I have dataframe like below :-
x<-c(3,2,1,8,7,11,10,9,7,5,4)
y<-c(\"a\",\"a\",\"a\", \"b\",\"b\",\"c\",\"c\",\"c\",\"c\",\"c\",\"c\")
z<-c(2,2,2,1,1
A solution with base R:
# df is split according to y, then we keep only the top "z" value (after ordering x)
# and rbind everything back together:
do.call(rbind,
lapply(split(df, df$y),
function(df1) df1[order(df1$x, decreasing=TRUE), ][1:unique(df1$z), ]))
# x y z
#a.1 3 a 2
#a.2 2 a 2
#b 8 b 1
#c.6 11 c 3
#c.7 10 c 3
#c.8 9 c 3
EDIT:
A much more direct way (still in base R
) provided in comment by @mt1022:
df[ave(1:nrow(df), df$y, FUN = seq_along) <= df$z, ]
# x y z
#1 3 a 2
#2 2 a 2
#4 8 b 1
#6 11 c 3
#7 10 c 3
#8 9 c 3
One approach with data.table
:
library(data.table)
setDT(df)
df[,.(inc=seq_len(.N)<=z,x,z),by=.(y)][inc==T ,-2]
# y x z
#1: a 3 2
#2: a 2 2
#3: b 8 1
#4: c 11 3
#5: c 10 3
#6: c 9 3
A solution with dplyr
that uses do
:
df %>%
group_by(y) %>%
do(head(.,as.numeric(unique(.$z))))
I'm posting the solution I was looking for using dplyr. It is based on @HNSKD:
library(dplyr)
x<-c(3,2,1,8,7,11,10,9,7,5,4)
y<-c("a","a","a", "b","b","c","c","c","c","c","c")
z<-c(2,2,2,1,1,3,3,3,3,3,3)
df<-data.frame(x,y,z)
df %>% group_by(y) %>% slice(1:2)
Which returns the first two elements for each y
:
# A tibble: 6 x 3
# Groups: y [3]
x y z
<dbl> <fct> <dbl>
1 3 a 2
2 2 a 2
3 8 b 1
4 7 b 1
5 11 c 3
6 10 c 3