I have my DateTime like this
0 2021-02-01 00:01:02.327
1 2021-02-01 00:01:21.445
2 2021-02-01 00:01:31.912
3 2021-02-01 00:01:32.600
4 2021-02-01 00:02:08.920
However, I would like them in the format of %Y.%m.%d %H:%M:%S. I have tried with these methods, but they did not make any difference.
df['Date'] = pd.to_datetime(df['Date']).dt.time
or
df['Date'] = pd.to_datetime(df['Date'],format='%Y.%m.%d %H:%M:%S')
Another quick question is if I would like to add a new column named 'Time', could I please have them only have the information only in %H:%M:%S from Date column?
0 00:01:02.327
1 00:01:21.445
2 00:01:31.912
3 00:01:32.600
4 00:02:08.920
Use Series.dt.floor by seconds:
df['Date'] = pd.to_datetime(df['Date']).dt.floor('S')
Related
I have a dataframe with two date columns (format: YYYY-MM-DD). I want to create one row for each year between those two dates. The rows would be identical with a new column which specifies the year. For example, if the dates are 2018-01-01 and 2020-01-01 then there would be three rows with same data and a new column with values 2018, 2019, and 2020.
You can use a custom function to compute the range then explode the column:
# Ensure to have datetime
df['date1'] = pd.to_datetime(df['date1'])
df['date2'] = pd.to_datetime(df['date2'])
# Create the new column
date_range = lambda x: range(x['date1'].year, x['date2'].year+1)
df = df.assign(year=df.apply(date_range, axis=1)).explode('year', ignore_index=True)
Output:
>>> df
date1 date2 year
0 2018-01-01 2020-01-01 2018
1 2018-01-01 2020-01-01 2019
2 2018-01-01 2020-01-01 2020
This should work for you:
import pandas
# some sample data
df = pandas.DataFrame(data={
'foo': ['bar', 'baz'],
'date1':['2018-01-01', '2022-01-01'],
'date2':['2020-01-01', '2017-01-01']
})
# cast date columns to datetime
for col in ['date1', 'date2']:
df[col] = pandas.to_datetime(df[col])
# reset index to ensure that selection by length of index works
df = df.reset_index(drop=True)
# the range of years between the two dates, and iterate through the resulting
# series to unpack the range of years and add a new row with the original data and the year
for i, years in df.apply(
lambda x: range(
min(x.date1, x.date2).year,
max(x.date1, x.date2).year + 1
),
axis='columns'
).iteritems():
for year in years:
new_index = len(df.index)
df.loc[new_index] = df.loc[i].values
df.loc[new_index, 'year'] = int(year)
output:
>>> df
foo date1 date2 year
0 bar 2018-01-01 2020-01-01 NaN
1 baz 2022-01-01 2017-01-01 NaN
2 bar 2018-01-01 2020-01-01 2018.0
3 bar 2018-01-01 2020-01-01 2019.0
4 bar 2018-01-01 2020-01-01 2020.0
5 baz 2022-01-01 2017-01-01 2017.0
6 baz 2022-01-01 2017-01-01 2018.0
7 baz 2022-01-01 2017-01-01 2019.0
8 baz 2022-01-01 2017-01-01 2020.0
9 baz 2022-01-01 2017-01-01 2021.0
10 baz 2022-01-01 2017-01-01 2022.0
I have a DataFrame with multiple formats as shown below
0 07-04-2021
1 06-03-1991
2 12-10-2020
3 07/04/2021
4 05/12/1996
What I want is to have one format after applying the Pandas function to the entire column so that all the dates are in the format
date/month/year
What I tried is the following
date1 = pd.to_datetime(df['Date_Reported'], errors='coerce', format='%d/%m/%Y')
But it is not working out. Can this be done? Thank you
try with dayfirst=True:
date1=pd.to_datetime(df['Date_Reported'], errors='coerce',dayfirst=True)
output of date1:
0 2021-04-07
1 1991-03-06
2 2020-10-12
3 2021-04-07
4 1996-12-05
Name: Date_Reported, dtype: datetime64[ns]
If needed:
date1=date1.dt.strftime('%d/%m/%Y')
output of date1:
0 07/04/2021
1 06/03/1991
2 12/10/2020
3 07/04/2021
4 05/12/1996
Name: Date_Reported, dtype: object
I cannot find a solution for this problem. I would like to add future dates to a datetime indexed Pandas dataframe for model prediction purposes.
Here is where I am right now:
new_datetime = df2.index[-1:] # current end of datetime index
increment = '1 days' # string for increment - eventually will be in a for loop to add add'l days
new_datetime = new_datetime+pd.Timedelta(increment)
And this is where I am stuck. The append examples online only seem always seem to show examples with ignore_index=True , and in my case, I want to use the proper datetime indexing.
Suppose you have this df:
date value
0 2020-01-31 00:00:00 1
1 2020-02-01 00:00:00 2
2 2020-02-02 00:00:00 3
then an alternative for adding future days is
df.append(pd.DataFrame({'date': pd.date_range(start=df.date.iloc[-1], periods=6, freq='D', closed='right')}))
which returns
date value
0 2020-01-31 00:00:00 1.0
1 2020-02-01 00:00:00 2.0
2 2020-02-02 00:00:00 3.0
0 2020-02-03 00:00:00 NaN
1 2020-02-04 00:00:00 NaN
2 2020-02-05 00:00:00 NaN
3 2020-02-06 00:00:00 NaN
4 2020-02-07 00:00:00 NaN
where the frequency is D (days) day and the period is 6 days.
I think I was making this more difficult than necessary because I was using a datetime index instead of the typical integer index. By leaving the 'date' field as a regular column instead of an index adding the rows is straightforward.
One thing I did do was add a reindex command so I did not end up with wonky duplicate index values:
df = df.append(pd.DataFrame({'date': pd.date_range(start=df.date.iloc[-1], periods=21, freq='D', closed='right')}))
df = df.reset_index() # resets index
i also needed this and i solve merging the code that you share with the code on this other response add to a dataframe as I go with datetime index and end out with the following code that work for me.
data=raw.copy()
new_datetime = data.index[-1:] # current end of datetime index
increment = '1 days' # string for increment - eventually will be in a for loop to add add'l days
new_datetime = new_datetime+pd.Timedelta(increment)
today_df = pd.DataFrame({'value': 301.124},index=new_datetime)
data = data.append(today_df)
data.tail()
here 'value' is the header of your own dataframe
I have a data frame with two columns Date and value.
I want to add new column named week_number that basically is how many weeks back from the given date
import pandas as pd
df = pd.DataFrame(columns=['Date','value'])
df['Date'] = [ '04-02-2019','03-02-2019','28-01-2019','20-01-2019']
df['value'] = [10,20,30,40]
df
Date value
0 04-02-2019 10
1 03-02-2019 20
2 28-01-2019 30
3 20-01-2019 40
suppose given date is 05-02-2019.
Then I need to add a column week_number in a way such that how many weeks back the Date column date is from given date.
The output should be
Date value week_number
0 04-02-2019 10 1
1 03-02-2019 20 1
2 28-01-2019 30 2
3 20-01-2019 40 3
how can I do this in pandas
First convert column to datetimes by to_datetime with dayfirst=True, then subtract from right side by rsub, convert timedeltas to days, get modulo by 7 and add 1:
df['Date'] = pd.to_datetime(df['Date'], dayfirst=True)
df['week_number'] = df['Date'].rsub(pd.Timestamp('2019-02-05')).dt.days // 7 + 1
#alternative
#df['week_number'] = (pd.Timestamp('2019-02-05') - df['Date']).dt.days // 7 + 1
print (df)
Date value week_number
0 2019-02-04 10 1
1 2019-02-03 20 1
2 2019-01-28 30 2
3 2019-01-20 40 3
i have the following data frame :
correction = ['2.0','-2.5','4.5','-3.0']
date = ['2015-05-19 20:45:00','2017-04-29 17:15:00','2011-05-09 10:40:00','2016-12-18 16:10:00']
i want to convert correction as hours and add it to the date. i tried the following code, but it get the error.
df['correction'] = pd.to_timedelta(df['correction'],unit='h')
df['date'] =pd.DatetimeIndex(df['date'])
df['date'] = df['date'] + df['correction']
I get the error in converting correction to timedelta as:
ValueError: no units specified
For me works cast to float column correction:
df['correction'] = pd.to_timedelta(df['correction'].astype(float),unit='h')
df['date'] = pd.DatetimeIndex(df['date'])
df['date'] = df['date'] + df['correction']
print (df)
correction date
0 02:00:00 2015-05-19 22:45:00
1 -1 days +21:30:00 2017-04-29 14:45:00
2 04:30:00 2011-05-09 15:10:00
3 -1 days +21:00:00 2016-12-18 13:10:00