I want to add empty rows at specific positions of a dataframe. Let’s say we have this dataframe:
df <- data.frame(var1 = c(1,2,3,4,5,6,7,8,9),
var2 = c(9,8,7,6,5,4,3,2,1))
In which I want to add an empty row after rows 1, 3 and 5 (I know that this is not best practice in most cases, ultimately I want to create a table using flextable here). These row numbers are saved in a vector:
rows <- c(1,3,5)
Now I want to use a for loop that loops through the rows vector to add an empty row after each row using add_row():
for (i in rows) {
df <- add_row(df, .after = i)
}
The problem is, that while the first iteration works flawlessly, the other empty rows get misplaced, since the dataframe gets obviously longer. To fix this I tried adding 1 to the vector after each iteration:
for (i in rows) {
df <- add_row(df, .after = i)
rows <- rows+1
}
Which does not work. I assume the rows vector does only get evaluated once. Anyone got any ideas?
>Solution :
Do it all at once, no need for looping. Make a sequence of row numbers, add the new rows in, sort, then replace the duplicated row numbers with NA:
s <- sort(c(seq_len(nrow(df)), rows))
out <- df[s,]
out[duplicated(s),] <- NA
# var1 var2
#1 1 9
#1.1 NA NA
#2 2 8
#3 3 7
#3.1 NA NA
#4 4 6
#5 5 5
#5.1 NA NA
#6 6 4
#7 7 3
#8 8 2
#9 9 1
This will be much more efficient than looping, for even moderately sized data:
df <- df[rep(1:9,1e4),]
rows <- seq(1,9e4,100)
system.time({
for (i in rev(rows)) {
df <- tibble::add_row(df, .after = i)
}
})
# user system elapsed
# 23.94 0.05 23.98
df <- df[rep(1:9,1e4),]
rows <- seq(1,9e4,100)
system.time({
s <- sort(c(seq_len(nrow(df)), rows))
out <- df[s,]
out[duplicated(s),] <- NA
})
# user system elapsed
# 0.00 0.01 0.02