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Convert date type object to datetime

I have a dataframe with Date_Birth in the following format: July 1, 1991 (as type object).
How can I change the entire column to datetime?
Thanks

Use the pandas.to_datetime function. You can write out a format specifier in the strptime format or have python guess the format. In your case, guessing works.
>>> import pandas as pd
>>> df=pd.DataFrame({"Date":["July 1, 1991"]})
>>> pd.to_datetime(df["Date"], format="%B %d, %Y")
0 1991-07-01
Name: Date, dtype: datetime64[ns]
>>> pd.to_datetime(df["Date"], infer_datetime_format=True)
0 1991-07-01
Name: Date, dtype: datetime64[ns]

Identifying a dateformat and change it into another

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

How to compare pandas date column with hardcoded date

This is my code:
print (df.loc[df.DATE == '2016-02-05'])
I am trying to compare this date with date of pandas. It returns empty dataframe. What should I do ?
Edit: Original dataframe:

Just convert your string to datetime (I suppose that you dataframe also contains datetimes, rather than strings) and do the comparison you wanted to do:
from datetime import datetime
if __name__ == "__main__":
t = datetime.strptime('2016-02-05', '%Y-%m-%d')
print(t)
Hope the answer will help, feel free to ask questions.
If your DATE df column is not datetimes, but just strings, convert them to datetimes the same way.

You need to convert the string to a datetime object as well.
print(df)
datetime_str = '2015/02/04'
print("({}){}".format(type(datetime_str), datetime_str))
datetime_object = datetime.strptime(datetime_str, '%Y/%m/%d')
print("({}){}".format(type(datetime_object), datetime_object))
value = df.loc[df.DATE == datetime_object]
print("value =", value)
OUTPUT:
year month day DATE
0 2015 2 4 2015-02-04
1 2016 3 5 2016-03-05
(<class 'str'>)2015/02/04
(<class 'datetime.datetime'>)2015-02-04 00:00:00
value = year month day DATE
0 2015 2 4 2015-02-04

pandas to_datetime converts non-zero padded month and day into datetime

I am using pd.to_datetime to convert strings into datetime;
df = pd.DataFrame(data={'id':['DD-83']})
pd.to_datetime(df['id'].str.replace(r'\D+', ''), errors='coerce', format='%d%m')
%d%m defines zero-padded day and month, but the code still converts the above string into
0 1900-03-08
Name: id, dtype: datetime64[ns]
I am wondering how to avoid it being converted into datetime (e.g. convert to NaT in this case), if the month and day in a string are not 0-padded. So
DD0306
DD0706
DD-83
will convert to
1900-06-03
1900-06-07
NaT

You need to look for - and only pass strings without -.
Setup:
df = pd.DataFrame(data={'id':['DD-83', 'DD0706', 'DD0306']})
Code:
df['date'] = pd.to_datetime(df['id'].loc[~df['id'].str.contains('-')].str.replace(r'\D+', ''), errors='coerce', format='%d%m')
Output:
id date
0 DD-83 NaT
1 DD0706 1900-06-07
2 DD0306 1900-06-03

Pandas Dataframe convert string to data without time

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()