Im learning python (3.6 with anaconda) for my studies.
Im using pandas to import a xls file with 2 columns : Date (dd-mm-yyyy) and price.
But pandas changes the date format :
xls_file = pd.read_excel('myfile.xls')
print(xls_file.iloc[0, 0])
Im getting :
2010-01-04 00:00:00
instead of :
04-01-2010 or at least : 2010-01-04
I dont know why hh:mm:ss is added, I get the same result for each row from the Date column. I tried also different things using to_datetime but it didnt fix it.
Any idea ?
Thanks
What you need is to define the format that the datetime values get printed. There might be a more elegant way to do it but something like that will work:
In [11]: df
Out[11]:
id date
0 1 2017-09-12
1 2 2017-10-20
# Specifying the format
In [16]: print(pd.datetime.strftime(df.iloc[0,1], "%Y-%m-%d"))
2017-09-12
If you want to store the date as string in your specific format then you can also do something like:
In [17]: df["datestr"] = pd.datetime.strftime(df.iloc[0,1], "%Y-%m-%d")
In [18]: df
Out[18]:
id date datestr
0 1 2017-09-12 2017-09-12
1 2 2017-10-20 2017-09-12
In [19]: df.dtypes
Out[19]:
id int64
date datetime64[ns]
datestr object
dtype: object
Related
Hello there stackoverflow community,
I would like to change the datetime format of a column, but I doesn't work and I don't know what I'am doing wrong.
After executing the following code:
df6['beginn'] = pd.to_datetime(df6['beginn'], unit='s', errors='ignore')
I got this output, and thats fine, but i would like to take out the hour to have only %m/%d/%Y left.
ID DATE
91060 2017-11-10 00:00:00
91061 2022-05-01 00:00:00
91062 2022-04-01 00:00:00
Name: beginn, Length: 91063, dtype: object
I've tried this one and many others
df6['beginn'] = df6['beginn'].dt.strftime('%m/%d/%Y')
and get the following output:
AttributeError: Can only use .dt accessor with datetimelike values.
But I don't understand why, I've transformed the data with pd.to_datetime or not?
Appreciate any hint you can give me! Thanks a lot!
The reason you have to use errors="ignore" is because not all the dates you are parsing are in the correct format. If you use errors="coerce" like #phi has mentioned then any dates that cannot be converted will be set to NaT. The columns datatype will still be converted to datatime64 and you can then format as you like and deal with the NaT as you want.
Example
A dataframe with one item in Date not written as Year/Month/Day (25th Month is wrong):
>>> df = pd.DataFrame({'ID': [91060, 91061, 91062, 91063], 'Date': ['2017/11/10', '2022/05/01', '2022/04/01', '2055/25/25']})
>>> df
ID Date
0 91060 2017/11/10
1 91061 2022/05/01
2 91062 2022/04/01
3 91063 2055/25/25
>>> df.dtypes
ID int64
Date object
dtype: object
Using errors="ignore":
>>> df['Date'] = pd.to_datetime(df['Date'], errors='ignore')
>>> df
ID Date
0 91060 2017/11/10
1 91061 2022/05/01
2 91062 2022/04/01
3 91063 2055/25/25
>>> df.dtypes
ID int64
Date object
dtype: object
Column Date is still an object because not all the values have been converted. Running df['Date'] = df['Date'].dt.strftime("%m/%d/%Y") will result in the AttributeError
Using errors="coerce":
>>> df['Date'] = pd.to_datetime(df['Date'], errors='coerce')
>>> df
ID Date
0 91060 2017-11-10
1 91061 2022-05-01
2 91062 2022-04-01
3 91063 NaT
>>> df.dtypes
ID int64
Date datetime64[ns]
dtype: object
Invalid dates are set to NaT and the column is now of type datatime64 and you can now format it:
>>> df['Date'] = df['Date'].dt.strftime("%m/%d/%Y")
>>> df
ID Date
0 91060 11/10/2017
1 91061 05/01/2022
2 91062 04/01/2022
3 91063 NaN
Note: When formatting datatime64, it is converted back to type object so NaT's are changed to NaN. The issue you are having is a case of some dirty data not in the correct format.
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)
I have this column in pandas df:
'''
full_date
2020-12-02T08:11:30-0600
2020-12-02T02:11:50-0600
2020-12-03T08:56:29-0600
'''
I only need the date, hoping to have this column:
'''
date
2020-12-02
2020-12-02
2020-12-03
'''
I have tried to find the solution from previous questions, but still failed. If anyone can help, I will appreciate that a lot. thanks.
In case your column is not a datetime type, you can convert it to that and then use the .dt accessor to get just the date:
>>> df["date"] = df["full_date"].pipe(pd.to_datetime, utc=True).dt.date
>>> print(df)
full_date date
0 2020-12-02T08:11:30-0600 2020-12-02
1 2020-12-02T02:11:50-0600 2020-12-02
2 2020-12-03T08:56:29-0600 2020-12-03
You can convert the datetime very easily using this python code, if suitable.
from dateutil.parser import parse
var = "2020-12-02T08:11:30-0600"
parseddate = parse(var).date()
I have a dataframe in pandas called 'munged_data' with two columns 'entry_date' and 'dob' which i have converted to Timestamps using pd.to_timestamp.I am trying to figure out how to calculate ages of people based on the time difference between 'entry_date' and 'dob' and to do this i need to get the difference in days between the two columns ( so that i can then do somehting like round(days/365.25). I do not seem to be able to find a way to do this using a vectorized operation. When I do munged_data.entry_date-munged_data.dob i get the following :
internal_quote_id
2 15685977 days, 23:54:30.457856
3 11651985 days, 23:49:15.359744
4 9491988 days, 23:39:55.621376
7 11907004 days, 0:10:30.196224
9 15282164 days, 23:30:30.196224
15 15282227 days, 23:50:40.261632
However i do not seem to be able to extract the days as an integer so that i can continue with my calculation.
Any help appreciated.
Using the Pandas type Timedelta available since v0.15.0 you also can do:
In[1]: import pandas as pd
In[2]: df = pd.DataFrame([ pd.Timestamp('20150111'),
pd.Timestamp('20150301') ], columns=['date'])
In[3]: df['today'] = pd.Timestamp('20150315')
In[4]: df
Out[4]:
date today
0 2015-01-11 2015-03-15
1 2015-03-01 2015-03-15
In[5]: (df['today'] - df['date']).dt.days
Out[5]:
0 63
1 14
dtype: int64
You need 0.11 for this (0.11rc1 is out, final prob next week)
In [9]: df = DataFrame([ Timestamp('20010101'), Timestamp('20040601') ])
In [10]: df
Out[10]:
0
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [11]: df = DataFrame([ Timestamp('20010101'),
Timestamp('20040601') ],columns=['age'])
In [12]: df
Out[12]:
age
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [13]: df['today'] = Timestamp('20130419')
In [14]: df['diff'] = df['today']-df['age']
In [16]: df['years'] = df['diff'].apply(lambda x: float(x.item().days)/365)
In [17]: df
Out[17]:
age today diff years
0 2001-01-01 00:00:00 2013-04-19 00:00:00 4491 days, 00:00:00 12.304110
1 2004-06-01 00:00:00 2013-04-19 00:00:00 3244 days, 00:00:00 8.887671
You need this odd apply at the end because not yet full support for timedelta64[ns] scalars (e.g. like how we use Timestamps now for datetime64[ns], coming in 0.12)
Not sure if you still need it, but in Pandas 0.14 i usually use .astype('timedelta64[X]') method
http://pandas.pydata.org/pandas-docs/stable/timeseries.html (frequency conversion)
df = pd.DataFrame([ pd.Timestamp('20010101'), pd.Timestamp('20040605') ])
df.ix[0]-df.ix[1]
Returns:
0 -1251 days
dtype: timedelta64[ns]
(df.ix[0]-df.ix[1]).astype('timedelta64[Y]')
Returns:
0 -4
dtype: float64
Hope that will help
Let's specify that you have a pandas series named time_difference which has type
numpy.timedelta64[ns]
One way of extracting just the day (or whatever desired attribute) is the following:
just_day = time_difference.apply(lambda x: pd.tslib.Timedelta(x).days)
This function is used because the numpy.timedelta64 object does not have a 'days' attribute.
To convert any type of data into days just use pd.Timedelta().days:
pd.Timedelta(1985, unit='Y').days
84494
I have a dataframe in pandas called 'munged_data' with two columns 'entry_date' and 'dob' which i have converted to Timestamps using pd.to_timestamp.I am trying to figure out how to calculate ages of people based on the time difference between 'entry_date' and 'dob' and to do this i need to get the difference in days between the two columns ( so that i can then do somehting like round(days/365.25). I do not seem to be able to find a way to do this using a vectorized operation. When I do munged_data.entry_date-munged_data.dob i get the following :
internal_quote_id
2 15685977 days, 23:54:30.457856
3 11651985 days, 23:49:15.359744
4 9491988 days, 23:39:55.621376
7 11907004 days, 0:10:30.196224
9 15282164 days, 23:30:30.196224
15 15282227 days, 23:50:40.261632
However i do not seem to be able to extract the days as an integer so that i can continue with my calculation.
Any help appreciated.
Using the Pandas type Timedelta available since v0.15.0 you also can do:
In[1]: import pandas as pd
In[2]: df = pd.DataFrame([ pd.Timestamp('20150111'),
pd.Timestamp('20150301') ], columns=['date'])
In[3]: df['today'] = pd.Timestamp('20150315')
In[4]: df
Out[4]:
date today
0 2015-01-11 2015-03-15
1 2015-03-01 2015-03-15
In[5]: (df['today'] - df['date']).dt.days
Out[5]:
0 63
1 14
dtype: int64
You need 0.11 for this (0.11rc1 is out, final prob next week)
In [9]: df = DataFrame([ Timestamp('20010101'), Timestamp('20040601') ])
In [10]: df
Out[10]:
0
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [11]: df = DataFrame([ Timestamp('20010101'),
Timestamp('20040601') ],columns=['age'])
In [12]: df
Out[12]:
age
0 2001-01-01 00:00:00
1 2004-06-01 00:00:00
In [13]: df['today'] = Timestamp('20130419')
In [14]: df['diff'] = df['today']-df['age']
In [16]: df['years'] = df['diff'].apply(lambda x: float(x.item().days)/365)
In [17]: df
Out[17]:
age today diff years
0 2001-01-01 00:00:00 2013-04-19 00:00:00 4491 days, 00:00:00 12.304110
1 2004-06-01 00:00:00 2013-04-19 00:00:00 3244 days, 00:00:00 8.887671
You need this odd apply at the end because not yet full support for timedelta64[ns] scalars (e.g. like how we use Timestamps now for datetime64[ns], coming in 0.12)
Not sure if you still need it, but in Pandas 0.14 i usually use .astype('timedelta64[X]') method
http://pandas.pydata.org/pandas-docs/stable/timeseries.html (frequency conversion)
df = pd.DataFrame([ pd.Timestamp('20010101'), pd.Timestamp('20040605') ])
df.ix[0]-df.ix[1]
Returns:
0 -1251 days
dtype: timedelta64[ns]
(df.ix[0]-df.ix[1]).astype('timedelta64[Y]')
Returns:
0 -4
dtype: float64
Hope that will help
Let's specify that you have a pandas series named time_difference which has type
numpy.timedelta64[ns]
One way of extracting just the day (or whatever desired attribute) is the following:
just_day = time_difference.apply(lambda x: pd.tslib.Timedelta(x).days)
This function is used because the numpy.timedelta64 object does not have a 'days' attribute.
To convert any type of data into days just use pd.Timedelta().days:
pd.Timedelta(1985, unit='Y').days
84494