I've a dataframe of this format -
var1 date
A 2017/01/01
A 2017/01/02
...
I want the date to be converted into YYYY-MM format but the df['date'].dtype is object.
How can I remove the day part from date while keeping the data type as datetime?
Expected Output -
A - 2017/01
Thanks
You can't have custom representation for the datetime dtype. But you have the following options:
use strings - you might have any representation (as you wish), but all datetime methods and attributes get lost
use datetime, but set the day part to 1 (as #Kopytok) has already shown.
use period dtype, which still allows you to use some date arithmetic
Demo:
In [207]: df
Out[207]:
var1 date
0 A 2018-12-31
1 A 2017-09-07
2 B 2016-02-29
In [208]: df['new'] = df['date'].dt.to_period('M')
In [209]: df
Out[209]:
var1 date new
0 A 2018-12-31 2018-12
1 A 2017-09-07 2017-09
2 B 2016-02-29 2016-02
In [210]: df.dtypes
Out[210]:
var1 object
date datetime64[ns]
new object
dtype: object
In [211]: df['new'] + 8
Out[211]:
0 2019-08
1 2018-05
2 2016-10
Name: new, dtype: object
It is possible replace every date with the first day of month:
pd.to_datetime(d["date"], format="%Y/%m/%d").apply(lambda x: x.replace(day=1))
Result:
0 2017-01-01
1 2017-01-01
I am working with python 3.5.2, pandas 0.18.1 and sqlite3.
In my data base, I have a column unix_time with INT for seconds since 1970. Ideally I want to read my dataframe from sqlite, and then create a time column which would correspond to the datetime or pandas.tslib.Timestamp conversion of the unix_time column that I woul only use for some processing and then drop before saving the dataframe back.
The issue is that when parsing the unix_time column using :
df = pd.read_from_sql_query("SELECT * FROM test", con, parse_dates=['unix_time'])
I obtain pandas.tslib.Timestamp types which is fine for my processing, but then I have to recreate my original unix_time column using :
df['unix_time'][i] = (df['unix_time'][i] - datetime(1970,1,1)).total_seconds()
which is really 'dirty'
First question : Do you have a better way?
I thought about giving up the unix time format and only use datetime format but the to_datetime method from pandas returns in fact pandas.tslib.Timestamp ... And anyway, doing so would force me to iterate over all rows which is a bad solution. (It is impossible to apply to_datetime on something else than a view over a single cell of the dataframe
Second question : Is it possible to apply it on a series?
My last try was with directly using df['time'] = datetime.datetime.fromtimestamp(df['unix_time']) but surprisingly, it also returns pandas.tslib.Timestamp.
In the end, knowing that I can only save unix timestamps or datetimes, my only choices for the moment are :
parsing but then having to convert them back to unix timestamp one by
one.
Or not parse it but have to convert them to pandas.tslib.Timestamp
one by one.
It would be great if I could convert a whole series.
Last question : Is there a way to convert a unix timestamps series to datetime (or at least pandas.tslib.Timestamp), or a pandas.tslib.Timestamp (or datetime) series to unix timestamps?
Thanks
EDIT:
During my processing, I extract a row that I want to append to my dataset. Apparently, the coversion to pandas.tslib.Timestamp appends implicitly when passing from dataframe to serie :
df = pd.DataFrame({'UNX':pd.date_range('2016-01-01', freq='9999S', periods=10).astype(np.int64)//10**9})
df['Date'] = pd.to_datetime(df.UNX, unit='s')
print(df.Date.dtypes)
print(type(df['Date'][0]))
test = df.iloc[0]
print(type(test.Date))
new_df = test.to_frame().transpose() #from here, impossible to do : new_df.to_sql("test", con) because the type for 'Date' is not supported
print(new_df.Date.dtypes)
returns
datetime64[ns]
<class 'pandas.tslib.Timestamp'>
<class 'pandas.tslib.Timestamp'>
object
Is there a way to convert the 'Date' in new_df from pandas.tslib.Timestamp to datetime64[ns] or datetime.datetime (or simply str) ?
IIUC you can do it this way:
In [96]: df = pd.DataFrame({'UNX':pd.date_range('2016-01-01', freq='9999S', periods=10).astype(np.int64)//10**9})
In [97]: df
Out[97]:
UNX
0 1451606400
1 1451616399
2 1451626398
3 1451636397
4 1451646396
5 1451656395
6 1451666394
7 1451676393
8 1451686392
9 1451696391
Convert UNIX epoch to Python datetime:
In [98]: df['Date'] = pd.to_datetime(df.UNX, unit='s')
In [99]: df
Out[99]:
UNX Date
0 1451606400 2016-01-01 00:00:00
1 1451616399 2016-01-01 02:46:39
2 1451626398 2016-01-01 05:33:18
3 1451636397 2016-01-01 08:19:57
4 1451646396 2016-01-01 11:06:36
5 1451656395 2016-01-01 13:53:15
6 1451666394 2016-01-01 16:39:54
7 1451676393 2016-01-01 19:26:33
8 1451686392 2016-01-01 22:13:12
9 1451696391 2016-01-02 00:59:51
Convert datetime to UNIX epoch:
In [100]: df['UNX2'] = df.Date.astype('int64')//10**9
In [101]: df
Out[101]:
UNX Date UNX2
0 1451606400 2016-01-01 00:00:00 1451606400
1 1451616399 2016-01-01 02:46:39 1451616399
2 1451626398 2016-01-01 05:33:18 1451626398
3 1451636397 2016-01-01 08:19:57 1451636397
4 1451646396 2016-01-01 11:06:36 1451646396
5 1451656395 2016-01-01 13:53:15 1451656395
6 1451666394 2016-01-01 16:39:54 1451666394
7 1451676393 2016-01-01 19:26:33 1451676393
8 1451686392 2016-01-01 22:13:12 1451686392
9 1451696391 2016-01-02 00:59:51 1451696391
Check:
In [102]: df.UNX.eq(df.UNX2).all()
Out[102]: True
Round trip between Pandas Timestamp and Unix Seconds (since 1970-01-01):
date_in = pd.to_datetime("2022-04-07")
# type(date_in) is: pandas._libs.tslibs.timestamps.Timestamp
unix_seconds = date_in.value//10**9
date_out = pd.to_datetime(unix_seconds, unit="s")
Output:
date_in
Out[1]: Timestamp('2021-04-07 00:00:00')
unix_seconds
Out[2]: 1617753600
date_out
Out[3]: Timestamp('2021-04-07 00:00:00')
I read a csv file into pandas dataframe df and I get the following:
df.columns
Index([u'TDate', u'Hour', u'SPP'], dtype='object')
>>> type(df['TDate'][0])
<class 'pandas.tslib.Timestamp'>
type(df['Hour'][0])
<type 'numpy.int64'>
>>> type(df['TradingDate'])
<class 'pandas.core.series.Series'>
>>> type(df['Hour'])
<class 'pandas.core.series.Series'>
Both the Hour and TDate columns have 100 elements. I want to add the corresponding elements of Hour to TDate.
I tried the following:
import pandas as pd
from datetime import date, timedelta as td
z3 = pd.DatetimeIndex(df['TDate']).to_pydatetime() + td(hours = df['Hour'])
But I get error as it seems td doesn't take array as argument. How do I add each element of Hour to corresponding element of TDate.
I think you can add to column TDate column Hour converted to_timedelta with unit='h':
df = pd.DataFrame({'TDate':['2005-01-03','2005-01-04','2005-01-05'],
'Hour':[4,5,6]})
df['TDate'] = pd.to_datetime(df.TDate)
print (df)
Hour TDate
0 4 2005-01-03
1 5 2005-01-04
2 6 2005-01-05
df['TDate'] += pd.to_timedelta(df.Hour, unit='h')
print (df)
Hour TDate
0 4 2005-01-03 04:00:00
1 5 2005-01-04 05:00:00
2 6 2005-01-05 06:00:00
I have a pandas dataframe with a date column the data type is datetime64[ns]. there are over 1000 observations in the dataframe. I want to transform the following column:
date
2013-05-01
2013-05-01
to
date
05/2013
05/2013
or
date
05-2013
05-2013
EDIT//
this is my sample code as of now
test = pd.DataFrame({'a':['07/2017','07/2017',pd.NaT]})
a
0 2017-07-13
1 2017-07-13
2 NaT
test['a'].apply(lambda x: x if pd.isnull(x) == True else x.strftime('%Y-%m'))
0 2017-07-01
1 2017-07-01
2 NaT
Name: a, dtype: datetime64[ns]
why did only the date change and not the format?
You can convert datetime64 into whatever string format you like using the strftime method. In your case you would apply it like this:
df.date = df.date[df.date.notnull()].map(lambda x: x.strftime('%m/%Y'))
df.date
Out[111]:
0 05/2013
1 05/2013
I have a Pandas data frame, one of the column contains date strings in the format YYYY-MM-DD
For e.g. '2013-10-28'
At the moment the dtype of the column is object.
How do I convert the column values to Pandas date format?
Essentially equivalent to #waitingkuo, but I would use pd.to_datetime here (it seems a little cleaner, and offers some additional functionality e.g. dayfirst):
In [11]: df
Out[11]:
a time
0 1 2013-01-01
1 2 2013-01-02
2 3 2013-01-03
In [12]: pd.to_datetime(df['time'])
Out[12]:
0 2013-01-01 00:00:00
1 2013-01-02 00:00:00
2 2013-01-03 00:00:00
Name: time, dtype: datetime64[ns]
In [13]: df['time'] = pd.to_datetime(df['time'])
In [14]: df
Out[14]:
a time
0 1 2013-01-01 00:00:00
1 2 2013-01-02 00:00:00
2 3 2013-01-03 00:00:00
Handling ValueErrors
If you run into a situation where doing
df['time'] = pd.to_datetime(df['time'])
Throws a
ValueError: Unknown string format
That means you have invalid (non-coercible) values. If you are okay with having them converted to pd.NaT, you can add an errors='coerce' argument to to_datetime:
df['time'] = pd.to_datetime(df['time'], errors='coerce')
Use astype
In [31]: df
Out[31]:
a time
0 1 2013-01-01
1 2 2013-01-02
2 3 2013-01-03
In [32]: df['time'] = df['time'].astype('datetime64[ns]')
In [33]: df
Out[33]:
a time
0 1 2013-01-01 00:00:00
1 2 2013-01-02 00:00:00
2 3 2013-01-03 00:00:00
I imagine a lot of data comes into Pandas from CSV files, in which case you can simply convert the date during the initial CSV read:
dfcsv = pd.read_csv('xyz.csv', parse_dates=[0]) where the 0 refers to the column the date is in.
You could also add , index_col=0 in there if you want the date to be your index.
See https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.read_csv.html
Now you can do df['column'].dt.date
Note that for datetime objects, if you don't see the hour when they're all 00:00:00, that's not pandas. That's iPython notebook trying to make things look pretty.
If you want to get the DATE and not DATETIME format:
df["id_date"] = pd.to_datetime(df["id_date"]).dt.date
Another way to do this and this works well if you have multiple columns to convert to datetime.
cols = ['date1','date2']
df[cols] = df[cols].apply(pd.to_datetime)
It may be the case that dates need to be converted to a different frequency. In this case, I would suggest setting an index by dates.
#set an index by dates
df.set_index(['time'], drop=True, inplace=True)
After this, you can more easily convert to the type of date format you will need most. Below, I sequentially convert to a number of date formats, ultimately ending up with a set of daily dates at the beginning of the month.
#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)
#Convert to monthly dates
df.index = df.index.to_period(freq='M')
#Convert to strings
df.index = df.index.strftime('%Y-%m')
#Convert to daily dates
df.index = pd.DatetimeIndex(data=df.index)
For brevity, I don't show that I run the following code after each line above:
print(df.index)
print(df.index.dtype)
print(type(df.index))
This gives me the following output:
Index(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='object', name='time')
object
<class 'pandas.core.indexes.base.Index'>
DatetimeIndex(['2013-01-01', '2013-01-02', '2013-01-03'], dtype='datetime64[ns]', name='time', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>
PeriodIndex(['2013-01', '2013-01', '2013-01'], dtype='period[M]', name='time', freq='M')
period[M]
<class 'pandas.core.indexes.period.PeriodIndex'>
Index(['2013-01', '2013-01', '2013-01'], dtype='object')
object
<class 'pandas.core.indexes.base.Index'>
DatetimeIndex(['2013-01-01', '2013-01-01', '2013-01-01'], dtype='datetime64[ns]', freq=None)
datetime64[ns]
<class 'pandas.core.indexes.datetimes.DatetimeIndex'>
For the sake of completeness, another option, which might not be the most straightforward one, a bit similar to the one proposed by #SSS, but using rather the datetime library is:
import datetime
df["Date"] = df["Date"].apply(lambda x: datetime.datetime.strptime(x, '%Y-%d-%m').date())
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 startDay 110526 non-null object
1 endDay 110526 non-null object
import pandas as pd
df['startDay'] = pd.to_datetime(df.startDay)
df['endDay'] = pd.to_datetime(df.endDay)
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 startDay 110526 non-null datetime64[ns]
1 endDay 110526 non-null datetime64[ns]
Try to convert one of the rows into timestamp using the pd.to_datetime function and then use .map to map the formular to the entire column