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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 Apr 11 at 14:08
jakesjakes
481315
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 Apr 11 at 14:08
jakesjakes
481315
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 Apr 11 at 14:08
jakesjakes
481315
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 Apr 11 at 14:08
jakesjakes
481315
asked Apr 11 at 14:08
jakesjakes
481315
asked Apr 11 at 14:08
jakesjakes
481315
asked Apr 11 at 14:08
jakesjakes
481315
asked Apr 11 at 14:08
jakesjakes
481315
481315
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 Apr 11 at 14:24
mgiormentimgiormenti
444211
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
Apr 11 at 14:27
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
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 Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
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 Apr 11 at 14:30
RushabhRushabh
1,332222
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 Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
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 Apr 11 at 14:24
mgiormentimgiormenti
444211
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
Apr 11 at 14:27
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
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 Apr 11 at 14:24
mgiormentimgiormenti
444211
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
Apr 11 at 14:27
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
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 Apr 11 at 14:24
mgiormentimgiormenti
444211
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered Apr 11 at 14:24
mgiormentimgiormenti
444211
answered Apr 11 at 14:24
mgiormentimgiormenti
444211
answered Apr 11 at 14:24
mgiormentimgiormenti
444211
answered Apr 11 at 14:24
mgiormentimgiormenti
444211
444211
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
Apr 11 at 14:27
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
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
Apr 11 at 14:27
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
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
Apr 11 at 14:27
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
Apr 11 at 14:27
3
3
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
Apr 11 at 14:27
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
Apr 11 at 14:27
@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
Apr 11 at 14:28
@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
Apr 11 at 14:28
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 Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
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 Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
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 Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
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 Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
404k1017951041
404k1017951041
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 Apr 11 at 14:30
RushabhRushabh
1,332222
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 Apr 11 at 14:30
RushabhRushabh
1,332222
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 Apr 11 at 14:30
RushabhRushabh
1,332222
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 Apr 11 at 14:30
RushabhRushabh
1,332222
answered Apr 11 at 14:30
RushabhRushabh
1,332222
answered Apr 11 at 14:30
RushabhRushabh
1,332222
answered Apr 11 at 14:30
RushabhRushabh
1,332222
1,332222
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 Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
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 Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
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 Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
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 Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.5k104369
48.5k104369
add a comment |
add a comment |
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