将多行追加到数据框

Question

I have a dataframe with timepoints nested in persons (unequal rows per person and missings). For each individual I want to add a new time point with NA's on all variables.

Here is an example of my data:

data_long      <- data.frame(id = factor(rep(1:3,each=4)), DV1 = c(1, 2, NA, 2), DV2 = c(2, 1, 2, 1), time = c(1989, 1995, 2003, 2010))
data_long$DV1       <- c(rnorm(12,0,1))
data_long$DV2       <- c(rnorm(12,0,1))
data_long$DV1[4]    <- NA
data_long$DV2[8]    <- NA
data_long[5,2:3]    <- NA
data_long[12,2:3]   <- NA
data_long       <- data_long[-9,]

T0 <- 1980 # new time point

This is what I want:

for (i in min(as.numeric(data_long$id)):max(as.numeric(data_long$id))){temp <- rbind(c(data_long[data_long$id == i,]$id[1], rep(NA,ncol(data_long[data_long$id == i,])-2), T0), data_long[data_long$id == i,])
write.table(temp, "test.dat", sep="\t", append=T, row.names=F, col.names=FALSE)}

data_long2 <- read.table("test.dat")

However, there must be a simpler way without actually saving the data in order to append differing numbers of rows. I apologize for this simple question and would be happy to be enlightened.

Answer 1

This doesn't exactly match what you share as your desired output, but it does seem to better match what you describe:

Use expand.grid to create a data.frame to merge with your original data.frame . The "id" will be just the existing unique "id" values in your source data.frame , and the "time" value will have the new value appended to it.

## set.seed(1) was used for this
X <- expand.grid(id = unique(data_long$id), 
                 time = c(1980, unique(data_long$time)))
merge(data_long, X, all.y = TRUE)
#    id time        DV1         DV2
# 1   1 1980         NA          NA
# 2   1 1989 -0.6264538 -0.62124058
# 3   1 1995  0.1836433 -2.21469989
# 4   1 2003 -0.8356286  1.12493092
# 5   1 2010         NA -0.04493361
# 6   2 1980         NA          NA
# 7   2 1989         NA          NA
# 8   2 1995 -0.8204684  0.94383621
# 9   2 2003  0.4874291  0.82122120
# 10  2 2010  0.7383247          NA
# 11  3 1980         NA          NA
# 12  3 1989         NA          NA  <---- This row is not there in your approach
# 13  3 1995 -0.3053884  0.78213630
# 14  3 2003  1.5117812  0.07456498
# 15  3 2010         NA          NA

Answer 2

One more approach

newrows <- data.frame(id=unique(data_long$id), DV1=NA, DV2=NA, time=T0)
res <- merge(newrows, data_long, all.x=T, all.y=T)
res <- res[with(res, order(id, time)), ]

The result is:

> res
   id        DV1         DV2 time
5   1         NA          NA 1980
2   1 -0.6264538 -0.62124058 1989
3   1  0.1836433 -2.21469989 1995
1   1 -0.8356286  1.12493092 2003
4   1         NA -0.04493361 2010
9   2         NA          NA 1980
10  2         NA          NA 1989
6   2 -0.8204684  0.94383621 1995
7   2  0.4874291  0.82122120 2003
8   2  0.7383247          NA 2010
13  3         NA          NA 1980
11  3 -0.3053884  0.78213630 1995
12  3  1.5117812  0.07456498 2003
14  3         NA          NA 2010

Hope it helps,

alex