this is with the time module not datetime module but might be what your looking for…

import time
X = time.strftime("%H:%M")
print(X)

or you with seconds

import time
X = time.strftime("%H:%M:%S")
print(X)

Give this a shot with:

df.index = df.index.map(lambda t: t.strftime('%Y-%m-%d %H:%M'))

As written in one of the comments, the above apply to the case where the dates are not strings. If they, however, are strings, you can simply slice the last three characters from each list in the list:

import pandas as pd

df = pd.DataFrame({'date': ["2016-05-19 08:25:00"]})

print(df['date'].map(lambda t: t[:-3]))

The above will output:

0    2016-05-19 08:25
Name: date, dtype: object

Set seconds to 0

pd.to_datetime will return datetime objects, which have second as attribute : there’s not much you can do about it. You can set second to 0, but the attribute will still be here and the standard representation will still include a trailing ':00'.

You need to apply replace on each element of df:

import pandas as pd

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])
df['start_date_time'] = df['start_date_time'].apply(lambda t: t.replace(second=0))

print(df)
#       start_date_time
# 0 2016-05-19 08:25:00
# 1 2016-05-19 16:00:00
# 2 2016-05-20 07:45:00
# 3 2016-05-24 12:50:00
# 4 2016-05-25 23:00:00
# 5 2016-05-26 19:45:00

:23 and :45 from the first times have been replaced by :00, but they are still printed.

Remove ':00' from the strings

If you just want a string representation of those times and only parse the strings to datetime objects in order to remove ':00' at the end of the string, you could just remove the last 3 characters :

>>> "2016-05-19 08:25:00"[:-3]
'2016-05-19 08:25'

You could apply this to every element in your list, before initializing df['start_date_time']:

>>> start_date_time = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]
>>> map(lambda s: s[:-3], start_date_time)
['2016-05-19 08:25', '2016-05-19 16:00', '2016-05-20 07:45', '2016-05-24 12:50', '2016-05-25 23:00', '2016-05-26 19:45']

Display datetimes without seconds

If you want to work with datetime objects but don’t want to show seconds :

print(df['start_date_time'].apply(lambda t: t.strftime('%Y-%m-%d %H:%M')))
# 0    2016-05-19 08:25
# 1    2016-05-19 16:00
# 2    2016-05-20 07:45
# 3    2016-05-24 12:50
# 4    2016-05-25 23:00
# 5    2016-05-26 19:45
# Name: start_date_time, dtype: object

As you mentioned removed so I assumed you don’t want the seconds or microsecond in the result.If this is the case then following might help:

datetime_variable.strftime("'%Y-%m-%d %H:%M'")

If you have datetime in string then you can convert it in datetime obj:

from dateutil import parser

datetime_variable = parser.parse(str_datetime_var)

datetime_variable.strftime("'%Y-%m-%d %H:%M'")

You can subtract the seconds using a timedelta:

import datetime    
d = datetime.datetime.now() #datetime including seconds
without_seconds = d - datetime.timedelta(seconds=d.second)

Convert String to datetime object first, then you can use the replace method.

from _datetime import *


df = dict()
df['start_date_time'] = ["2016-05-19 08:25:00",
                         "2016-05-19 16:00:00",
                         "2016-05-20 07:45:00",
                         "2016-05-24 12:50:00",
                         "2016-05-25 23:00:00",
                         "2016-05-26 19:45:00"]

for dt in df['start_date_time']:
    cur_dt = datetime.strptime(dt, '%Y-%m-%d %H:%M:%S')
    cur_dt = cur_dt.replace(second=0)
    print(cur_dt)

    cur_dt_without_second = cur_dt.strftime('%Y-%m-%d %H:%M')
    print(cur_dt_without_second)

-------------------
2016-05-19 08:25:00
2016-05-19 08:25
2016-05-19 16:00:00
2016-05-19 16:00
2016-05-20 07:45:00
2016-05-20 07:45
2016-05-24 12:50:00
2016-05-24 12:50
2016-05-25 23:00:00
2016-05-25 23:00
2016-05-26 19:45:00
2016-05-26 19:45

Convert the string to a datetime object and then manipulate that

>>> x = ["2016-05-19 08:25:00","2016-05-19 16:00:00","2016-05-20 07:45:00","2016-05-24 12:50:00","2016-05-25 23:00:00","2016-05-26 19:45:00"]
>>> for i in x:
...  y = datetime.datetime.strptime(i, '%Y-%m-%d %H:%M:%S')
...  z = datetime.datetime.strftime(y, '%Y-%m-%d %H:%M')
...  print (y, type(y))
...  print (z, type(z))
... 
(datetime.datetime(2016, 5, 19, 8, 25), <type 'datetime.datetime'>)
('2016-05-19 08:25', <type 'str'>)
(datetime.datetime(2016, 5, 19, 16, 0), <type 'datetime.datetime'>)
('2016-05-19 16:00', <type 'str'>)
(datetime.datetime(2016, 5, 20, 7, 45), <type 'datetime.datetime'>)
('2016-05-20 07:45', <type 'str'>)
(datetime.datetime(2016, 5, 24, 12, 50), <type 'datetime.datetime'>)
('2016-05-24 12:50', <type 'str'>)
(datetime.datetime(2016, 5, 25, 23, 0), <type 'datetime.datetime'>)
('2016-05-25 23:00', <type 'str'>)
(datetime.datetime(2016, 5, 26, 19, 45), <type 'datetime.datetime'>)
('2016-05-26 19:45', <type 'str'>)

HTML Code:

< input type="time" class="form-control" name="meeting_time" required /> 

Python Django Code:

meeting_time = request.POST['meeting_time'] #Like your_time = "12:35:00"
get_time = meeting_time.strftime("%H:%M")

Result is:

get_time = "12:35"

Here is a very simple way to remove seconds from datetime:

from datetime import datetime 
print(str(datetime.today())[:16])

Output:

2021-02-14 21:30

It effectively transforms the timestamp into text and leaves only the first 16 symbols.
Just don’t lose yourself in all those brackets 😉