Brtdku

I have a Python data frame containing a column with Date Time like this
2019-01-02 09:00:00 (which means January 2, 2019 9 AM)

There may be a bunch of rows which have the same date in the Date Time column.

In other words, I can have 2019-01-02 09:00:00 or 2019-01-02 09:15:00 or 2019-01-02 09:30:00 and so on.

Now I need to find the row index of the first occurrence of the date 2019-01-02 in the Python data frame.

I obviously do this using a loop, but am wondering if there is a better way.

With the df['Date Time'].str.contains() method, I can get that all the rows that match a given date, but I need the index.

The generic question is that how do we find the index of a first occurrence of a match in a cell in Python data frame that matches a given string pattern.

The more specific question is that how do we find the index of a first occurrence of a match in a cell in Python data frame that matches a given date in a cell that contains date Time assuming that the Python data frame is sorted in chronologically ascending order of date Time , i.e.
2019-01-02 09:00:00 occurs at an index earlier than 2019-01-02 09:15:00 followed by 2019-01-03 09:00:00 and so on.

Thank you for any inputs

You can use next with iter for first index value matched condition for prevent failed if no matched values:

df = pd.DataFrame({'dates':pd.date_range(start='2018-01-01 20:00:00',

                                         end='2018-01-02 02:00:00', freq='H')})

print (df)

                dates

0 2018-01-01 20:00:00

1 2018-01-01 21:00:00

2 2018-01-01 22:00:00

3 2018-01-01 23:00:00

4 2018-01-02 00:00:00

5 2018-01-02 01:00:00

6 2018-01-02 02:00:00



date = '2018-01-02'

mask = df['dates'] >= date

idx = next(iter(mask.index[mask]), 'not exist')

print (idx)

4





date = '2018-01-08'

mask = df['dates'] >= date

idx = next(iter(mask.index[mask]), 'not exist')

print (idx)

not exist

If performance is important, see Efficiently return the index of the first value satisfying condition in array.

Yep you can use .loc and a condition to slice the df, and then return the index using .iloc.

import pandas as pd

df = pd.DataFrame({'time':pd.date_range(start='2018-01-01 00:00:00',end='2018-12-31 00:00:00', freq='H')}, index=None).reset_index(drop=True)



# then use conditions and .iloc to get the first instance

df.loc[df['time']>'2018-10-30 01:00:00'].iloc[[0,]].index[0]



# if you specify a coarser condition, for instance without time,

# it will also return the first instance

df.loc[df['time']>'2018-10-30'].iloc[[0,]].index[0]

I do not know, if it is optimal, but it works

(df['Date Time'].dt.strftime('%Y-%m-%d') == '2019-01-02').idxmax()