how to calculate the median for groups separately in R - r

Question

Little example of data
df=structure(list(Dt = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L,
9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L,
22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L,
35L, 36L, 37L, 38L, 39L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L,
10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L,
23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L,
36L, 37L, 38L, 39L), .Label = c("2018-02-20 00:00:00.000", "2018-02-21 00:00:00.000",
"2018-02-22 00:00:00.000", "2018-02-23 00:00:00.000", "2018-02-24 00:00:00.000",
"2018-02-25 00:00:00.000", "2018-02-26 00:00:00.000", "2018-02-27 00:00:00.000",
"2018-02-28 00:00:00.000", "2018-03-01 00:00:00.000", "2018-03-02 00:00:00.000",
"2018-03-03 00:00:00.000", "2018-03-04 00:00:00.000", "2018-03-05 00:00:00.000",
"2018-03-06 00:00:00.000", "2018-03-07 00:00:00.000", "2018-03-08 00:00:00.000",
"2018-03-09 00:00:00.000", "2018-03-10 00:00:00.000", "2018-03-11 00:00:00.000",
"2018-03-12 00:00:00.000", "2018-03-13 00:00:00.000", "2018-03-14 00:00:00.000",
"2018-03-15 00:00:00.000", "2018-03-16 00:00:00.000", "2018-03-17 00:00:00.000",
"2018-03-18 00:00:00.000", "2018-03-19 00:00:00.000", "2018-03-20 00:00:00.000",
"2018-03-21 00:00:00.000", "2018-03-22 00:00:00.000", "2018-03-23 00:00:00.000",
"2018-03-24 00:00:00.000", "2018-03-25 00:00:00.000", "2018-03-26 00:00:00.000",
"2018-03-27 00:00:00.000", "2018-03-28 00:00:00.000", "2018-03-29 00:00:00.000",
"2018-03-30 00:00:00.000"), class = "factor"), ItemRelation = c(158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 158043L, 158043L, 158043L, 158043L,
158043L, 158043L, 158043L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L,
234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L, 234L
), stuff = c(200L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3600L,
0L, 0L, 0L, 0L, 700L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 1000L, 2600L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 400L, 700L,
200L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 3600L, 0L, 0L, 0L,
0L, 700L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1000L,
2600L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 400L, 700L), num = c(1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L, 1459L,
1459L, 1459L, 1459L, 1459L, 1459L), year = c(2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L,
2018L, 2018L, 2018L), action = c(0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 0L, 0L, 1L, 1L, 1L, 1L)), .Names = c("Dt", "ItemRelation",
"stuff", "num", "year", "action"), class = "data.frame", row.names = c(NA,
-78L))
now for each group
ItemRelation +num +year i have to calculate the median.
If i use this solution
# df with action 0 and stuff > 0
v <- df$stuff[intersect(which(df$action == 0),
which(df$stuff > 0))]
# df with action 1 and stuff > 0
w <- df$stuff[intersect(which(df$action == 1),
which(df$stuff > 0))]
# calulating the median of v for the last 5 observations
l <- length(v)
m0 <- median(v[(l-4):l]) # taking the median of the last 5 observations
# computing the final difference
m <- median(w) - m0
i calculate median for all group at once, but i have to calculate for
each group separately.
How can i perform it?
here expected output
ItemRelation num year value
158043 1459 2018 45
158043 234 2018 67
post edited. Note that value are not real, the medians will another, i just wanted to show what i wan as output
Edit
The action column has only two values 0 and 1. i must calculate median by stuff for 1 category of action, then median by stuff of zero category of action, using last five integer values before one category. I just take the last 5 observations, It is necessary to take the last 5 observations in the zero category of action, but only the integer value, and not calculate the median by all values of zero category. In our case this is
200
3600
700
1000
2600
then substract median of zero category from median of one category.
The number of observations by stuff in the zero category of action can vary from 0-10. If we have 10 integer values of zero category, we take last five. If there is only 1,2,3,4,5 values integer, we subtract median of real number of integer values. If we have only 0 without integer , we just substact 0.
But code must calculate the median by zero category, but 5 last obs before one category.
Note, instead of 0, there may be other values for the zero category of action.

Answer 1

One solution can be achieved using dplyr and following below mentioned steps. Please find comments in code below for approach.
Note: It seems that sample data from OP is not very meaningful as such.
library(dplyr)
df %>% filter(stuff > 0) %>% #First filter out for stuff > 0 which of our interest
group_by(ItemRelation, num, year) %>%
mutate(m = median(stuff[action==1]),
m0 = median(tail(stuff[action==0], 5))) %>% # Calculate m and m0 for all rows
filter(action == 1) %>% # Now keep only rows with action == 1
mutate(m = m-m0) %>%
select(-Dt,-m0,-action)
# # A tibble: 4 x 5
# # Groups: ItemRelation, num, year [2]
# ItemRelation stuff num year m
# <int> <int> <int> <int> <dbl>
# 1 158043 400 1459 2018 -450
# 2 158043 700 1459 2018 -450
# 3 234 400 1459 2018 -450
# 4 234 700 1459 2018 -450

Answer 2

The easiest way to do this is to use group_by and summarize from the dplyr package:
library(dplyr)
# median of groups
medians <- df %>%
group_by(ItemRelation, num, year) %>%
summarize(med = median(stuff, na.rm = T))
# median of nonzero values in each group
medians <- df %>%
filter(stuff>0) %>%
group_by(ItemRelation, num, year) %>%
summarize(med = median(stuff, na.rm = T))
subtract <- function(x){return(x[1]-x[2])}
median_diffs <- medians %>%
group_by(ItemRelation, num, year) %>%
mutate(med_diff = subtract(med))