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Date: 2025-05-02 16:06:01
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# Setting Day-of-Year to the Oldest Leap Year in Pandas

For your use case of converting day-of-year values (1-366) to dates using an unlikely leap year, here's the most robust approach:

## The Oldest Leap Year in Pandas

Pandas can handle dates starting from **1678 AD** (the earliest leap year in its supported range). However, for practical purposes and to ensure full datetime functionality, I recommend using **1972** - the first leap year in the Unix epoch era (1970-01-01 onward).

```python

import pandas as pd

# Example with day-of-year values (1-366)

day_of_year = pd.Series([60, 366, 100]) # Example values

# Convert to dates using 1972 (first Unix epoch leap year)

dates = pd.to_datetime('1972') + pd.to_timedelta(day_of_year - 1, unit='D')

print(dates)

# Output:

# 0 1972-02-29

# 1 1972-12-31

# 2 1972-04-09

# dtype: datetime64[ns]

```

## Why 1972?

1. **Unix Epoch Compatibility**: 1972 is the first leap year after 1970 (Unix epoch start)

2. **Modern Calendar**: Uses the current Gregorian calendar rules

3. **Pandas Optimization**: Works efficiently with pandas' datetime operations

4. **Unlikely in Time Series**: Very old year that won't conflict with modern data

## Alternative: Using the Minimum Pandas Leap Year

If you truly need the oldest possible leap year that pandas supports:

```python

min_leap_year = 1678 # Earliest pandas-supported leap year

dates = pd.to_datetime(str(min_leap_year)) + pd.to_timedelta(day_of_year - 1, unit='D')

```

## For Your Existing Datetime Series

If you're modifying existing datetime objects (as in your example):

```python

dates = pd.Series(pd.to_datetime(['2023-05-01', '2021-12-15', '2019-07-20']))

new_dates = dates.dt.dayofyear # Extract day-of-year

new_dates = pd.to_datetime('1972') + pd.to_timedelta(new_dates - 1, unit='D')

```

This approach is more efficient than using `apply` with `replace`.

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Posted by: Aditya attrish