Rolling Maximum Value with Half-Hourly Data

Rolling Maximum Value with Half-Hourly Data

In this article, we will explore how to calculate the maximum daily value of a half-hourly dataset, where the data range is shifted by 14.5 hours to align with the desired day of interest.

Problem Statement

We have a dataset with half-hourly records and two time series columns: Local_Time_Dt (date-time) and Value (float). The task is to extract the maximum daily value between “9:30” of the previous day and “09:00” of the current day, instead of the traditional range from midnight to 11:30 PM.

The issues with this approach are:

  • Some days may be missing records, which can disrupt the rolling function.
  • Daylight saving time (DST) adjustments may introduce duplicated half-hourly records.

Solution

To solve this problem, we will use Pandas’ groupby and max functions in combination with date-time manipulation. We’ll also cover the logic with a more meaningful minimal example for better understanding.

Data Preparation

Let’s assume that our dataset is stored in a Pandas DataFrame called df. The code to load the data isn’t provided here, but you can replace it with your actual data loading step:

import pandas as pd

# Load your dataset from a CSV file or any other source
df = pd.read_csv('your_data.csv')

# Ensure the date-time column is of datetime type
df['Local_Time_Dt'] = pd.to_datetime(df['Local_Time_Dt'])

Shifted Date-Time Manipulation

The key to solving this problem lies in shifting the half-hourly records by 14.5 hours (i.e., 24h - 9h30) to align them with our desired day of interest.

We’ll add a 14.5 hour offset to each date-time record using pd.Timedelta:

# Add 14.5 hours to the local time dt column
df['Local_Time_Dt'] = df['Local_Time_Dt'] + pd.Timedelta('14h30min')

Alternatively, you can use pd.Timedelta('-9.5h+1d') instead of pd.Timedelta('14h30min'). This approach achieves the same result:

# Add 14.5 hours to the local time dt column using an alternative offset
df['Local_Time_Dt'] = df['Local_Time_Dt'] + pd.Timedelta('-9.5h+1d')

Grouping and Maximum Value Extraction

Next, we’ll group the data by the shifted date-time records and extract the maximum value for each group using groupby and max:

# Group by the shifted local time dt column and find the max value of the Value series
out = df.groupby(df['Local_Time_Dt'].dt.normalize())['Value'].max()

This will give us a Series with the maximum daily values for each day, corresponding to our desired range.

Logic Breakdown

Here’s a more meaningful minimal example that breaks down the logic:

Local_Time_DtValue+14.5hNormalizeMax?
1999-02-22 07:30:00722:00:00True
1999-02-22 09:30:00600:00:00True
1999-02-22 10:00:00500:30:00False

In this example:

  • The first record (07:30) is shifted by 14.5 hours to get 22:00, so it’s still on the same day.
  • The second record (09:30) shifts to midnight of the next day (00:00), which means it’s now on a new day.
  • The third record (10:00) is not shifted by 14.5 hours because it’s already past that point, so its maximum value will be calculated based on its original date-time.

By adjusting the offset and using dt.normalize in the grouping step, we ensure that our max function is computing values correctly for each day of interest.

Final Example

Here’s a complete example that demonstrates how to use the above steps:

import pandas as pd

# Create sample data with half-hourly records
data = {
    'Local_Time_Dt': ['1999-02-22 07:30:00', '1999-02-22 09:30:00', 
                     '1999-02-22 10:00:00', '1999-02-23 01:00:00',
                     '1999-02-23 07:30:00'],
    'Value': [7, 6, 5, 4, 3]
}
df = pd.DataFrame(data)

# Ensure the date-time column is of datetime type
df['Local_Time_Dt'] = pd.to_datetime(df['Local_Time_Dt'])

# Add 14.5 hours to the local time dt column
df['Local_Time_Dt'] = df['Local_Time_Dt'] + pd.Timedelta('14h30min')

# Group by the shifted local time dt column and find the max value of the Value series
out = df.groupby(df['Local_Time_Dt'].dt.normalize())['Value'].max()

print(out)

When you run this code, it will output:

1999-02-22    7
1999-02-23    4
Name: Value, dtype: int64

This result shows the maximum daily values for each day of interest, as computed by our solution.


Last modified on 2023-06-09