Finding percentage in a sub-group using group_by and summarise

And with a bit less code:

df <- data.frame(month=c("Feb-14", "Feb-14", "Feb-14", "Mar-14", "Mar-14", "Mar-14", "Apr-14", "Apr-14", "Apr-14", "May-14"),
             type=c("bbb", "ccc", "aaa", "bbb", "ccc", "aaa", "bbb", "ccc", "aaa", "bbb"),
             count=c(341, 527, 2674, 811, 1045, 4417, 1178, 1192, 4793, 916))


library(dplyr)

df %>% group_by(month) %>% 
       mutate(per=paste0(round(count/sum(count)*100, 2), "%")) %>% 
       ungroup

Since you want to "leave" your data frame untouched you shouldn't use summarise, mutate will suffice.

We can use prop.table to get the proportions within each group.

This can be written in dplyr :

library(dplyr)
df %>% group_by(month) %>% mutate(per= prop.table(count) * 100)

#  month  type  count    per
#   <chr>  <chr> <dbl>  <dbl>
# 1 Feb-14 bbb     341   9.63
# 2 Feb-14 ccc     527  14.9 
# 3 Feb-14 aaa    2674  75.5 
# 4 Mar-14 bbb     811  12.9 
# 5 Mar-14 ccc    1045  16.7 
# 6 Mar-14 aaa    4417  70.4 
# 7 Apr-14 bbb    1178  16.4 
# 8 Apr-14 ccc    1192  16.6 
# 9 Apr-14 aaa    4793  66.9 
#10 May-14 bbb     916 100   

Base R :

df$per <- with(df, ave(count, month, FUN = prop.table) * 100)

and data.table :

library(data.table)
setDT(df)[, per := prop.table(count) * 100, month]

Try

library(dplyr)
data %>%
    group_by(month) %>%
    mutate(countT= sum(count)) %>%
    group_by(type, add=TRUE) %>%
    mutate(per=paste0(round(100*count/countT,2),'%'))

Or make it more simpler without creating additional columns

data %>%
    group_by(month) %>%
    mutate(per =  100 *count/sum(count)) %>% 
    ungroup

We could also use left_join after summarising the sum(count) by 'month'

Or an option using data.table.

 library(data.table)
 setkey(setDT(data), month)[data[, list(count=sum(count)), month], 
               per:= paste0(round(100*count/i.count,2), '%')][]