Drop Python Pandas dataframe rows based on date in index

Question

I have a Python Dataframe that looks like this:

                        Facility   PUE   PUEraw   Servers
2016-11-14 00:00:00     6.0        NaN   1.2      5.0 
2016-11-14 00:30:00     6.0        NaN   1.2      5.0 
2016-11-14 01:00:00     6.0        NaN   1.2      5.0

etc.

As you can see, the index is date/time. The dataframe is updated with a new value every half hour.

I'm trying to write a script that removes all rows except those that correspond to TODAY's date, for which I am utilising date = dt.datetime.today(). However, I am struggling, partly perhaps because the index also contains the time.

Does anyone have any suggestions? Alternatively, a script that removes all but the last 48 rows would also work for me (the last 48 x half hourly values = the latest day's data).

akuiper · Accepted Answer · 2016-11-16 14:30:01Z

2

Here are two options you can use to extract data on a specific day:

df['2016-11-16']
#                     Facility  PUE  PUEraw  Servers
# 2016-11-16 01:00:00      6.0  NaN     1.2      5.0

import datetime
df[df.index.date == datetime.datetime.today().date()]
#                     Facility  PUE  PUEraw  Servers
# 2016-11-16 01:00:00      6.0  NaN     1.2      5.0

answered Nov 16, 2016 at 14:30

akuiper

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pottolom Over a year ago

Thank you. I haven't tried this as I've gone with the other method, but it looks good to me.

schlump · Accepted Answer · 2016-11-16 14:24:11Z

1

You can always access the last rows in a DataFrame with df.tail()

df = df.tail(48)

For further information:

Pandas Documentation

answered Nov 16, 2016 at 14:24

schlump

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1 Comment

pottolom Over a year ago

Thanks. This works perfectly and is nice and simple.

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