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How to obtain a position of last non-zero element
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
The Ask Question Wizard is Live!
Data science time! April 2019 and salary with experienceHow 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 yesterday
jakesjakes
431315
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 yesterday
jakesjakes
431315
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 yesterday
jakesjakes
431315
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 yesterday
jakesjakes
431315
asked yesterday
jakesjakes
431315
asked yesterday
jakesjakes
431315
asked yesterday
jakesjakes
431315
asked yesterday
jakesjakes
431315
431315
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 yesterday
mgiormentimgiormenti
394111
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
yesterday
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
yesterday
@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
yesterday
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 yesterday
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 yesterday
RushabhRushabh
1,300221
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 yesterday
Ronak ShahRonak Shah
46.7k104269
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 yesterday
mgiormentimgiormenti
394111
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
yesterday
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
yesterday
@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
yesterday
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 yesterday
mgiormentimgiormenti
394111
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
yesterday
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
yesterday
@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
yesterday
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 yesterday
mgiormentimgiormenti
394111
Taking advantage of the fact that you have a binary vector, the following gives your desired output:
cummax(seq_along(event) * event)
answered yesterday
mgiormentimgiormenti
394111
answered yesterday
mgiormentimgiormenti
394111
answered yesterday
mgiormentimgiormenti
394111
answered yesterday
mgiormentimgiormenti
394111
394111
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
yesterday
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
yesterday
@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
yesterday
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
yesterday
3
or without multiplicationcummax(ifelse(event, seq_along(event), 0))
– jogo
yesterday
@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
yesterday
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
yesterday
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
yesterday
3
3
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
yesterday
or without multiplication
cummax(ifelse(event, seq_along(event), 0))– jogo
yesterday
@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
yesterday
@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
yesterday
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 yesterday
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 yesterday
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 yesterday
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 yesterday
Konrad RudolphKonrad Rudolph
404k1017921041
answered yesterday
Konrad RudolphKonrad Rudolph
404k1017921041
answered yesterday
Konrad RudolphKonrad Rudolph
404k1017921041
answered yesterday
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 yesterday
RushabhRushabh
1,300221
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 yesterday
RushabhRushabh
1,300221
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 yesterday
RushabhRushabh
1,300221
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 yesterday
RushabhRushabh
1,300221
answered yesterday
RushabhRushabh
1,300221
answered yesterday
RushabhRushabh
1,300221
answered yesterday
RushabhRushabh
1,300221
1,300221
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 yesterday
Ronak ShahRonak Shah
46.7k104269
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 yesterday
Ronak ShahRonak Shah
46.7k104269
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 yesterday
Ronak ShahRonak Shah
46.7k104269
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 yesterday
Ronak ShahRonak Shah
46.7k104269
answered yesterday
Ronak ShahRonak Shah
46.7k104269
answered yesterday
Ronak ShahRonak Shah
46.7k104269
answered yesterday
Ronak ShahRonak Shah
46.7k104269
46.7k104269
add a comment |
add a comment |
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-base, r, tidyverse
