How to remove seconds from datetime?

Question

I have the following date and I tried the following code,

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"]
df['start_date_time'] = pd.to_datetime([df['start_date_time']).replace(second = 0)

I get the following error:

TypeError: replace() got an unexpected keyword argument 'second'

Answer 1

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'")

Answer 2

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

Answer 3

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'>)

Answer 4

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"

Answer 5

Solutions if need datetimes in output:

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])
print (df)
       start_date_time
0  2016-05-19 08:25:23
1  2016-05-19 16:00:45

Use Series.dt.floor by minutes T or Min:

df['start_date_time'] = df['start_date_time'].dt.floor('T')

df['start_date_time'] = df['start_date_time'].dt.floor('Min')

---

You can use convert to numpy values first and then truncate seconds by cast to <M8[m], but this solution remove possible timezones:

df['start_date_time'] = df['start_date_time'].values.astype('<M8[m]')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

Another solution is create timedelta Series from second and substract:

print (pd.to_timedelta(df['start_date_time'].dt.second, unit='s'))
0   00:00:23
1   00:00:45
Name: start_date_time, dtype: timedelta64[ns]

df['start_date_time'] = df['start_date_time'] - 
                        pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
print (df)
      start_date_time
0 2016-05-19 08:25:00
1 2016-05-19 16:00:00

Timings:

df = pd.DataFrame({'start_date_time': ["2016-05-19 08:25:23","2016-05-19 16:00:45"]})
df['start_date_time'] = pd.to_datetime(df['start_date_time'])

#20000 rows
df = pd.concat([df]*10000).reset_index(drop=True)


In [28]: %timeit df['start_date_time'] = df['start_date_time'] - pd.to_timedelta(df['start_date_time'].dt.second, unit='s')
4.05 ms ± 130 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [29]: %timeit df['start_date_time1'] = df['start_date_time'].values.astype('<M8[m]')
1.73 ms ± 117 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [30]: %timeit df['start_date_time'] = df['start_date_time'].dt.floor('T')
1.07 ms ± 116 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [31]: %timeit df['start_date_time2'] = df['start_date_time'].apply(lambda t: t.replace(second=0))
183 ms ± 19.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

---

Solutions if need strings repr of datetimes in output

Use Series.dt.strftime:

print(df['start_date_time'].dt.strftime('%Y-%m-%d %H:%M'))
0    2016-05-19 08:25
1    2016-05-19 16:00
Name: start_date_time, dtype: object

And if necessary set :00 to seconds:

print(df['start_date_time'].dt.strftime('%Y-%m-%d %H:%M:00'))
0    2016-05-19 08:25:00
1    2016-05-19 16:00:00
Name: start_date_time, dtype: object

Answer 6

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