Subsetting a Dataframe in R on last two month's - r

Question

Hi I have a dataframe for example:
Order Number Date
4378345 2020-01-02
4324375 2020-02-03
Now I want to subset this Dataframe only on orders with Date greater than last 2 month from today's Date, So that when I put it on automation the code it automatically take's last two month to till date.
Any help would be appreciated.
EDIT: My apologies i think i Should have been more clear as by last two month i mean if the date is today 2020-03-16, I wound want my data to be from 2020-02-01 to till date

Answer 1

library(lubridate)
subset(your_data, Date > today() - months(2))
This assumes your date column is of class Date already.
In general, months are not super well-defined... you may want to use a more deterministic criterion. For example, what is 2 months before April 28, April 29, April 30, May 1? Keep in mind that February had 29 days this year. You can see lubridate's opinion with (as.Date("2020-04-28") + 0:3) - months(2), which is NA in the case of 2020-04-30. Doing 60 days before (Date > today() - days(60)) or some other better-defined criteria will give you more consistency.
To go from the first day of the previous month, use the following code. This is well-defined, as all months have a first day.
subset(your_data, Date >= floor_date(today(), unit = "month") - months(1))

Answer 2

You can use the package lubridate with something like:
library(lubridate)
subset(yourDF,
Date >= (today() - months(2)))
[edit: Ouch, someone was faster)

Answer 3

library(dplyr)
library(lubridate)
x <- tibble(Date = as_date(c("2020-01-02", "2020-02-03")))
x %>% filter(Date >= today() - months(2))

Answer 4

time <- read.table(textConnection("
OrderNumber Date
4378345 2020-01-02
4324375 2020-02-03"), header = TRUE)
time <- as.data.frame(time)
library(lubridate)
library(dplyr)
time2 <- time %>%
filter(Date >= today() - months(2))

Answer 5

In base R you can find the first of last month using seq (credits to #G.Grothendieck), and replacing the day with the first using strftime.
last.1st <-
as.Date(paste0(strftime(seq(today, length=2, by="-1 month")[2], format="%Y-%m"), "-01"))
The first of current month can be obtained using gsub and regular expressions.
curr.1 <- as.Date(gsub("\\d{2}$", "01", Sys.Date()))
Then subsetting as usual is straightforward. The whole period:
dat[dat$date >= last.1, ]
# order.num date
# 18 432174 2020-02-01
# 19 432175 2020-02-03
# 20 432176 2020-02-05
# 21 432177 2020-02-07
# 22 432178 2020-02-09
# 23 432179 2020-02-11
# 24 432180 2020-02-13
# 25 432181 2020-02-15
# 26 432182 2020-02-17
# 27 432183 2020-02-19
# 28 432184 2020-02-21
# 29 432185 2020-02-23
# 30 432186 2020-02-25
# 31 432187 2020-02-27
# 32 432188 2020-02-29
# 33 432189 2020-03-02
# 34 432190 2020-03-04
# 35 432191 2020-03-06
# 36 432192 2020-03-08
# 37 432193 2020-03-10
# 38 432194 2020-03-12
# 39 432195 2020-03-14
# 40 432196 2020-03-16
And just the last month:
dat[dat$date >= last.1 & dat$date <= curr.1, ]
# order.num date
# 18 432174 2020-02-01
# 19 432175 2020-02-03
# 20 432176 2020-02-05
# 21 432177 2020-02-07
# 22 432178 2020-02-09
# 23 432179 2020-02-11
# 24 432180 2020-02-13
# 25 432181 2020-02-15
# 26 432182 2020-02-17
# 27 432183 2020-02-19
# 28 432184 2020-02-21
# 29 432185 2020-02-23
# 30 432186 2020-02-25
# 31 432187 2020-02-27
# 32 432188 2020-02-29
Toy data
dat <- structure(list(order.num = c(432157, 432158, 432159, 432160,
432161, 432162, 432163, 432164, 432165, 432166, 432167, 432168,
432169, 432170, 432171, 432172, 432173, 432174, 432175, 432176,
432177, 432178, 432179, 432180, 432181, 432182, 432183, 432184,
432185, 432186, 432187, 432188, 432189, 432190, 432191, 432192,
432193, 432194, 432195, 432196), date = structure(c(18259, 18261,
18263, 18265, 18267, 18269, 18271, 18273, 18275, 18277, 18279,
18281, 18283, 18285, 18287, 18289, 18291, 18293, 18295, 18297,
18299, 18301, 18303, 18305, 18307, 18309, 18311, 18313, 18315,
18317, 18319, 18321, 18323, 18325, 18327, 18329, 18331, 18333,
18335, 18337), class = "Date")), class = "data.frame", row.names = c(NA,
-40L))