I am working with the following piece of data which has a different format of dates and which creates confusion later in the process. The data is like:
S. No DateTime Area
1 03/05/2019 6:33 A
2 06/03/2019 07:23:45 AM B
The first row is the format %m/%d/%Y h: mm and the second row is the format of %d/%m/%Y hh:mm: ss AM/PM. The first date value can be confusing though, is it 5th march or 3rd May. So to make everything of the same format, I want that my code detects the date format and changes in the desired format.
I have tried doing this:
df['Detection Date'] = pd.to_datetime(df['Detection Date & Time'], errors = 'coerce').dt.datetime
col = df['Detection Date'].apply(str)
for i in df.index:
if datetime.datetime.strptime(col, '%m/%d/%Y h:mm'):
ColDate = datetime.datetime.strftime(col, '%d/%m/%Y hh:mm:ss AM/PM')
But i am getting an error saying:
TypeError: strptime() argument 1 must be str, not Series
How it should be conducted.
Thanks
If it is ok to install a dependency then you can use dateparser link
import pandas as pd
import dateparser
df = pd.DataFrame({'Detection Date & Time': ['03/05/2019 6:33', '06/03/2019 07:23:45 AM']})
df['Date & time'] = df['Detection Date & Time'].apply(dateparser.parse)
You can specify both possible formats in to_datetime, so if format not match is returned missing values, so is possible use Series.fillna:
date1 = pd.to_datetime(df['DateTime'], errors = 'coerce', format='%m/%d/%Y %H:%M')
date2 = pd.to_datetime(df['DateTime'], errors = 'coerce', format='%d/%m/%Y %H:%M:%S %p')
df['DateTime'] = date1.fillna(date2)
print (df)
S. No DateTime Area
0 1 2019-03-05 06:33:00 A
1 2 2019-03-06 07:23:45 B
Last if want specify new format add Series.dt.strftime - advanatage of solution are verify both formats:
df['DateTime'] = date1.fillna(date2).dt.strftime('%d/%m/%Y %H:%M:%S %p')
print (df)
S. No DateTime Area
0 1 05/03/2019 06:33:00 AM A
1 2 06/03/2019 07:23:45 AM B
Details:
print (date1)
0 2019-03-05 06:33:00
1 NaT
Name: DateTime, dtype: datetime64[ns]
print (date2)
0 NaT
1 2019-03-06 07:23:45
Name: DateTime, dtype: datetime64[ns]
Another possible solution without verify another formats - only repalaced format %m/%d/%Y %H:%M to %d/%m/%Y %H:%M:%S %p:
date1 = pd.to_datetime(df['DateTime'], errors = 'coerce', format='%m/%d/%Y %H:%M').dt.strftime('%d/%m/%Y %H:%M:%S %p')
df['DateTime'] = date1.replace('NaT', df['DateTime'])
print (df)
S. No DateTime Area
0 1 05/03/2019 06:33:00 AM A
1 2 06/03/2019 07:23:45 AM B
Related
I'm trying to covert the next date column (str) to datetime64 and say that format doesn't match, can anyone help me pleas :)
Column:
df["Date"]
0 15/7/21
...
2541 13/9/21
dtype: object
What I try:
pd.to_datetime(df["Date"], format = "%d/%m/%Y")
ValueError: time data '15/7/21' does not match format '%d/%m/%Y' (match)
I also try:
pd.to_datetime(df["Date"].astype("datetime64"), format='%d/%m/%Y')
And it convert it as datetime but there is some date the day is in the month.
Anyone know what to do ?
%Y expects a 4-digit year. Use %y for a 2-digit year (See the docs):
>>> import pandas as pd
>>> df = pd.DataFrame({'Date':['15/7/21','13/9/21']})
>>> df['Date']
0 15/7/21
1 13/9/21
Name: Date, dtype: object
>>> pd.to_datetime(df['Date'].astype('datetime64'),format='%d/%m/%y')
0 2021-07-15
1 2021-09-13
Name: Date, dtype: datetime64[ns]
Note that pandas is pretty good at guessing the format:
>>> pd.to_datetime(df['Date'])
0 2021-07-15
1 2021-09-13
I have month column with values formatted as: 2019M01
To find the seasonality I need this formatted into Pandas DateTime format.
How to format 2019M01 into datetime so that I can use it for my seasonality plotting?
Thanks.
Use to_datetime with format parameter:
print (df)
date
0 2019M01
1 2019M03
2 2019M04
df['date'] = pd.to_datetime(df['date'], format='%YM%m')
print (df)
date
0 2019-01-01
1 2019-03-01
2 2019-04-01
I have a dataframe where one of the columns called 'date', containing objects, looks like:
df =
|Date
|Mar-24
|Aug-22
|Sep-25
|...
I want to convert that column into date so for example Mar-24 would look like 2024-03-01. So far i have tried
df['Date'] = pd.to_datetime(df['Date'], format= '%b-%y')
which i think should work but from the few thousand rows i've found that there are rows which contain the full year such as 'Apr 2023' which won't be picked up by %y, is there a way i could find those rows in the column and change them into the short year before applying the above code or just giving the code both %y and %Y arguments?
Use the parameter errors='coerce' in combination with combine_first:
Minimal example:
import pandas as pd
series = pd.Series(['Mar-24', 'Aug-22', 'Sep-2025'], [0, 1, 2])
date1 = pd.to_datetime(series, format='%b-%y', errors='coerce')
date2 = pd.to_datetime(series, format='%b-%Y', errors='coerce')
date1.combine_first(date2)
Output:
0 2024-03-01
1 2022-08-01
2 2025-09-01
dtype: datetime64[ns]
Or for your specific case in one line:
df['Date'] = pd.to_datetime(df['Date'], format='%b-%y', errors='coerce').combine_first(pd.to_datetime(df['Date'], format='%b-%Y', errors='coerce'))
I have a dataframe and the Date column has two different types of date formats going on.
eg. 1983-11-10 00:00:00 and 10/11/1983
I want them all to be the same type, how can I iterate through the Date column of my dataframe and convert the dates to one format?
I believe you need parameter dayfirst=True in to_datetime:
df = pd.DataFrame({'Date': {0: '1983-11-10 00:00:00', 1: '10/11/1983'}})
print (df)
Date
0 1983-11-10 00:00:00
1 10/11/1983
df['Date'] = pd.to_datetime(df.Date, dayfirst=True)
print (df)
Date
0 1983-11-10
1 1983-11-10
because:
df['Date'] = pd.to_datetime(df.Date)
print (df)
Date
0 1983-11-10
1 1983-10-11
Or you can specify both formats and then use combine_first:
d1 = pd.to_datetime(df.Date, format='%Y-%m-%d %H:%M:%S', errors='coerce')
d2 = pd.to_datetime(df.Date, format='%d/%m/%Y', errors='coerce')
df['Date'] = d1.combine_first(d2)
print (df)
Date
0 1983-11-10
1 1983-11-10
General solution for multiple formats:
from functools import reduce
def convert_formats_to_datetimes(col, formats):
out = [pd.to_datetime(col, format=x, errors='coerce') for x in formats]
return reduce(lambda l,r: pd.Series.combine_first(l,r), out)
formats = ['%Y-%m-%d %H:%M:%S', '%d/%m/%Y']
df['Date'] = df['Date'].pipe(convert_formats_to_datetimes, formats)
print (df)
Date
0 1983-11-10
1 1983-11-10
I want them all to be the same type, how can I iterate through the
Date column of my dataframe and convert the dates to one format?
Your input data is ambiguous: is 10 / 11 10th November or 11th October? You need to specify logic to determine which is appropriate. A function is useful if you with to try multiple date formats sequentially:
def date_apply_formats(s, form_lst):
s = pd.to_datetime(s, format=form_lst[0], errors='coerce')
for form in form_lst[1:]:
s = s.fillna(pd.to_datetime(s, format=form, errors='coerce'))
return s
df['Date'] = date_apply_formats(df['Date'], ['%Y-%m-%d %H:%M:%S', '%d/%m/%Y'])
Priority is given to the first item in form_lst. The solution is extendible to an arbitrary number of provided formats.
Input date is
NSECODE Date Close
1 NSE500 20000103 1291.5500
2 NSE500 20000104 1335.4500
3 NSE500 20000105 1303.8000
history_nseindex_df["Date"] = pd.to_datetime(history_nseindex_df["Date"])
history_nseindex_df["Date"] = history_nseindex_df["Date"].dt.strftime("%Y-%m-%d")
ouput is now
NSECode Date Close
1 NSE500 2000-01-03 1291.5500
2 NSE500 2000-01-04 1335.4500
3 NSE500 2000-01-05 1303.8000
I have a Pandas Dataframe df:
a date
1 2014-06-29 00:00:00
df.types return:
a object
date object
I want convert column data to data without time but:
df['date']=df['date'].astype('datetime64[s]')
return:
a date
1 2014-06-28 22:00:00
df.types return:
a object
date datetime64[ns]
But value is wrong.
I'd have:
a date
1 2014-06-29
or:
a date
1 2014-06-29 00:00:00
I would start by putting your dates in pd.datetime:
df['date'] = pd.to_datetime(df.date)
Now, you can see that the time component is still there:
df.date.values
array(['2014-06-28T19:00:00.000000000-0500'], dtype='datetime64[ns]')
If you are ok having a date object again, you want:
df['date'] = [x.strftime("%y-%m-%d") for x in df.date]
Here would be ending with a datetime:
df['date'] = [x.date() for x in df.date]
df.date
datetime.date(2014, 6, 29)
Here you go. Just use this pattern:
df.to_datetime().date()