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Scan for outdated or missing drivers - takes under a minuteDriver Scan →Repair Windows errors before they cause bigger problemsFix Now →Fix the driver behind crashes, sound loss and screen glitchesFind Drivers →Subtract Python date objects and read the result’s .days attribute. For example, (date(2026, 10, 4) - date(2026, 10, 1)).days returns 3: the dates are three days apart, not four days inclusive.
Find the difference between calendar dates
For dates with no relevant time of day, use datetime.date. Subtracting one date from another produces a timedelta; its .days attribute gives the signed difference in whole calendar days. The Python documentation describes this behavior in its datetime reference.
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from datetime import date
start = date.fromisoformat("2026-10-01")
end = date.fromisoformat("2026-10-04")
days_between = (end - start).days
print(days_between) # 3
The order matters: end - start is positive if end is later, zero when both dates are the same, and negative if the dates are reversed. Ordinary subtraction measures the difference between date labels; it does not count both endpoints. For an inclusive count when the dates are ordered from earlier to later, add one: (end - start).days + 1.
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ISO dates: YYYY-MM-DD
For strings such as 2026-10-01, date.fromisoformat() parses directly to a date:
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from datetime import date
start = date.fromisoformat("2026-10-01")
end = date.fromisoformat("2026-10-04")
print((end - start).days) # 3
Other known formats
Use datetime.strptime() with a format string that matches the input, then call .date() when only the calendar date matters. For example, for month/day/year strings:
from datetime import datetime
start = datetime.strptime("10/01/2026", "%m/%d/%Y").date()
end = datetime.strptime("10/04/2026", "%m/%d/%Y").date()
print((end - start).days) # 3
Parsing makes the intended format explicit and rejects invalid calendar values. Avoid subtracting raw strings: their lexical order is not generally the same as chronological order unless the format guarantees it.
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Use datetimes when time of day matters
If the question is about elapsed time between timestamps rather than the difference between calendar dates, use datetime objects. A timedelta can include sub-day components, so delta.days is only its normalized whole-day component—not a fractional total-days value. To express an elapsed duration as fractional days, divide by a one-day timedelta:
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from datetime import datetime, timedelta
a = datetime(2026, 10, 1, 12, 0)
b = datetime(2026, 10, 2, 18, 0)
delta = b - a
fractional_days = delta / timedelta(days=1)
print(fractional_days) # 1.25
Use delta.total_seconds() when you need a duration in seconds rather than days. The Python datetime documentation defines the timedelta fields and arithmetic.
Handle time zones and daylight-saving changes deliberately
Python does not allow subtraction between a naive datetime (one without time-zone information) and an aware datetime (one with it). If both values are aware but have different tzinfo attributes, subtraction follows the documented UTC-equivalent rule. If both aware values share the same tzinfo object, the documented arithmetic rule ignores time-zone adjustments. That distinction matters across daylight-saving transitions: a local calendar interval and elapsed time between instants may not be the same.
When the desired result is elapsed time between instants, convert both aware datetimes to UTC before subtracting. For named local zones, Python’s zoneinfo module provides IANA time-zone support; see the zoneinfo documentation for its behavior and daylight-saving examples. Decide whether your application needs calendar-date difference or elapsed UTC duration before choosing the calculation.
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