I have this type of date '20181115 0756' and also in a diffent dataframe in this format '2018-11-15'. I would like to know if there is any way to convert it to datetime without the hours and minutes
date['DATE']= pd.to_datetime(date.DATE)
this converts it to 218-11-15 00:00:00 and I'd like to avoid that
What I trying to do is to calcuate the time difference between the dates in the two dataframes that I have
Thank you in advance
You can use the following code
date['DATE'] = pd.to_datetime(date['DATE'], errors='coerce').dt.date
Related
I have a dataset which stores durations like 3 hours and 7 minutes in the format of, 3.11 as a string.
I want to convert the column containing these values into datetime in a way that I get: 03:07.
When I do:
df["ConnectedDuration"] = pd.to_datetime(df['ConnectedDuration'])
I get: 1970-01-01 00:00:00.000000003 which is obviousely not what I want.
When I do:
df["ConnectedDuration"] = pd.to_datetime(df['ConnectedDuration'], format='%H:%M')
I get the following error: ValueError: time data '3' does not match format '%H:%M' (match)
Any help is highly appreciated
You want to convert this values to timedelta instead of datetime. Thus you should use the pd.to_timedelta method, like:
pd.to_timedelta(df["ConnectedDuration"].astype('float'), unit='h')
I'm having trouble align two different dates. I have an excel import which I turn into a DateTime in pandas and I would like to compare this DateTime with the current DateTime. The troubles are in the formatting of the imported DateTime.
Excel format of the date:
2020-07-06 16:06:00 (which is yyyy-dd-mm hh:mm:ss)
When I add the DateTime to my DataFrame it creates the datatype Object. After I convert it with pd.to_datetime it creates the format yyyy-mm-dd hh:mm:ss. It seems that the month and the day are getting mixed up.
Example code:
df = pd.read_excel('my path')
df['Arrival'] = pd.to_datetime(df['Arrival'], format='%Y-%d-%m %H:%M:%S')
print(df.dtypes)
Expected result:
2020-06-07 16:06:00
Actual result:
2020-07-06 16:06:00
How do I resolve this?
Gr,
Sempah
An ISO-8601 date/time is always yyyy-MM-dd, not yyyy-dd-MM. You've got the month and date positions switched around.
While localized date/time strings are inconsistent about the order of month and date, this particular format where the year comes first always starts with the biggest units (years) and decreases in unit size going right (month, date, hour, etc.)
It's solved. I think that I misunderstood the results. It already was working without me knowledge. Thanks for the help anyway.
I have a pandas dataframe with a column of orderdates formatted like this: 2019-12-26.
However when I take the max of this date it will give 2019-12-12. While it is actually 2019-12-26. It makes sense because my dateformat is Dutch and the max() function uses the 'American' (correct me if I'm wrong) format.
This meas that my calculations aren't correct.
How I can change the way the function calculate? Or if thats not possible, change the format of my date column so the calculations are correct?
[In] df['orderdate'] = df['orderdate'].astype('datetime64[ns]')
print(df["orderdate"].max())
[Out] 2019-12-12 00:00:00
Thank you!
Currently, I have a series of Datetime Values that display as so
0 Datetime
1 20041001
2 20041002
3 20041003
4 20041004
they are within a series named
d['Datetime']
They were originally something like
20041001ABCDEF
But I split the end off just to leave them with the remaining numbers. How do I go about putting them into the following format?
2004-10-01
You can do the following,
df['Datetime'] = pd.to_datetime(df['Datetime'], format='%Y%m%d'))
I have a string object in this format 2014-12-08 09:30:00.066000 but I want to convert to datetime variable. I also want this to be less granular- I want it to be just in the order of second for example
2014-12-08 09:30:00.066000 to 2014-12-08 09:30:00
I am trying to use pd.to_datetime function but it's not working for me. Anyone know how to do this? Thanks!
See this:
How to round a Pandas `DatetimeIndex`?
from pandas.lib import Timestamp
def to_the_second(ts):
return Timestamp(long(round(ts.value, -9)))
df['My_Date_Column'].apply(to_the_second)