Insert a blank row before zero - r

Question

x<-c(0,1,1,0,1,1,1,0,1,1)
aaa<-data.frame(x)
How to insert a blank row before zero? When the first row is zero，do not add blank row. Thank you.
Result:
0
1
1
.
0
1
1
1
.
0
1
1

Answer 1

Below we used dot but you can replace "." with NA or "" or something else depending on what you want.
1) We can use Reduce and append:
Append <- function(x, y) append(x, ".", y - 1)
data.frame(x = Reduce(Append, setdiff(rev(which(aaa$x == 0)), 1), init = aaa$x))
2) gsub Another possibility is to convert to a character string, use gsub and convert back:
data.frame(x = strsplit(gsub("(.)0", "\\1.0", paste(aaa$x, collapse = "")), "")[[1]])
3) We can create a two row matrix in which the first row is dot before each 0 and NA otherwise. Then unravel it to a vector and use na.omit to remove the NA values.
data.frame(x = na.omit(c(rbind(replace(ifelse(aaa$x == 0, ".", NA), 1, NA), aaa$x))))
4) We can lapply over aaa$x[-1] outputting c(".", 9) or 1. Unlist that and insert aaa$x[1] back in. No packages are used.
repl <- function(x) if (!x) c(".", 0) else 1
data.frame(x = c(aaa$x[1], unlist(lapply(aaa$x[-1], repl))))
5) Create a list of all but the first element and replace the 0's in that list with c(".", 0) . Unlist that and insert the first element back in. No packages are used.
L <- as.list(aaa$x[-1])
L[x[-1] == 0] <- list(c(".", 0))
data.frame(x = c(aaa$x[1], unlist(L)))
6) Assuming aaa has two columns where the second column is character (NOT factor). Append a row of dots to aaa and then create an index vector using unlist and Map to access the appropriate row of the extended aaa.
aaa <- data.frame(x = c(0,1,1,0,1,1,1,0,1,1), y = letters[1:10],
stringsAsFactors = FALSE)
nr <- nrow(aaa); nc <- ncol(aaa)
fun <- function(ix, x) if (!is.na(x) & x == 0 & ix > 1) c(nr + 1, ix) else ix
rbind(aaa, rep(".", nc))[unlist(Map(fun, 1:nr, aaa$x)), ]
If we did want to have y be factor then note that we can't just add a dot to a factor if it is not a level of that factor so there is the question of what levels the factor can have. To get around that let us add an NA rather than a dot to the factor. Then we get the following which is the same except that aaa has been redefined so that y is a factor, we no longer need nc since we are assuming 2 columns and rep(...) in the last line is replaced with c(".", NA).
aaa <- data.frame(x = c(0,1,1,0,1,1,1,0,1,1), y = letters[1:10])
nr <- nrow(aaa)
fun <- function(ix, x) if (!is.na(x) & x == 0 & ix > 1) c(nr + 1, ix) else ix
rbind(aaa, c(".", NA))[unlist(Map(fun, 1:nr, aaa$x)), ]

Answer 2

One dplyr and tidyr possibility may be:
aaa %>%
uncount(ifelse(row_number() > 1 & x == 0, 2, 1)) %>%
mutate(x = ifelse(x == 0 & lag(x == 1, default = first(x)), NA_integer_, x))
x
1 0
2 1
3 1
4 NA
5 0
6 1
7 1
8 1
9 NA
10 0
11 1
12 1
It is not adding a blank row as you have a numeric vector. Instead, it is adding a row with NA. If you need a blank row, you can convert it into a character vector and then replace NA with blank.

Answer 3

ind = with(aaa, ifelse(x == 0 & seq_along(x) > 1, 2, 1))
d = aaa[rep(1:NROW(aaa), ind), , drop = FALSE]
transform(d, x = replace(x, sequence(ind) == 2, NA))

Answer 4

Here is an option with rleid
library(data.table)
setDT(aaa)[, .(x = if(x[.N] == 1) c(x, NA) else x), rleid(x)][-.N, .(x)]
# x
# 1: 0
# 2: 1
# 3: 1
# 4: NA
# 5: 0
# 6: 1
# 7: 1
# 8: 1
# 9: NA
#10: 0
#11: 1
#12: 1

Answer 5

data.frame(x = unname(unlist(by(aaa$x,cumsum(aaa==0),c,'.'))))
x
1 0
2 1
3 1
4 .
5 0
6 1
7 1
8 1
9 .
10 0
11 1
12 1
13 .

Answer 6

My solution is
aaa <- data.frame(x = c(0,1,1,0,1,1,1,0,1,1), y = letters[1:10])
aaa$ind = with(aaa, ifelse(x == 0 & seq_along(x) > 1, 2, 1))
aaa<-aaa[rep(1:nrow(aaa), aaa$ind), ,]
aaa[(aaa$ind== 2 & !grepl(".1",rownames(aaa))),]<-NA
aaa$ind<- NULL
aaa
x y
1 0 a
2 1 b
3 1 c
4 NA <NA>
4.1 0 d
5 1 e
6 1 f
7 1 g
8 NA <NA>
8.1 0 h
9 1 i
10 1 j