Combining scenario to Replace Medians by Groups in R

Question

I have dataset

mydat <- 
structure(list(code = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("52382MCK", 
"52499MCK"), class = "factor"), item = c(11709L, 11709L, 11709L, 
11709L, 11708L, 11708L, 11708L, 11710L, 11710L, 11710L, 11710L, 
11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 
11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 11710L, 
11710L, 11202L, 11203L, 11203L, 11204L, 11204L, 11205L, 11205L
), sales = c(30L, 10L, 20L, 15L, 2L, 10L, 3L, 30L, 10L, 20L, 
15L, 2L, 10L, 3L, 30L, 10L, 20L, 15L, 2L, 10L, 3L, 30L, 10L, 
20L, 15L, 2L, 10L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), action = c(0L, 
1L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 
1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 
1L, 1L)), row.names = c(NA, -35L), class = "data.frame")
# coerce to data.table
setDT(mydat)

with this dataset, several operations are performed.

1. selecting scenario by groups.

So there is action column. It can have only two values zero(0) or one(1).

The scenarios are the number of zero categories of action before first category of action and the number of zeros after one category of action.

For example
52382МСК    11709

it is scenario when we have 1 zero category of action col. before first category of action col , and two zeros after first category of action col. Note: maybe scenario when we have 2 zero category of action col. before first category of action col , and 1 zero after first category of action col.

mydat1

code    item    sales   action
52382МСК    11709   30  0
52382МСК    11709   10  1
52382МСК    11709   20  0
52382МСК    11709   15  0

to detect this scenario i use this script/ This script very well works, thank for @Uwe

library(data.table)
library(magrittr)

max_zeros <- 3
zeros <- sapply(0:max_zeros, stringr::str_dup, string = "0")
names(zeros) <- as.character(nchar(zeros))
sc <- CJ(zeros.before = zeros, zeros.after = zeros)[
  , scenario.name := paste(nchar(zeros.before), nchar(zeros.after), sep = "-")][
    , action.pattern := sprintf("%s1+(?=%s)", zeros.before, zeros.after)][]
# special case: all zero
sc0 <- data.table(
  zeros.before = NA,
  zeros.after = NA, 
  scenario.name = "no1", 
  action.pattern = "^0+$")
sc <- rbind(sc0, sc)

and then

setDT(mydat)
class <- mydat[, .(scenario.name = sc$scenario.name[
  paste(action, collapse = "") %>% 
    stringr::str_count(sc$action.pattern) %>%
    is_greater_than(0) %>% 
    which() %>% 
    max()
  ]),
  by = .(code, item)][]

class
mydat[class, on = .(code, item)]

So i get data with class of scenario.

2.operation it is replace median.

For each scenario median by zero category is calculated.

I need to calculate the median by 1 preceding zeros category by action column, i.e. which go before one category of action column, and by 2 zeros by action column that go after the one category. The median replacing performed only for first category of action column by sale column. if median is more than the sales, then do not replace it.

To do it i use the script

sales_action <- function(DF, zeros_before, zeros_after) {
  library(data.table)
  library(magrittr)
  action_pattern <- 
    do.call(sprintf, 
            c(fmt = "%s1+(?=%s)", 
              stringr::str_dup("0", c(zeros_before, zeros_after)) %>% as.list()
            ))
  message("Action pattern used: ", action_pattern)
  setDT(DF)[, rn := .I]
  tmp <- DF[, paste(action, collapse = "") %>% 
              stringr::str_locate_all(action_pattern) %>% 
              as.data.table() %>% 
              lapply(function(x) rn[x]),
            by = .(code, item)][
              , end := end + zeros_after]
  DF[tmp, on = .(code, item, rn >= start, rn <= end), 
     med := as.double(median(sales[action == 0])), by = .EACHI][
       , output := as.double(sales)][action == 1, output := pmin(sales, med)][
         , c("rn", "med") := NULL][]
}

and then

sales_action(mydat, 1L, 2L)

so i get the result.

the question is based on the following

Each time i must manually enter the scenario to replacing by median

sales_action(mydat, 1L, 2L)
sales_action(mydat, 3L, 1L)
sales_action(mydat, 2L, 2L)

and so on.

How to do that replacing median was perform for all possible scenarios automatically so that I do not write every time sales_action(mydat, .L, .L)

So example of output

code    i    tem    sales   action  output  pattern
52382MCK    11709   30        0       30    01+00
52382MCK    11709   10        1       10    01+00
52382MCK    11709   20        0       20    01+00
52382MCK    11709   15        0       15    01+00
52382MCK    1170    8         0        8    01+00
52382MCK    1170    10        1        8    01+00
52382MCK    1170    2         0        2    01+00
52382MCK    1170    15        0        15   01+00

Uwe · Accepted Answer

If I understand correctly, the OP wants to analyse the success of sales actions by comparing sales figures during actions with the median sales of the periods immediately before and after the sales action.

There are some challenges:

There might be more than one sales action per code, item group.
The available data might cover less than the requested 3 three days each before and after a sales action.

IMHO, the introduction of scenarios is a detour to handle issue 2.

The approach below

identifies the sales actions within each code, item group,
picks up to three zero action rows before and up to three rows after each sales action,
computes the median sales of those rows, and
updates output in case the sales figure within a sales action exceeds the median of the surrounding zero action rows.

The term category has been coined by the OP to distinguish between periods of sales actions (contiguous streaks of action == 1L) and the zero action periods before and after.

library(data.table)
# coerce to data.table and create categories
setDT(mydat)[, cat := rleid(action), by = .(code, item)][]

# extract action categories, identify preceeding & succeeding zero action categories
mycat <- mydat[, .(action = first(action)), by = .(code, item, cat = cat)][
  , `:=`(before = cat - 1L, after = cat + 1L)][action == 1L]

mycat

       code  item cat action before after
1: 52382MCK 11709   2      1      1     3
2: 52382MCK 11708   2      1      1     3
3: 52382MCK 11710   2      1      1     3
4: 52382MCK 11710   4      1      3     5
5: 52382MCK 11710   6      1      5     7
6: 52499MCK 11203   2      1      1     3
7: 52499MCK 11205   1      1      0     2

Note that group 52382MCK, 11710 includes three separate sales actions. before and after may point to non-existing cat but this will be rectified automatically during the subsequent joins.

# compute median of surrouding zero action categories
action_cat_median <- 
  rbind(
    # get sales from up to 3 zero action rows before action category
    mydat[mycat, on = .(code, item, cat = before), 
          .(sales = tail(sales, 3), i.cat), by =.EACHI],
    # get sales from up to 3 zero action rows after action category
    mydat[mycat, on = .(code, item, cat = after), 
          .(sales = head(sales, 3), i.cat), by =.EACHI]
  )[
    # remove empty groups
    !is.na(sales)][
      # compute median for each action category
      , .(med = as.double(median(sales))), by = .(code, item, cat = i.cat)]

action_cat_median

       code  item cat  med
1: 52382MCK 11709   2 20.0
2: 52382MCK 11708   2  2.5
3: 52382MCK 11710   2 10.0
4: 52382MCK 11710   4 10.0
5: 52382MCK 11710   6 10.0
6: 52499MCK 11203   2  2.0

# prepare result
mydat[, output := as.double(sales)][
  # update join
  action_cat_median, on = .(code, item, cat), output := pmin(sales, med)]

Edit: Alternatively, the call to pmin() can be replaced by a non-equi join which updates only rows where sales exceeds the median:

# prepare result, alternative approach
mydat[, output := as.double(sales)][
  # non-equi update join
  action_cat_median, on = .(code, item, cat, output > med), output := med]


mydat

        code  item sales action cat output
 1: 52382MCK 11709    30      0   1   30.0
 2: 52382MCK 11709    10      1   2   10.0
 3: 52382MCK 11709    20      0   3   20.0
 4: 52382MCK 11709    15      0   3   15.0
 5: 52382MCK 11708     2      0   1    2.0
 6: 52382MCK 11708    10      1   2    2.5
 7: 52382MCK 11708     3      0   3    3.0
 8: 52382MCK 11710    30      0   1   30.0
 9: 52382MCK 11710    10      0   1   10.0
10: 52382MCK 11710    20      0   1   20.0
11: 52382MCK 11710    15      1   2   10.0
12: 52382MCK 11710     2      0   3    2.0
13: 52382MCK 11710    10      0   3   10.0
14: 52382MCK 11710     3      0   3    3.0
15: 52382MCK 11710    30      0   3   30.0
16: 52382MCK 11710    10      0   3   10.0
17: 52382MCK 11710    20      0   3   20.0
18: 52382MCK 11710    15      1   4   10.0
19: 52382MCK 11710     2      0   5    2.0
20: 52382MCK 11710    10      0   5   10.0
21: 52382MCK 11710     3      0   5    3.0
22: 52382MCK 11710    30      0   5   30.0
23: 52382MCK 11710    10      0   5   10.0
24: 52382MCK 11710    20      0   5   20.0
25: 52382MCK 11710    15      1   6   10.0
26: 52382MCK 11710     2      0   7    2.0
27: 52382MCK 11710    10      0   7   10.0
28: 52382MCK 11710     3      0   7    3.0
29: 52499MCK 11202     2      0   1    2.0
30: 52499MCK 11203     2      0   1    2.0
31: 52499MCK 11203     2      1   2    2.0
32: 52499MCK 11204     2      0   1    2.0
33: 52499MCK 11204     2      0   1    2.0
34: 52499MCK 11205     2      1   1    2.0
35: 52499MCK 11205     2      1   1    2.0
        code  item sales action cat output

The following rows have been updated:

mydat[output != sales]

       code  item sales action cat output
1: 52382MCK 11708    10      1   2    2.5
2: 52382MCK 11710    15      1   2   10.0
3: 52382MCK 11710    15      1   4   10.0
4: 52382MCK 11710    15      1   6   10.0

Combining scenario to Replace Medians by Groups in R

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Answers (1)

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