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Define %--% 

Fixes #635
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Hadley Wickham 2022-08-09 15:57:18 -05:00
parent 4c3f5aa604
commit 776606f0a5
1 changed files with 9 additions and 6 deletions
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datetimes.qmd
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@ -519,24 +519,27 @@ It's obvious what `dyears(1) / ddays(365)` should return: one, because durations
What should `years(1) / days(1)` return? What should `years(1) / days(1)` return?
Well, if the year was 2015 it should return 365, but if it was 2016, it should return 366! Well, if the year was 2015 it should return 365, but if it was 2016, it should return 366!
There's not quite enough information for lubridate to give a single clear answer. There's not quite enough information for lubridate to give a single clear answer.
What it does instead is give an estimate, with a warning: What it does instead is give an estimate:
```{r} ```{r}
years(1) / days(1) years(1) / days(1)
``` ```
If you want a more accurate measurement, you'll have to use an **interval**. If you want a more accurate measurement, you'll have to use an **interval**.
An interval is a duration with a starting point: that makes it precise so you can determine exactly how long it is: An interval is a pair of starting and ending date times, or you can think of it as a duration with a starting point.
You can create an interval by writing `start %--% end`:
```{r} ```{r}
next_year <- today() + years(1) to_next_year <- today() %--% (today() + years(1))
(today() %--% next_year) / ddays(1) to_next_year
``` ```
To find out how many periods fall into an interval, you need to use integer division: You could then divide it by a duration or a period:
```{r} ```{r}
(today() %--% next_year) %/% days(1) to_next_year / ddays(1)
to_next_year / months(1)
``` ```
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