I am trying to get a date time field in Pandas in the below format
df['date'] = pd.to_datetime(df['date'])
The above code returns date time column in the below format
2021-11-27 03:30:00
I would like to get an output of 27/11/2021 (format is dd/mm/yyyy) and the data type of the column needs to be datetime and not object.
If your column is a string, you will need to first use pd.to_datetime,
df['Date'] = pd.to_datetime(df['Date'])
Then, use .dt datetime accessor with strftime:
df = pd.DataFrame({'Date':pd.date_range('2017-01-01', periods = 60, freq='D')})
df.Date.dt.strftime('%Y%m%d').astype(int)
Or use lambda function:
df.Date.apply(lambda x: x.strftime('%Y%m%d')).astype(int)
Related
I have column of dates in the format below in a pandas dataframe.
What is the most effective way to convert
2021-11-06T21:54:35.825Z
to
2021-11-6 21:54:35
pd.to_datetime(df['date'], format='%Y-%m-%d %H:%M:%S') only returns 2021-11-06 without the timestamp
You can use .dt accessor on Pandas Series followed by by .strftime property dt.strftime, to format datetime into desired string representation.
import pandas as pd
import datetime
df = pd.DataFrame({'date': ["2021-11-06T21:54:35.825Z"]})
fmt = '%Y-%m-%d %H:%M:%S'
pd.to_datetime(df['date']).dt.strftime(fmt)
returns
0 2021-11-06 21:54:35
Name: date, dtype: object
Or if you don't want to have zero padding before the day, you can use: fmt="%Y-%m-%-d %H:%M:%S" (notice the hyphen between % and d). This results in: 2021-11-6 21:54:35
I am looking to convert datetime to date for a pandas datetime series.
I have listed the code below:
df = pd.DataFrame()
df = pandas.io.parsers.read_csv("TestData.csv", low_memory=False)
df['PUDATE'] = pd.Series([pd.to_datetime(date) for date in df['DATE_TIME']])
df['PUDATE2'] = datetime.datetime.date(df['PUDATE']) #Does not work
Can anyone guide me in right direction?
You can access the datetime methods of a Pandas series by using the .dt methods (in a aimilar way to how you would access string methods using .str. For your case, you can extract the date of your datetime column as:
df['PUDATE'].dt.date
This is a simple way to get day of month, from a pandas
#create a dataframe with dates as a string
test_df = pd.DataFrame({'dob':['2001-01-01', '2002-02-02', '2003-03-03', '2004-04-04']})
#convert column to type datetime
test_df['dob']= pd.to_datetime(test_df['dob'])
# Extract day, month , year using dt accessor
test_df['DayOfMonth']=test_df['dob'].dt.day
test_df['Month']=test_df['dob'].dt.month
test_df['Year']=test_df['dob'].dt.year
I think you need to specify the format for example
df['PUDATE2']=datetime.datetime.date(df['PUDATE'], format='%Y%m%d%H%M%S')
So you just need to know what format you are using
I have dataset df with date column. I have dates from 2020-01-01 to 2021-03-30 in date column. Now i have a variable like a=20210130(which is actually a date). I need take take values from the df which is <=a.
First idea is convert a to datetimes and compare, then filter by boolean indexing:
df['date'] = pd.to_datetime(df['date'])
a = 20210130
df = df[df['date'] <= pd.to_datetime(a)]
Or convert column to integers and compare:
a = 20210130
df = df[df['date'].dt.strftime('%Y%m%d').astype(int) <= a]
I have dates in this format: 2015-02-02 14:19:00.
I use this code:
dateparse = lambda dates: pd.datetime.strptime(dates, '%Y/%m/%d %H:%M:%S')
df = pd.read_csv('3df_uniti.csv', parse_dates=True, index_col='date', date_parser=dateparse)
df.head()
but it doesn't work because it gives me the follow error:
time data does not match format
Can you help me to set the right format?
Your format uses / instead of -. Try changing it to %Y-%m-%d %H:%M:%S.
I am getting date data in string format in pandas like 10-Oct,11-Oct but i want to make it date data type like this format 2019-10-10,2019-10-11
is there any easy way available in pandas?
Use to_datetime with added year and parameter format:
df = pd.DataFrame({'date':['10-Oct', '11-Oct']})
df['date'] = pd.to_datetime(df['date'] + '-2019', format='%d-%b-%Y')
print (df)
date
0 2019-10-10
1 2019-10-11