Select first 80 observations for each level in R

Question

I have a data set that looks like this:

structure(list(A = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("1", 
"2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
"14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
"25"), class = "factor"), T = c(0.04, 0.08, 0.12, 0.16, 0.2, 
0.24), X = c(464.4, 464.4, 464.4, 464.4, 464.4, 464.4), Y = c(418.5, 
418.5, 418.5, 418.5, 418.5, 418.5), V = c(0, 0, 0, 0, 0, 0), 
    GD = c(0, 0, 0, 0, 0, 0), ND = c(NA, 0, 0, 0, 0, 0), ND2 = c(NA, 
    0, 0, 0, 0, 0), TID = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("t1", 
    "t10", "t11", "t12", "t13", "t14", "t15", "t16", "t17", "t18", 
    "t19", "t2", "t20", "t21", "t22", "t23", "t24", "t25", "t3", 
    "t4", "t5", "t6", "t7", "t8", "t9"), class = "factor")), .Names = c("A", 
"T", "X", "Y", "V", "GD", "ND", "ND2", "TID"), row.names = c(NA, 
6L), class = "data.frame")

I want to select the first 80 observations of all variables for each TID. So far, I can do this with the first TID only using the code:

sub.data1<-NM[1:80, ]

How can I do it for all my other TIDs?

Thanks!

Quadrennium

Answer 1

Using function ddply() from plyr you can split data by TID and then select forst 80 with head() and then put all again in one data frame,

library(plyr)
ddply(NM, .(TID), head, n = 80)

Boeotia

+1! Probably there is no need for the lambda function, ddply(NM, .(TID), head, n = 80) should work. – Knawel

Answer 2

I would do:

lapply(split(dat, dat$TID), head, 80)

It returns a list of data.frames with 80 (or less) rows. If instead you want everything into one data.frame:

do.call(rbind, lapply(split(dat, dat$TID), head, 80))

Dingdong

sorry I forgot to mention that I want to retain all the other variables too. – Quadrennium

Answer 3

Using data tables, I made a shorter example with just TIDs t1 and t2 that returns the first 2 rows of t1 and t2. It can be adjusted for your data.

library(data.table)
data<-structure(list(A = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = c("1", 
                "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", 
                "14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", 
                "25"), class = "factor"), T = c(0.04, 0.08, 0.12, 0.16, 0.2, 
                0.24), X = c(464.4, 464.4, 464.4, 464.4, 464.4, 464.4), Y = c(418.5, 
                        418.5, 418.5, 418.5, 418.5, 418.5), V = c(0, 0, 0, 0, 0, 0), 
                GD = c(0, 0, 0, 0, 0, 0), ND = c(NA, 0, 0, 0, 0, 0), ND2 = c(NA, 
                        0, 0, 0, 0, 0), TID = c("t1","t1","t1","t2","t2","t2")), .Names = c("A", 
                "T", "X", "Y", "V", "GD", "ND", "ND2", "TID"), row.names = c(NA, 
                6L), class = "data.frame")
dt<-data.table(data)
dt[,head(.SD,2),by=TID]

This results in:

   TID A    T     X     Y V GD ND ND2
1:  t1 1 0.04 464.4 418.5 0  0 NA  NA
2:  t1 1 0.08 464.4 418.5 0  0  0   0
3:  t2 1 0.16 464.4 418.5 0  0  0   0
4:  t2 1 0.20 464.4 418.5 0  0  0   0

and can be changed back to a data frame if desired by changing the last line to

as.data.frame(dt[,head(.SD,2),by=TID])

Dennet

Answer 4

Here is another solution in base:

do.call(rbind, by(NM, NM$TID, head, 80))

Coolish