataframe
I need to split it into 2 dataframe, all the row that have the date YYYY-MM-DD into the df1.
The data type is object.
All all the row that have the date DD/MM/YYYY into the df2.
Anyone know how to code it?

If dont need convert to datetimes use Series.str.contains with boolean indexing:
mask = df['date'].str.contains('-')
df1 = df[mask].copy()
df2 = df[~mask].copy()
If need datetimes you can use parameter errors='coerce' in to_datetime for missing values if not matching format, so last remove missing values:
df1 = (df.assign(date = pd.to_datetime(df['date'], format='%Y-%m-%d', errors='coerce')
.dropna(subset=['date']))
df2 = (df.assign(date = pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')
.dropna(subset=['date']))
EDIT: If need output column filled by correct datetimes you can replace missing values by another Series by Series.fillna:
date1 = pd.to_datetime(df['date'], format='%Y-%m-%d', errors='coerce')
date2 = pd.to_datetime(df['date'], format='%d/%m/%Y', errors='coerce')
df['date'] = date1.fillna(date2)

you can use the fact that the separation is different to find the dates.
If your dataframe is in this format:
df = pd.DataFrame({'id' : [1,1,2,2,3,3],
"Date": ["30/8/2020","30/8/2021","30/8/2022","2019-10-24","2019-10-25","2020-10-24"] })
With either "-" or "/" to separate the data
you can use a function that finds this element and apply it to the date column:
def find(string):
if string.find('/')==2:
return True
else:
return False
df[df['date'].apply(find)]

Convert '9999-12-31 00:00:00' to 'dd/mm/yyyy' in Pandas

I have a dataframe containing the column 'Date' with value as '9999-12-31 00:00:00'. I need to convert it to 'dd/mm/yyyy'.
import pandas as pd
data = (['9999-12-31 00:00:00'])
df = pd.DataFrame(data, columns=['Date'])

Use daily periods by custom function with remove times by split and change format by strftime:
df['Date'] = (df['Date'].str.split()
.str[0]
.apply(lambda x: pd.Period(x, freq='D'))
.dt.strftime('%d/%m/%Y'))
print (df)
Date
0 31/12/9999

Set new column from datetime on dataframe pandas

I am trying to set a new column(Day of year & Hour)
My date time consist of date and hour, i tried to split it up by using
data['dayofyear'] = data['Date'].dt.dayofyear
and
df['Various', 'Day'] = df.index.dayofyear
df['Various', 'Hour'] = df.index.hour
but it is always returning error, im not sure how i can split this up and get it to a new column.

I think problem is there is no DatetimeIndex, so use to_datetime first and then assign to new columns names:
df.index = pd.to_datetime(df.index)
df['Day'] = df.index.dayofyear
df['Hour'] = df.index.hour
Or use DataFrame.assign:
df.index = pd.to_datetime(df.index)
df = df.assign(Day = df.index.dayofyear, Hour = df.index.hour)