Replace a sequence in data frame column - r

Question

I have a data frame in R that looks somewhat like this:
A | B
0 0
1 0
0 0
0 0
0 1
0 1
1 0
1 0
1 0
I now want to replace all sequences of more than one "1" in the columns so that only the first "1" is kept and the others are replaced by "0", so that the result looks like this
A | B
0 0
1 0
0 0
0 0
0 1
0 0
1 0
0 0
0 0
I hope you understood what I meant (English is not my mother tongue and especially the R-"vocabulary" is a bit hard for, which is probably why I couldn't find a solution through googling). Thank you in advance!

Answer 1

Try this solution:
Input data
df<-data.frame(
A=c(1,0,0,0,0,0,1,1,1,0),
B=c(1,1,0,1,0,0,1,1,0,0))
f<-function(X)
{
return(as.numeric((diff(c(0,X)))>0))
}
Your output
data.frame(lapply(df,f))
A B
1 1 1
2 0 0
3 0 0
4 0 1
5 0 0
6 0 0
7 1 1
8 0 0
9 0 0
10 0 0

Answer 2

You can use ave and create groups based on the difference of your values to capture the consecutives 1s and 0s as different groups and replace duplicates with 0, i.e.
df[] <- lapply(df, function(i)ave(i, cumsum(c(1, diff(i) != 0)),
FUN = function(i) replace(i, duplicated(i), 0)))
which gives,
A B
1 0 0
2 1 0
3 0 0
4 0 0
5 0 1
6 0 0
7 1 0
8 0 0
9 0 0

Answer 3

Here's a simple one line answer:
> df * rbind(c(0,0), sapply(df, diff))
A B
1 0 0
2 1 0
3 0 0
4 0 0
5 0 1
6 0 0
7 1 0
8 0 0
9 0 0
This takes advantage of the fact that all unwanted 1's in the original data will become 0's with the diff function.

Answer 4

Here is an option with rleid
library(data.table)
df1[] <- lapply(df1, function(x) +(x==1& !ave(x, rleid(x), FUN = duplicated)))
df1
# A B
#1 0 0
#2 1 0
#3 0 0
#4 0 0
#5 0 1
#6 0 0
#7 1 0
#8 0 0
#9 0 0
<

Answer 5

Here's a more functional approach. Though, I find shorter answers here, but it's good to know the possible implementation under the hood:
# helper function
make_zero <- function(val)
{
get_index <- c()
for(i in seq(val))
{
if(val[i] == 1) get_index <- c(get_index, i)
else if (val[i] != 1) get_index <- c()
if(all(diff(get_index)) == 1)
{
val[get_index[-1]] <- 0
}
}
# set values as 0
return (val)
}
df <- sapply(df, make_zero)
head(df)
A B
[1,] 0 0
[2,] 1 0
[3,] 0 0
[4,] 0 0
[5,] 0 1
[6,] 0 0
[7,] 1 0
[8,] 0 0
[9,] 0 0
Explanation:
1. We save the indexes of consecutive 1s in get_index.
2. Next, we check if the difference between indexes is 1.
3. If found, we update the value in the column.