Calculate proportion of positives values by group

Question

With this dataframe:

table <- \"
    trt rep ss  d1  d4  d5  d6  d7
    1   1   1   0   0   0   0   0
    1   1   2   0   0   0   0   0
    1   1   3   0          


        

Answer 1

Thanks to @A.Webb, here's a way in base R:

aggregate(d[,4:8]>0~d$trt, FUN = mean)

#   d$trt d1        d4        d5        d6        d7
# 1     1  0 0.2222222 0.4444444 0.5555556 0.5555556
# 2     2  0 0.0000000 0.0000000 0.3750000 0.5000000

---

Here was my original idea:

rowsum(+(d[-(1:3)] > 0), d$trt, na.rm=TRUE) / 
  rowsum(+!is.na(d[-(1:3)]), d$trt, na.rm=TRUE)

The + is there because rowsum only works with numbers, and not with logicals.

Answer 2

Using data.table, something like this:

library(data.table)
d <- data.table(d)
d[,lapply(.SD,function(x) sum(x>0,na.rm=T)/sum(!is.na(x))),
  .SDcols=grep("^d",names(d),val=T),
   by=trt]

   trt d1        d4        d5        d6        d7
1:   1  0 0.2222222 0.4444444 0.5555556 0.5555556
2:   2  0 0.0000000 0.0000000 0.3750000 0.5000000

Answer 3

We can use dplyr

library(dplyr)
d %>%
  group_by(trt) %>% 
  summarise_each( funs(round(mean(.>0, na.rm=TRUE),2)), d1:d7) 
#   trt    d1    d4    d5    d6    d7
#  (int) (dbl) (dbl) (dbl) (dbl) (dbl)
#1     1     0  0.22  0.44  0.56  0.56
#2     2     0  0.00  0.00  0.38  0.50