can i convert float to datetime using python?

Question

I have a date column with data type as float, can I get help on how to convert to datetime64

this is what i tried newdf["Year_Sold"] = pd.to_datetime(df["Year_Sold"]) n the Year_sold is a column with all values having float as data type but I want the data type as datetime64

Answer 1

To convert a column with float data type to datetime64 in pandas, you generally need to ensure that the float values represent valid dates. Since you're dealing with a column named "Year_Sold", I'm assuming these floats represent years, possibly with some additional decimal representation for more precise timing within the year. The problem with your code is that pd.to_datetime expects the input to be in a certain date format, like "YYYY-MM-DD", "DD/MM/YYYY", etc. for dates or similar for timestamps. In other words, if you try to add float values representing years to it, it won't know if those numbers represent years, days, months, or some other combination of years+dates+numbers like "2024-01-01" written as "20240101.0".

Therefore, given the two possible assumptions I explained above about the meaning of your "Year_Sold" column, here's how you could achieve your end result, for either case:

Option 1: If "Year_Sold" represent years as floats

If your "Year_Sold" column contains just the year as a float, you can first convert these floats to integers and then to string. Since a date contains not only information about the year, but month and day as well, you'll need to add a standard month and day to make it a complete date string. You can then use pd.to_datetime to convert these strings into datetime64 type values.

Here's a summary of the steps needed to convert the column:

1. Convert the float values to integers (to remove the decimal part, assuming it's just representing the year).
2. Convert these integers to strings and append a standard date to them (e.g., "-01-01" for January 1st) to form a complete date string.
3. Use pd.to_datetime to convert these strings into datetime64 format.

Here's the implementation:

import pandas as pd

# Sample DataFrame creation with float years
df = pd.DataFrame({'Year_Sold': [2020.0, 2021.5, 2022.0]})
# df looks like this:
#
#    Year_Sold
# 0     2020.0
# 1     2021.5
# 2     2022.0

# Step 1 & 2: Convert floats to strings representing full dates (assuming "-01-01" for simplicity)
df['Year_Sold'] = df['Year_Sold'].apply(lambda x: str(int(x)) + "-01-01")

# Step 3: Convert the string dates to datetime64
df['Year_Sold'] = pd.to_datetime(df['Year_Sold'])

print(df)
# Prints:
#
#    Year_Sold
# 0 2020-01-01
# 1 2021-01-01
# 2 2022-01-01

Option 2: If "Year_Sold" contains decimals for more precise timing

If the "Year_Sold" column represents years with additional decimal representation for more precise timing within the year, you'll need a more nuanced approach to convert these floats into datetime objects. The decimal part can represent a fraction of the year, which needs to be converted into the corresponding month and day. This is a bit more complex because:

1. You need to determine how many days correspond to the decimal part of the year. Since different years have different numbers of days (365 or 366 in a leap year), this calculation will vary slightly by year.
2. After calculating the number of days the decimal part represents, you'll add those days to the beginning of the year to get the precise date.

Here's how you could approach this:

1. Convert the year part of your float to an integer (this represents the year).
2. Multiply the decimal part by the number of days in that year to get the fraction of the year as days.
3. Add these days to the start of the year to get the exact datetime.

Let's implement this in Python:

import pandas as pd
from datetime import datetime, timedelta

def float_year_to_datetime(year_float):
    year = int(year_float)
    remainder = year_float - year
    start_of_year = datetime(year, 1, 1)
    days_in_year = 366 if (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0) else 365
    days_from_fraction = round(remainder * days_in_year)
    return start_of_year + timedelta(days=days_from_fraction)


# Example DataFrame
df = pd.DataFrame({'Year_Sold': [2020.25, 2021.5, 2022.75, 2022.98,
                                 2022.99, 2022.995, 2022.996, 2022.999]})
# df looks like this:
#
#    Year_Sold
# 0   2020.250
# 1   2021.500
# 2   2022.750
# 3   2022.980
# 4   2022.990
# 5   2022.995
# 6   2022.996
# 7   2022.999

# Convert the float years to datetime
df['Year_Sold'] = df['Year_Sold'].apply(float_year_to_datetime)
print(df)
# Prints:
#
#    Year_Sold
# 0 2020-04-02
# 1 2021-07-02
# 2 2022-10-02
# 3 2022-12-25
# 4 2022-12-28
# 5 2022-12-30
# 6 2022-12-31
# 7 2023-01-01

This function float_year_to_datetime does the following:

• It separates the year and the decimal part.
• It calculates the start of the year as a datetime object.
• It determines the number of days in the year to account for leap years.
• It converts the decimal part into a number of days and adds those days to the start of the year.