count distinct and calculate percent differences in two columns ifrom R dataframe

df %>%
   group_by(category) %>%
   summarise(perc_diff = 100 * mean(table(tag_number) == 1))

# A tibble: 3 x 2
  category perc_diff
  <chr>        <dbl>
1 GT             100
2 TB             100
3 YT               0

Try this solution, but I am not sure whether it can be generalized to a larger case. The approach is to create a wide table then calculate difference percentage.

library(dplyr)
library(tidyr)
library(purrr)

df <- data.frame(
  stringsAsFactors = FALSE,
  month = c("jan","jan","jan","jan",
            "jan","feb","feb","feb","feb"),
  category = c("TB", "GT", "TB", "YT", "GT", "TB", "GT", "TB", "YT"),
  tag_number = c(101L, 101L, 223L, 223L, 223L, 345L, 345L, 655L, 223L)
 )

df2 <- df %>% 
  pivot_wider(names_from = "month", values_from = "tag_number") %>% 
  mutate(both_months = map2(jan, feb, ~intersect(.x,.y))) %>%  #find intersect
  mutate(all_lenght = map2(jan,feb, ~union(.x,.y))) %>%  #find all length
  mutate(percent_diff = map2(both_months, all_lenght, 
                             ~(100 - length(.x)*100/length(.y))) %>% 
                        unlist()) # calculate difference percentage

df2

#> # A tibble: 3 x 6
#>   category jan       feb       both_months all_lenght percent_diff
#>   <chr>    <list>    <list>    <list>      <list>            <dbl>
#> 1 TB       <int [2]> <int [2]> <int [0]>   <int [4]>           100
#> 2 GT       <int [2]> <int [1]> <int [0]>   <int [3]>           100
#> 3 YT       <int [1]> <int [1]> <int [1]>   <int [1]>             0

You can try the base R code below

transform(
  reshape(
    aggregate(
      tag_number ~ .,
      df,
      toString
    ),
    direction = "wide",
    idvar = "category",
    timevar = "month"
  ),
  pct_diff = mapply(
    function(...) {
      sprintf("%s%%", mean(table(unlist(c(...))) == 1) * 100)
    },
    strsplit(tag_number.feb, ", "),
    strsplit(tag_number.jan, ", ")
  )
)

which gives

  category tag_number.feb tag_number.jan pct_diff
1       GT            345       101, 223     100%
3       TB       345, 655       101, 223     100%
5       YT            223            223       0%

I tried to come up with a solution that will show you the percentage of unie values based on month-pairs that scales to more than two months.

The code can be shortened by combining rows, but I kept it like this so you can easily inspect the results of each step, to help you understand what is actually happening.

library(data.table)
# make mydata a data.table
setDT(mydata)
mydata[, month_f := factor(month, labels = c("jan", "feb"))]
#build list columns 'tags', containing a list of all tag-values for that month + category
tags <- mydata[, .(alltags = as.list(.SD)), by = .(month_f, category)]
# build lookup table with all month >> month combinations
ans <- CJ(month  = unique(mydata$month_f), 
   month2 = unique(mydata$month_f),
   category = unique(mydata$category))
# join in the tags for month and month2
ans[tags, tags  := i.alltags, on = .(month = month_f, category)]
ans[tags, tags2 := i.alltags, on = .(month2 = month_f, category)]
# find overlapping values between tags and tags2 list columns
ans[, tags_overlap := Map(intersect, tags, tags2)][]
# calculate the percentage of the overlap
myfun <- function(x, y) {
  paste0(100 * (1 - (length(x) / length(y))), "%")
}
ans[, unique := Map(myfun, tags_overlap, tags)][]
# cast to wide format
dcast(ans, category + month ~ month2, value.var = "unique")
#    category month  jan  feb
# 1:       GT   jan   0% 100%
# 2:       GT   feb 100%   0%
# 3:       TB   jan   0% 100%
# 4:       TB   feb 100%   0%
# 5:       YT   jan   0%   0%
# 6:       YT   feb   0%   0%

A possible solution, based on dplyr:

library(dplyr)

df <- data.frame(
  stringsAsFactors = FALSE,
  month = c("jan","jan","jan","jan",
            "jan","feb","feb","feb","feb"),
  category = c("TB", "GT", "TB", "YT", "GT", "TB", "GT", "TB", "YT"),
  tag_number = c(101L, 101L, 223L, 223L, 223L, 345L, 345L, 655L, 223L)
)

df %>% 
  group_by(category) %>% 
  summarise(
    pct_diff = 100 * (1 - 2*length(
    intersect(tag_number[month == "jan"], tag_number[month == "feb"]))
    / length(tag_number)), .groups = "drop")

#> # A tibble: 3 × 2
#>   category pct_diff
#>   <chr>       <dbl>
#> 1 GT            100
#> 2 TB            100
#> 3 YT              0

Super interesting puzzle. First thing I did was to expand the test data a little and use actual dates rather than character strings so I could utilize the natural ordering. I added a few more test cases just for laughs.

library(dplyr)
library(purrr)
library(tidyr)

df <- data.frame(
  stringsAsFactors = FALSE,
  month = c(as.Date("2019-01-01"),as.Date("2019-01-01"),as.Date("2019-01-01"),as.Date("2019-01-01"),
            as.Date("2019-01-01"),as.Date("2019-01-01"),as.Date("2019-01-01"),as.Date("2019-02-01"),
            as.Date("2019-02-01"),as.Date("2019-02-01"),as.Date("2019-02-01"),as.Date("2019-02-01"),
            as.Date("2019-02-01"),as.Date("2019-03-01"),as.Date("2019-03-01"),as.Date("2019-03-01"),
            as.Date("2019-03-01"),as.Date("2019-04-01"),as.Date("2019-04-01"),as.Date("2019-04-01"),
            as.Date("2019-04-01"),as.Date("2019-04-01")),
  category = c("TB", "GT", "TB", "YT", "GT", "TB", "GT", "ZW", "YT", "TB", "GT", "TB", "YT", "GT", 
               "TB", "GT", "TB", "YT", "TB", "YT", "TB", "GT"),
  tag_number = c(101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 655L, 345L, 345L, 345L, 345L, 
                 223L, 655L, 345L, 223L, 345L, 655L, 101L, 379L, 345L))

Then I considered your question from the set theory perspective. The question you are asking is essentially "What is the percentage new tags occurring this month (setdiff) of the total tags over this and the preceding month (union)?" Here is a little function to answer that question along with minimal error handling for the first month.

g <- function(t)
  {
    r <- c()
    for (i in seq_along(t))
      {
        if (i == 1 & length(t[[i]] > 0))             { r <- c(1) } 
          else if (i == 1 & length(t[[i]]) == 0)     { r <- c(0) }
          else { r <- c(r, length(Reduce(setdiff, list(t[[i]], t[[i-1]]))) 
                              / length(union(t[[i-1]], t[[i]]))) }
      }
    return(r)
  }

Then, I used pivot_wider to put the data into a form with the date as a unique identifier and each category in a column, and finally applied the function to each of the columns. Et voilà.

s <- df %>% group_by(month) %>%
            unique() %>%
            pivot_wider(id_cols = month, 
                     names_from = category, 
                    values_from = tag_number, 
                      values_fn = list) %>%
            map_if(is.list, ~ g(.x))

List of 5
 $ month: Date[1:4], format: "2019-01-01" "2019-02-01" ...
 $ TB   : num [1:4] 1 0.5 0.667 0.333
 $ GT   : num [1:4] 1 0.5 0.5 0
 $ YT   : num [1:4] 1 0.667 0 1
 $ ZW   : num [1:4] 0 1 0 NaN