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How to obtain a position of last non-zero element
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!How to trim leading and trailing whitespace?How do I replace NA values with zeros in an R dataframe?data.table vs dplyr: can one do something well the other can't or does poorly?Calculate days since last event in REfficient way of taking the max date within groupsTurn off verbose messages when loading tidyverse using library() functionColumn name of last non-NA row per row; using tidyverse solution?Filtering data relative to first and last occurance of an eventdplyr approach to get the last row number with a positive valueA code to imput missing values with linear dependency
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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
486315
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
486315
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
486315
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
486315
asked Apr 11 at 14:08
jakesjakes
486315
asked Apr 11 at 14:08
jakesjakes
486315
asked Apr 11 at 14:08
jakesjakes
486315
asked Apr 11 at 14:08
jakesjakes
486315
486315
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
405k1017951041
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
1,345322
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
48.7k104370
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
405k1017951041
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
405k1017951041
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
405k1017951041
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
405k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
405k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
405k1017951041
answered Apr 11 at 14:23
Konrad RudolphKonrad Rudolph
405k1017951041
405k1017951041
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
1,345322
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
1,345322
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
1,345322
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
1,345322
answered Apr 11 at 14:30
RushabhRushabh
1,345322
answered Apr 11 at 14:30
RushabhRushabh
1,345322
answered Apr 11 at 14:30
RushabhRushabh
1,345322
1,345322
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
48.7k104370
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
48.7k104370
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
48.7k104370
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
48.7k104370
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.7k104370
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.7k104370
answered Apr 11 at 14:31
Ronak ShahRonak Shah
48.7k104370
48.7k104370
add a comment |
add a comment |
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