R data.table：如何计算组内向量列的值总和与表中列中其余值的比率？

Question

Let say we have a DT:

A	B	C
1	1	1
1	2	3
2	3	5
2	4	7
3	5	9
3	6	11

Let's say we want to obtain the ratio of mean value of B and C within each group A c(1,2,3) to mean values of B and C but for all other groups A (=exluding the group A)...

Let's say the "columns to calculate mean" = _vars = c(B,C).

What if I am grouping by A and C and _vars = c(B) only ?
Update:

How to do it automatically for all column names in a given vector, ie c("B","C")

Answer 1

If I understood your question correctly:

cols<-c('A','C')
cols.remaining <- setdiff(colnames(dt),cols) 

global.means <- paste0("GlobalMean",cols.remaining)
group.means <- paste0("GroupMean",cols.remaining)
group.ratios <- paste0("GroupRatio",cols.remaining)


dt[, (global.means):= lapply(.SD, mean) ,.SDcols=c(cols.remaining)][
   ,(group.means) := lapply(.SD, mean), by = cols,.SDcols=cols.remaining][
   ,(group.ratios):= mapply(group.means,global.means,FUN = function(m,g) get(m)/get(g),SIMPLIFY = F)][]

   A B  C GlobalMeanB GroupMeanB GroupRatioB
1: 1 1  1         3.5          1   0.2857143
2: 1 2  3         3.5          2   0.5714286
3: 2 3  5         3.5          3   0.8571429
4: 2 4  7         3.5          4   1.1428571
5: 3 5  9         3.5          5   1.4285714
6: 3 6 11         3.5          6   1.7142857

and with cols <- c('B') :

   A B  C GlobalMeanA GlobalMeanC GroupMeanA GroupMeanC GroupRatioA GroupRatioC
1: 1 1  1           2           6          1          1         0.5   0.1666667
2: 1 2  3           2           6          1          3         0.5   0.5000000
3: 2 3  5           2           6          2          5         1.0   0.8333333
4: 2 4  7           2           6          2          7         1.0   1.1666667
5: 3 5  9           2           6          3          9         1.5   1.5000000
6: 3 6 11           2           6          3         11         1.5   1.8333333
6: 3 11 0.28571429

Answer 2

Here is one possible way to solve your problem.

# ratios of means related to column B and C grouped by A
cols = c("B", "C")
DT[, Map(`/`, lapply(.SD, mean), lapply(DT[-.I, cols, with=FALSE], mean)), by=.(A), .SDcols=cols]
#        A         B     C
# 1:     1 0.3333333  0.25
# 2:     2 1.0000000  1.00
# 3:     3 2.2000000  2.50

# alternative solution (gives the same result)
DT[, Map(`/`, lapply(.SD, mean), lapply(DT[!.BY, cols, with=FALSE, on=.(A)], mean)), by=.(A), .SDcols=cols]

lapply(.SD, mean) computes the groups' means.
lapply(DT[-.I, cols, with=FALSE], mean) : computes the means excluding the current group.
Map function then uses the division operator, / , to compute the ratio between the groups' means (calculated by lapply(.SD, mean) ) and the means excluding the current group (calculated by lapply(DT[-.I, cols, with=FALSE], mean) ) element-wise.

For other scenarios, you just adapt the .SDcols and by arguments in an appropriate way.

# ratios of means related to column B grouped by A and C.
cols = "B"
DT[, Map(`/`, lapply(.SD, mean), lapply(DT[-.I, cols, with=FALSE], mean)), by=.(A, C), .SDcols=cols]
#        A     C         B
# 1:     1     1 0.2500000
# 2:     1     3 0.5263158
# 3:     2     5 0.8333333
# 4:     2     7 1.1764706
# 5:     3     9 1.5625000
# 6:     3    11 2.0000000