Compare day in current month to same day previous month PostgreSQL - date

Question

I'm trying to compare values of current month's data to previous months using PostgreSQL. So if today is 4/23/2018, I want the data for 3/23/2018.
I've tried current_date - interval '1 month' but it is problematic for months with 31 days.
My table is structured as simply as
date, value

Answer 1

Check this example query:
WITH dates AS (SELECT date::date FROM generate_series('2018-01-01'::date, '2018-12-31'::date, INTERVAL '1 day') AS date)
SELECT
start_dates.date AS start_date,
end_dates.date AS end_date
FROM
dates AS start_dates
RIGHT JOIN dates AS end_dates
ON ( start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date);
It will output all end_dates and corresponding start_dates. The corresponding dates are defined by interval '1 month' and checked in both ways:
start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date
The output looks like this:
....
2018-02-26 2018-03-26
2018-02-27 2018-03-27
2018-02-28 2018-03-28
2018-03-29
2018-03-30
2018-03-31
2018-03-01 2018-04-01
2018-03-02 2018-04-02
2018-03-03 2018-04-03
2018-03-04 2018-04-04
....
Note, that there are 'gaps' for days without corresponding dates.
Back to your table, join the table with itself (giving aliases) and use given join condition, so the query would look like this:
SELECT
start_dates.value - end_dates.value AS change,
start_dates.date AS start_date,
end_dates.date AS end_date
FROM
_your_table_name_ AS start_dates
RIGHT JOIN _your_table_name_ AS end_dates
ON ( start_dates.date + interval '1 month' = end_dates.date AND
end_dates.date - interval '1 month' = start_dates.date);

Answer 2

Given the following table structure:
create table t (
d date,
v int
);
After populating with some dates and values, there is a way to find the value of the previous month using simple calculations and the LAG function, without resorting to joins. I am not sure how it compares from a performance perspective, so please run your own tests before selecting which solution to use.
select
*,
lag(v, day_of_month) over (order by d) as v_end_of_last_month,
lag(v, last_day_of_previous_month + day_of_month - cast(extract(day from d - interval '1 month') as int)) over (order by d) as v_same_day_last_month
from (
select
*,
lag(day_of_month, day_of_month) over (order by d) as last_day_of_previous_month
from (
select
*,
cast(extract(day from d) as int) as day_of_month
from
t
) t_dom
) t_dom_ldopm;
You may note that between the 29th and 31st of March, the comparison will be made against the 28th of February, since the same day does not exist in February for those particular dates. The same logic applies to other months with different number of days.