I have a data frame with "Date" column in UTC format.
Date
2021-10-14T06:57:00.000+0000
2021-09-05T08:30:00.000+0000
2021-10-20T04:34:00.000+0000
2021-10-19T21:49:00.000+0000
2021-09-30T20:53:00.000+0000
Tried this but didnt work;
df['Date'] = df['Date'].substr(replace(to_iso8601(from_iso8601_timestamp(Date) AT TIME ZONE 'Australia/Melbourne'), 'T', ' '), 1, 16) Date_local
I am unable to converte the UTC time to the local time zone (Australia/Melbourne).
Any help would be highly appreciated.
use pandas functionality; pd.to_datetime and then tz_convert.
# input strings to datetime data type:
df['Date'] = pd.to_datetime(df['Date'])
# UTC is already set (aware datetime); just convert:
df['Date'] = df['Date'].dt.tz_convert('Australia/Melbourne')
df['Date']
Out[2]:
0 2021-10-14 17:57:00+11:00
1 2021-09-05 18:30:00+10:00
2 2021-10-20 15:34:00+11:00
3 2021-10-20 08:49:00+11:00
4 2021-10-01 06:53:00+10:00
Name: Date, dtype: datetime64[ns, Australia/Melbourne]
Related
I have a dataframe with columns:
time: time in UTC format
timezone: the corresponding timezone.
time timezone
0 2022-12-28T20:16:31.373Z Europe/Athens
1 2022-07-28T20:16:31.373Z Europe/Athens
2 2022-11-01T21:35:35.865Z Europe/Dublin
3 2022-08-03T19:44:07.611Z America/Los_Angeles
4 2022-08-02T12:44:44.360Z Europe/Minsk
I want to:
Convert UTC time to Local time (using timezone)
Remove the Timezone and just keep the datetime
It seems to me that this solution works, but want to make sure that I am not missing something (eg. doesn't deal with dailight saving or something)
import pandas as pd
# example dataframe
df = pd.DataFrame({
'time' : ['2022-12-28T20:16:31.373Z', '2022-07-28T20:16:31.373Z', '2022-11-01T21:35:35.865Z', '2022-08-03T19:44:07.611Z', '2022-08-02T12:44:44.360Z'],
'timezone': ['Europe/Athens', 'Europe/Athens', 'Europe/Dublin', 'America/Los_Angeles', 'Europe/Minsk']
})
# function
def get_local_time (timestamp: pd.Timestamp, timezone: str) -> pd.Timestamp:
timestamp = pd.to_datetime(timestamp).tz_convert(timezone).replace(tzinfo=None)
return timestamp
df['local_time'] = df.apply(lambda row: get_local_time(row['time'], row['timezone']), axis = 1).dt.round(freq='S')
print (df)
---
OUT:
time timezone local_time
0 2022-12-28T20:16:31.373Z Europe/Athens 2022-12-28 22:16:31
1 2022-07-28T20:16:31.373Z Europe/Athens 2022-07-28 23:16:31
2 2022-11-01T21:35:35.865Z Europe/Dublin 2022-11-01 21:35:36
3 2022-08-03T19:44:07.611Z America/Los_Angeles 2022-08-03 12:44:08
4 2022-08-02T12:44:44.360Z Europe/Minsk 2022-08-02 15:44:44
How to remove T00:00:00+05:30 after year, month and date values in pandas? I tried converting the column into datetime but also it's showing the same results, I'm using pandas in streamlit. I tried the below code
df['Date'] = pd.to_datetime(df['Date'])
The output is same as below :
Date
2019-07-01T00:00:00+05:30
2019-07-01T00:00:00+05:30
2019-07-02T00:00:00+05:30
2019-07-02T00:00:00+05:30
2019-07-02T00:00:00+05:30
2019-07-03T00:00:00+05:30
2019-07-03T00:00:00+05:30
2019-07-04T00:00:00+05:30
2019-07-04T00:00:00+05:30
2019-07-05T00:00:00+05:30
Can anyone help me how to remove T00:00:00+05:30 from the above rows?
If I understand correctly, you want to keep only the date part.
Convert date strings to datetime
df = pd.DataFrame(
columns={'date'},
data=["2019-07-01T02:00:00+05:30", "2019-07-02T01:00:00+05:30"]
)
date
0 2019-07-01T02:00:00+05:30
1 2019-07-02T01:00:00+05:30
2 2019-07-03T03:00:00+05:30
df['date'] = pd.to_datetime(df['date'])
date
0 2019-07-01 02:00:00+05:30
1 2019-07-02 01:00:00+05:30
Remove the timezone
df['datetime'] = df['datetime'].dt.tz_localize(None)
date
0 2019-07-01 02:00:00
1 2019-07-02 01:00:00
Keep the date only
df['date'] = df['date'].dt.date
0 2019-07-01
1 2019-07-02
Don't bother with apply to Python dates or string changes. The former will leave you with an object type column and the latter is slow. Just round to the day frequency using the library function.
>>> pd.Series([pd.Timestamp('2000-01-05 12:01')]).dt.round('D')
0 2000-01-06
dtype: datetime64[ns]
If you have a timezone aware timestamp, convert to UTC with no time zone then round:
>>> pd.Series([pd.Timestamp('2019-07-01T00:00:00+05:30')]).dt.tz_convert(None) \
.dt.round('D')
0 2019-07-01
dtype: datetime64[ns]
Pandas doesn't have a builtin conversion to datetime.date, but you could use .apply to achieve this if you want to have date objects instead of string:
import pandas as pd
import datetime
df = pd.DataFrame(
{"date": [
"2019-07-01T00:00:00+05:30",
"2019-07-01T00:00:00+05:30",
"2019-07-02T00:00:00+05:30",
"2019-07-02T00:00:00+05:30",
"2019-07-02T00:00:00+05:30",
"2019-07-03T00:00:00+05:30",
"2019-07-03T00:00:00+05:30",
"2019-07-04T00:00:00+05:30",
"2019-07-04T00:00:00+05:30",
"2019-07-05T00:00:00+05:30"]})
df["date"] = df["date"].apply(lambda x: datetime.datetime.fromisoformat(x).date())
print(df)
import pandas as pd
import datetime
dictt={'s_time': ["06:30:00", "07:30:00","16:30:00"], 'f_time': ["10:30:00", "23:30:00","23:30:00"]}
df=pd.DataFrame(dictt)
in this case i want to convert them times in to datetime object so i can later on use it for calculation or others.
when i command df['s_time']=pd.to_datetime(df['s_time'],format='%H:%M:%S').dt.time
it gives error:
time data '24:00:00' does not match format '%H:%M:%S' (match)
so i dont know how to fix this
"24:00:00" means "00:00:00"
If it's just "24:00:00" that's causing trouble, you can replace the "24:" prefix with "00:":
import pandas as pd
df = pd.DataFrame({'time': ["06:30:24", "07:24:00", "24:00:00"]})
# replace prefix "24:" with "00:"
df['time'] = df['time'].str.replace('^24:', '00:', regex=True)
# now to_datetime
df['time'] = pd.to_datetime(df['time'])
df['time']
0 2021-04-17 06:30:24
1 2021-04-17 07:24:00
2 2021-04-17 00:00:00
Name: time, dtype: datetime64[ns]
1 to 24 hour clock (instead of 0 to 23)
If however your time notation goes from 1 to 24 hours (instead of 0 to 23), you can parse string to timedelta, subtract one hour and then cast to datetime:
df = pd.DataFrame({'time': ["06:30:24", "07:24:00", "24:00:00"]})
# to timedelta and subtract one hour
df['time'] = pd.to_timedelta(df['time']) - pd.Timedelta(hours=1)
# to string and then datettime:
df['time'] = pd.to_datetime(df['time'].astype(str).str.split(' ').str[-1])
df['time']
0 2021-04-17 05:30:24
1 2021-04-17 06:24:00
2 2021-04-17 23:00:00
Name: time, dtype: datetime64[ns]
Note: the underlying assumption here is that the date is irrelevant. If there also is a date, see the related question I linked in the comments section.
Can anyone solve this problem! I am trying to convert a Date object column to Datetime string format with the help of python. From 'YY-mm-dd' to 'YY/mm/dd 00:00' format. Dataset is given below. I have tried every options like energy_df['Date']= pd.to_datetime(energy_df['Date']),
energy_df['Date'] = pd.to_datetime(energy_df['Date'])
energy_df['month'] = energy_df['Date'].dt.month.astype(int)
energy_df['day_of_month'] = energy_df['Date'].dt.day.astype(int)
energy_df['day_of_week'] = energy_df['Date'].dt.dayofweek.astype(int)
energy_df['hour_of_day'] = energy_df['Hours']
selected_columns = ['Date', 'day_of_week', 'hour_of_day', 'Avg Specific Humidity[g/Kg]']
energy_df = energy_df[selected_columns]
Dataset image:
Convert the 'date' column to dtype datetime, the 'hour' column to dtype timedelta, add them together, and format to string.
Ex:
import pandas as pd
# some dummy input...
df = pd.DataFrame({'date': ['2015-01-01', '2015-01-01', '2015-01-01'],
'hour': [1, 2, 3]})
# to datetime / timedelta...
df['datetime'] = pd.to_datetime(df['date']) + pd.to_timedelta(df['hour'], unit='h')
# and format to string...
df['timestamp'] = df['datetime'].dt.strftime('%Y/%m/%d %H:%M')
# will give you:
df
date hour datetime timestamp
0 2015-01-01 1 2015-01-01 01:00:00 2015/01/01 01:00
1 2015-01-01 2 2015-01-01 02:00:00 2015/01/01 02:00
2 2015-01-01 3 2015-01-01 03:00:00 2015/01/01 03:00
I have a dataset where I have 2 columns in a data frame - Date in YYYY-MM-DD format and another column with Hour in format 0100 (for 1am) until 2300 (for 12pm).
Date Hour
2017-01-01 0200
2017-01-01 0400
etc
In order to get it ready for Time series mode, I want to convert these into datetime objects and concatenate these columns. Example output desired: 2017-01-01 01:00:00, etc
I have tried df['Date'] = pd.to_datetime(df['Date']) and converted this into datetime object, But I'm struggling with the Hour column. Please help
This is one way. The trick is to note that pd.to_datetime is actually quite flexible: it accepts strings of the format "YYYY-MM-DD HHMM".
I assume here that your Hour is given as a string (otherwise leading zeros are not possible).
import pandas as pd
df = pd.DataFrame({'Date': ['2017-01-01', '2017-01-01'],
'Hour': ['0200', '0400']})
# as per #COLDSPEED's suggestion
df['DateTime'] = pd.to_datetime(df['Date'] + ' ' + df['Hour'])
print(df)
# Date Hour DateTime
# 0 2017-01-01 0200 2017-01-01 02:00:00
# 1 2017-01-01 0400 2017-01-01 04:00:00
print(df.dtypes)
# Date object
# Hour object
# DateTime datetime64[ns]
# dtype: object
Previous version with pd.DataFrame.apply is possible but inefficient:
df['DateTime'] = df.apply(lambda x: x['Date'] + ' ' + x['Hour'], axis=1)\
.apply(pd.to_datetime)