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How to obtain a position of last non-zero element
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How to obtain a position of last non-zero element
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I've got a binary variable representing if event happened or not:
event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)
I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:
last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)
How can I obtain that with base R, tidyverse or any other way?
r tidyverse base
asked 14 hours ago
jakesjakes
409315
add a comment |
I've got a binary variable representing if event happened or not:
event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)
I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:
last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)
How can I obtain that with base R, tidyverse or any other way?
r tidyverse base
asked 14 hours ago
jakesjakes
409315
add a comment |
I've got a binary variable representing if event happened or not:
event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)
I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:
last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)
How can I obtain that with base R, tidyverse or any other way?
r tidyverse base
asked 14 hours ago
jakesjakes
409315
I've got a binary variable representing if event happened or not:
event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)
I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:
last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)
How can I obtain that with base R, tidyverse or any other way?
r tidyverse base
r tidyverse base
asked 14 hours ago
jakesjakes
409315
asked 14 hours ago
jakesjakes
409315
asked 14 hours ago
jakesjakes
409315
asked 14 hours ago
jakesjakes
409315
asked 14 hours ago
jakesjakes
409315
409315
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered 14 hours ago
mgiormentimgiormenti
369111
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
14 hours ago
@jogo That solution makes sense if the type ofeventislogical. It does work even for a numeric vector due to R’s implicit conversions but … eh.
– Konrad Rudolph
14 hours ago
add a comment |
Whenever you need to fill repetitions with a value, think run-length encoding.
In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:
lengths = rle(event == 0)$lengths
nonzeros = which(event != 0)
runs = c(0, rep(nonzeros, each = 2))
result = rep(runs, lengths)
Alternative, substitute the runs in the RLE and then inverse it:
rle = rle(event == 0)
nonzeros = which(event != 0)
rle$values = c(0, rep(nonzeros, each = 2))
result = inverse.rle(rle)
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
add a comment |
You can also do somthing like this-
> zero.locf <- function(x) {
v <- x!=0
c(0, x[v])[cumsum(v)+1]
}
> zero.locf(1:length(event)*event)
[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
RushabhRushabh
1,190221
add a comment |
Another option is to find the index where event == 1 and repeat it based on length.
rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
#[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered 14 hours ago
mgiormentimgiormenti
369111
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
14 hours ago
@jogo That solution makes sense if the type ofeventislogical. It does work even for a numeric vector due to R’s implicit conversions but … eh.
– Konrad Rudolph
14 hours ago
add a comment |
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered 14 hours ago
mgiormentimgiormenti
369111
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
14 hours ago
@jogo That solution makes sense if the type ofeventislogical. It does work even for a numeric vector due to R’s implicit conversions but … eh.
– Konrad Rudolph
14 hours ago
add a comment |
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered 14 hours ago
mgiormentimgiormenti
369111
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered 14 hours ago
mgiormentimgiormenti
369111
answered 14 hours ago
mgiormentimgiormenti
369111
answered 14 hours ago
mgiormentimgiormenti
369111
answered 14 hours ago
mgiormentimgiormenti
369111
369111
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
14 hours ago
@jogo That solution makes sense if the type ofeventislogical. It does work even for a numeric vector due to R’s implicit conversions but … eh.
– Konrad Rudolph
14 hours ago
add a comment |
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
14 hours ago
@jogo That solution makes sense if the type ofeventislogical. It does work even for a numeric vector due to R’s implicit conversions but … eh.
– Konrad Rudolph
14 hours ago
6
6
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.
– Konrad Rudolph
14 hours ago
3
3
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
14 hours ago
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
14 hours ago
@jogo That solution makes sense if the type of
event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.– Konrad Rudolph
14 hours ago
@jogo That solution makes sense if the type of
event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.– Konrad Rudolph
14 hours ago
add a comment |
Whenever you need to fill repetitions with a value, think run-length encoding.
In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:
lengths = rle(event == 0)$lengths
nonzeros = which(event != 0)
runs = c(0, rep(nonzeros, each = 2))
result = rep(runs, lengths)
Alternative, substitute the runs in the RLE and then inverse it:
rle = rle(event == 0)
nonzeros = which(event != 0)
rle$values = c(0, rep(nonzeros, each = 2))
result = inverse.rle(rle)
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
add a comment |
Whenever you need to fill repetitions with a value, think run-length encoding.
In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:
lengths = rle(event == 0)$lengths
nonzeros = which(event != 0)
runs = c(0, rep(nonzeros, each = 2))
result = rep(runs, lengths)
Alternative, substitute the runs in the RLE and then inverse it:
rle = rle(event == 0)
nonzeros = which(event != 0)
rle$values = c(0, rep(nonzeros, each = 2))
result = inverse.rle(rle)
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
add a comment |
Whenever you need to fill repetitions with a value, think run-length encoding.
In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:
lengths = rle(event == 0)$lengths
nonzeros = which(event != 0)
runs = c(0, rep(nonzeros, each = 2))
result = rep(runs, lengths)
Alternative, substitute the runs in the RLE and then inverse it:
rle = rle(event == 0)
nonzeros = which(event != 0)
rle$values = c(0, rep(nonzeros, each = 2))
result = inverse.rle(rle)
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
Whenever you need to fill repetitions with a value, think run-length encoding.
In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:
lengths = rle(event == 0)$lengths
nonzeros = which(event != 0)
runs = c(0, rep(nonzeros, each = 2))
result = rep(runs, lengths)
Alternative, substitute the runs in the RLE and then inverse it:
rle = rle(event == 0)
nonzeros = which(event != 0)
rle$values = c(0, rep(nonzeros, each = 2))
result = inverse.rle(rle)
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
answered 14 hours ago
Konrad RudolphKonrad Rudolph
404k1017921041
404k1017921041
add a comment |
add a comment |
You can also do somthing like this-
> zero.locf <- function(x) {
v <- x!=0
c(0, x[v])[cumsum(v)+1]
}
> zero.locf(1:length(event)*event)
[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
RushabhRushabh
1,190221
add a comment |
You can also do somthing like this-
> zero.locf <- function(x) {
v <- x!=0
c(0, x[v])[cumsum(v)+1]
}
> zero.locf(1:length(event)*event)
[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
RushabhRushabh
1,190221
add a comment |
You can also do somthing like this-
> zero.locf <- function(x) {
v <- x!=0
c(0, x[v])[cumsum(v)+1]
}
> zero.locf(1:length(event)*event)
[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
RushabhRushabh
1,190221
You can also do somthing like this-
> zero.locf <- function(x) {
v <- x!=0
c(0, x[v])[cumsum(v)+1]
}
> zero.locf(1:length(event)*event)
[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
RushabhRushabh
1,190221
answered 14 hours ago
RushabhRushabh
1,190221
answered 14 hours ago
RushabhRushabh
1,190221
answered 14 hours ago
RushabhRushabh
1,190221
1,190221
add a comment |
add a comment |
Another option is to find the index where event == 1 and repeat it based on length.
rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
#[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
add a comment |
Another option is to find the index where event == 1 and repeat it based on length.
rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
#[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
add a comment |
Another option is to find the index where event == 1 and repeat it based on length.
rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
#[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
Another option is to find the index where event == 1 and repeat it based on length.
rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
#[1] 0 0 0 0 5 5 5 5 5 5 5 5 13 13 13 13
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
answered 14 hours ago
Ronak ShahRonak Shah
46.3k104268
46.3k104268
add a comment |
add a comment |
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