I'm curious about how to use pandas to deal with this sort of info in a .csv file:
2022-08-11 11:50:01 America/Los_Angeles
My goal is to extract the date, hour and minute, and the timezone info for further analysis.
I have tried to lift out the date and time using:
df['Date'] = pd.to_datetime(df['datetime']).dt.date
but got an error because of the string at the end. Other than extracting the date and time using specific indices, is there any better and quicker way? Thank you so much.
pandas cannot handle a datetime column with different timezones. You can start by splitting the datetime and timezone in separate columns:
df[['datetime', 'timezone']] = df['datetime'].str.rsplit(' ', n=1, expand=True)
df['datetime'] = pd.to_datetime(df['datetime']) # this column now has the datetime64[ns] type
Now you are able to do the following:
df['date_only'] = df['datetime'].dt.date
If you want to express all local date/times in America/Los_Angeles time:
df['LA_datetime'] = df.apply(lambda x: x['datetime'].tz_localize(tz=x['timezone']).tz_convert('America/Los_Angeles'), axis = 1)
You can change America/Los_Angeles to the timezone of your liking.
Related
I have a column of years from the sunspots dataset.
I want to convert column 'year' in integer e.g. 1992 to datetime format then find the time delta and eventually compute total seconds (cumulative) to represent the time index column of a time series.
I am trying to use the following code but I get the error
TypeError: dtype datetime64[ns] cannot be converted to timedelta64[ns]
sunspots_df['year'] = pd.to_timedelta(pd.to_datetime(sunspots_df['year'], format='%Y') ).dt.total_seconds()
pandas.Timedelta "[r]epresents a duration, the difference between two dates or times." So you're trying to get Python to tell you the difference between a particular datetime and...nothing. That's why it's failing.
If it's important that you store your index this way (and there may be better ways), then you need to pick a start datetime and compute the difference to get a timedelta.
For example, this code...
import pandas as pd
df = pd.DataFrame({'year': [1990,1991,1992]})
diff = (pd.to_datetime(df['year'], format='%Y') - pd.to_datetime('1990', format='%Y'))\
.dt.total_seconds()
...returns a series whose values are seconds from January 1st, 1990. You'll note that it doesn't invoke pd.to_timedelta(), because it doesn't need to: the result of the subtraction is automatically a pd.timedelta column.
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)
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 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 dealing with timestamps that, according to Google's documentation, are:
A timestamp in RFC3339 UTC "Zulu" format, accurate to nanoseconds. Example: "2014-10-02T15:01:23.045123456Z".
So, for example, if the string is '2019-11-06T06:24:42.558008Z', then pd.to_datetime('2019-11-06T06:24:42.558008Z',infer_datetime_format=True) works and returns Timestamp('2019-11-06 06:24:42.558008').
However, letting Pandas infer the format is slow, and I have many rows of data. What would I pass the format parameter to help speed up the processing?
You could use to_datetime with utc=True + tz_convert:
import pandas as pd
utc = pd.to_datetime('2019-11-06T06:24:42.558008Z', utc=True).tz_convert(None)
inferred = pd.to_datetime('2019-11-06T06:24:42.558008Z', infer_datetime_format=True)
print(utc == inferred)
Output
True
From the documentation on tz_convert:
A tz of None will convert to UTC and remove the timezone information.
Note that only doing:
utc = pd.to_datetime('2019-11-06T06:24:42.558008Z', utc=True) # or pd.to_datetime('2019-11-06T06:24:42.558008Z')
throws a TypeError exception when comparing with inferred:
TypeError: Cannot compare tz-naive and tz-aware timestamps