How to subset a pandas dataframe column by a single date in a datetime column?

user_7841401 0
Viewed 72

So I thought this should have been a simpler task but I could not find it. I have a pandas dataframe as follows

Company Id      DateTime               col1  col2       col3    col4   col5     col6
0   25502921    2018-08-16 10:23:36     0   175.000     0.0     0.0     0.0     0
1   25502921    2018-08-16 10:33:55     0   155.557     0.0     0.0     0.0     0
2   25502921    2018-08-16 10:43:55     0   153.615     0.0     0.0     0.0     0

type(df['DateTime'][0]) outputs pandas._libs.tslibs.timestamps.Timestamp

How can i subset the dataframe based on a date?

df_tmp = df[df['DateTime'].dt.date=='16-08-2018'] does not seem to work nor does df_tmp = df[df['DateTime']=='16-08-2018']

What are the other intuitive methods i can try?

pandas subset python-3.x timestamp datetime
2 Answers

use .dt.normalize() which converts the time to 00:00:00 but returns a datetime object.

df['DateTime'].dt.normalize().apply(type)
pandas._libs.tslibs.timestamps.Timestamp

print(df['DateTime'].dt.normalize())

0   2018-08-16
1   2018-08-16
2   2018-08-16
Name: DateTime, dtype: datetime64[ns]

df[df['DateTime'].dt.normalize() == '2018-08-16']

   Company_Id            DateTime  col1     col2  col3  col4  col5  col6
0    25502921 2018-08-16 10:23:36     0  175.000   0.0   0.0   0.0     0
1    25502921 2018-08-16 10:33:55     0  155.557   0.0   0.0   0.0     0
2    25502921 2018-08-16 10:43:55     0  153.615   0.0   0.0   0.0     0
user_93751020

I believe u'll get better performance and easier processing when u set the DateTime as the index : 

df = pd.read_clipboard(sep = '\s{2,}',parse_dates=['DateTime'])
df.set_index("DateTime").loc["2018-08-16"]
user_71757130
Related