Add empty rows at specific positions of dataframe

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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?

dataframe r
2 Answers

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 or loop-like code, for even moderately sized data:

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 
#   0.01    0.00    0.02 
 
df <- df[rep(1:9,1e4),]
rows <- seq(1,9e4,100)
 
system.time({
Reduce(function(x, y) tibble::add_row(x, .after = y), rev(rows), init = df)
})
#   user  system elapsed 
#  26.03    0.00   26.03 
 
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 
#  25.05    0.00   25.04
user_4968030

You could achieve your result by looping in the reverse direction:

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)
)
rows <- c(1, 3, 5)

for (i in rev(rows)) {
  df <- tibble::add_row(df, .after = i)
}
df
#>    var1 var2
#> 1     1    9
#> 2    NA   NA
#> 3     2    8
#> 4     3    7
#> 5    NA   NA
#> 6     4    6
#> 7     5    5
#> 8    NA   NA
#> 9     6    4
#> 10    7    3
#> 11    8    2
#> 12    9    1
user_129938610
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