Calculating time

[exsell-jc]

In R with lubridate, it would look like this:

one_year_before = some_date - years(1)
one_year_before = some_date - months(12)

But in tidypolars functions list, there doesn't seem to be a years or months function: https://tidypolars.readthedocs.io/en/latest/reference.html

[markfairbanks]

Hmm I'll have to look into this more. At first glance there isn't a "simple translation". There might have to be a tidypolars helper for this.

[markfairbanks]

For now I'd recommend using col().dt.offset_by() straight from polars.

FYI this method is built into the col() expression, so you don't need to import polars for this to work.

import tidypolars as tp
from tidypolars import col

df = tp.Tibble(date = ['2021-01-01', '2021-10-01']).mutate(date = col('date').str.strptime(tp.Date))

(
    df
    .mutate(minus_two_months = col('date').dt.offset_by("-2mo"),
            add_year = col('date').dt.offset_by("1y"))
)

┌────────────┬──────────────────┬────────────┐
│ date       ┆ minus_two_months ┆ add_year   │
│ ---        ┆ ---              ┆ ---        │
│ date       ┆ date             ┆ date       │
╞════════════╪══════════════════╪════════════╡
│ 2021-01-01 ┆ 2020-11-01       ┆ 2022-01-01 │
├╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌╌┼╌╌╌╌╌╌╌╌╌╌╌╌┤
│ 2021-10-01 ┆ 2021-08-01       ┆ 2022-10-01 │
└────────────┴──────────────────┴────────────┘

[markfairbanks]

Here's the documentation in case you need more options.

[exsell-jc]

Workaround (not exactly a solution) -- use polars inbetween, not as elegant, but works

I'm having problem with filtering/date comparison. I just added the filter() pipe to your code.

monthss = 6
(
    df
    .mutate(minus_two_months = col('date').dt.offset_by("-2mo"),
            add_year = col('date').dt.offset_by("1y"))
    .filter(col('date') > max(col('date')).dt.offset_by(f'-{monthss}mo'))
)

Here is R's equivalent:

monthss = 6
df |>
  mutate(minus_two_months = date - months(2),
         add_year = date - years(1)) |>
  filter(date > max(date) - months(monthss))

[PathosEthosLogos]

(Link is dead) Reporting another issue related -- calculation of date - date = x days (as duration, so it would be ddays(1)) e.g. 2000-01-01 - 1999-12-31 = 1 day Compensating for leap years and other anomalies would be nice also.