R - filling a cell with the value from the previous column
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I have a dataframe like this
x <- c(1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0)
y <- c(10:20)
dat <- rbind(x, y)
I don't want the X row to contain 0s. Instead, I would like to replace the 0s with the value from the most recent non-zero column.
The expected output is
1 1 1 1 2 2 2 3 3 3 3
10 11 12 13 14 15 16 17 18 19 20
This is similar to the solution found here, but operating column-wise rather than row-wise.
Thanks!
r
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
add a comment |
I have a dataframe like this
x <- c(1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0)
y <- c(10:20)
dat <- rbind(x, y)
I don't want the X row to contain 0s. Instead, I would like to replace the 0s with the value from the most recent non-zero column.
The expected output is
1 1 1 1 2 2 2 3 3 3 3
10 11 12 13 14 15 16 17 18 19 20
This is similar to the solution found here, but operating column-wise rather than row-wise.
Thanks!
r
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
add a comment |
I have a dataframe like this
x <- c(1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0)
y <- c(10:20)
dat <- rbind(x, y)
I don't want the X row to contain 0s. Instead, I would like to replace the 0s with the value from the most recent non-zero column.
The expected output is
1 1 1 1 2 2 2 3 3 3 3
10 11 12 13 14 15 16 17 18 19 20
This is similar to the solution found here, but operating column-wise rather than row-wise.
Thanks!
r
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
I have a dataframe like this
x <- c(1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 0)
y <- c(10:20)
dat <- rbind(x, y)
I don't want the X row to contain 0s. Instead, I would like to replace the 0s with the value from the most recent non-zero column.
The expected output is
1 1 1 1 2 2 2 3 3 3 3
10 11 12 13 14 15 16 17 18 19 20
This is similar to the solution found here, but operating column-wise rather than row-wise.
Thanks!
r
r
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
asked Nov 26 '18 at 16:46
EmKayDeeEmKayDee
413
413
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Here is an option with tidyverse, where we transpose the data, and use fill on all the column, transpose it back
library(tidyverse)
dat %>%
t %>%
as.data.frame %>%
na_if(., 0) %>%
fill(!!! rlang::syms(names(.))) %>%
t
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
add a comment |
step 1: replace all zeros with NA because, ...
dat[1, dat[1,] == 0] <- NA
step 2: ... you can then use a function that is esp. designed to do what you want with NA-values
dat[1, ] <- zoo::na.locf(unlist(dat[1,]))
# dat[1, ] <- zoo::na.locf(dat[1, ])
result:
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#x 1 1 1 1 2 2 2 3 3 3 3
#y 10 11 12 13 14 15 16 17 18 19 20
Since your example was a matrix and you possibly have a data.frame for your real data. Make sure you ?unlist your data.frame-row into a vector, so zoo::na.locf can function as desired.
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is an option with tidyverse, where we transpose the data, and use fill on all the column, transpose it back
library(tidyverse)
dat %>%
t %>%
as.data.frame %>%
na_if(., 0) %>%
fill(!!! rlang::syms(names(.))) %>%
t
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
add a comment |
Here is an option with tidyverse, where we transpose the data, and use fill on all the column, transpose it back
library(tidyverse)
dat %>%
t %>%
as.data.frame %>%
na_if(., 0) %>%
fill(!!! rlang::syms(names(.))) %>%
t
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
add a comment |
Here is an option with tidyverse, where we transpose the data, and use fill on all the column, transpose it back
library(tidyverse)
dat %>%
t %>%
as.data.frame %>%
na_if(., 0) %>%
fill(!!! rlang::syms(names(.))) %>%
t
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
Here is an option with tidyverse, where we transpose the data, and use fill on all the column, transpose it back
library(tidyverse)
dat %>%
t %>%
as.data.frame %>%
na_if(., 0) %>%
fill(!!! rlang::syms(names(.))) %>%
t
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
answered Nov 26 '18 at 18:12
akrunakrun
420k13207284
420k13207284
add a comment |
add a comment |
step 1: replace all zeros with NA because, ...
dat[1, dat[1,] == 0] <- NA
step 2: ... you can then use a function that is esp. designed to do what you want with NA-values
dat[1, ] <- zoo::na.locf(unlist(dat[1,]))
# dat[1, ] <- zoo::na.locf(dat[1, ])
result:
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#x 1 1 1 1 2 2 2 3 3 3 3
#y 10 11 12 13 14 15 16 17 18 19 20
Since your example was a matrix and you possibly have a data.frame for your real data. Make sure you ?unlist your data.frame-row into a vector, so zoo::na.locf can function as desired.
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
add a comment |
step 1: replace all zeros with NA because, ...
dat[1, dat[1,] == 0] <- NA
step 2: ... you can then use a function that is esp. designed to do what you want with NA-values
dat[1, ] <- zoo::na.locf(unlist(dat[1,]))
# dat[1, ] <- zoo::na.locf(dat[1, ])
result:
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#x 1 1 1 1 2 2 2 3 3 3 3
#y 10 11 12 13 14 15 16 17 18 19 20
Since your example was a matrix and you possibly have a data.frame for your real data. Make sure you ?unlist your data.frame-row into a vector, so zoo::na.locf can function as desired.
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
add a comment |
step 1: replace all zeros with NA because, ...
dat[1, dat[1,] == 0] <- NA
step 2: ... you can then use a function that is esp. designed to do what you want with NA-values
dat[1, ] <- zoo::na.locf(unlist(dat[1,]))
# dat[1, ] <- zoo::na.locf(dat[1, ])
result:
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#x 1 1 1 1 2 2 2 3 3 3 3
#y 10 11 12 13 14 15 16 17 18 19 20
Since your example was a matrix and you possibly have a data.frame for your real data. Make sure you ?unlist your data.frame-row into a vector, so zoo::na.locf can function as desired.
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
step 1: replace all zeros with NA because, ...
dat[1, dat[1,] == 0] <- NA
step 2: ... you can then use a function that is esp. designed to do what you want with NA-values
dat[1, ] <- zoo::na.locf(unlist(dat[1,]))
# dat[1, ] <- zoo::na.locf(dat[1, ])
result:
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
#x 1 1 1 1 2 2 2 3 3 3 3
#y 10 11 12 13 14 15 16 17 18 19 20
Since your example was a matrix and you possibly have a data.frame for your real data. Make sure you ?unlist your data.frame-row into a vector, so zoo::na.locf can function as desired.
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
edited Nov 27 '18 at 8:56
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
answered Nov 26 '18 at 16:58
Andre ElricoAndre Elrico
5,70911230
5,70911230
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
add a comment |
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Thanks! This seems like it's what I want. But when I use it on my actual dataset, I get "Error in x[[jj]][iseq] <- vjj : replacement has length zero", which I took to mean that the thing I'm trying to put in instead of an NA is empty (like I'm trying to sub an empty string in for an NA), and I'm not sure why that's the case since all the cells in that row are either a (non-empty) character string or NA. Any insight?
– EmKayDee
Nov 26 '18 at 17:31
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
Could you add data to your question that is more like your real data and where the solution fails?
– Andre Elrico
Nov 26 '18 at 21:32
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
@EmKayDee I have updated my solution. Please tell me if this now works for you.
– Andre Elrico
Nov 27 '18 at 8:50
add a comment |
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