How to remove seconds from datetime?

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

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

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

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