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I'd like to fill missing values with a "row distance" to the nearest non-NA value. In other words, how would I convert column x in this sample dataframe into column y?

#    x y
#1   0 0
#2  NA 1
#3   0 0
#4  NA 1
#5  NA 2
#6  NA 1
#7   0 0
#8  NA 1
#9  NA 2
#10 NA 3
#11 NA 2
#12 NA 1
#13  0 0

I can't seem to find the right combination of dplyr group_by and mutate row_number() statements to do the trick. The various imputation packages that I've investigated are designed for more complicated scenarios where imputation is performed using statistics and other variables.

d<-data.frame(x=c(0,NA,0,rep(NA,3),0,rep(NA,5),0),y=c(0,1,0,1,2,1,0,1,2,3,2,1,0))
markus
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Dan Strobridge
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3 Answers3

5

We can use 

d$z = sapply(seq_along(d$x), function(z) min(abs(z - which(!is.na(d$x)))))
#     x y z
# 1   0 0 0
# 2  NA 1 1
# 3   0 0 0
# 4  NA 1 1
# 5  NA 2 2
# 6  NA 1 1
# 7   0 0 0
# 8  NA 1 1
# 9  NA 2 2
# 10 NA 3 3
# 11 NA 2 2
# 12 NA 1 1
# 13  0 0 0

If you want to do this in dplyr, you can just wrap the sapply part in a mutate. 

d %>%
   mutate(z = sapply(seq_along(x), function(z) min(abs(z - which(!is.na(x))))))

or, using also library(purrr) (thanks to @Onyambu):

d %>% mutate(m=map_dbl(1:n(),~min(abs(.x-which(!is.na(x))))))
dww
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  • very useful. however, in my effort to keep my question short and simple, i forgot to mention that i'm a big tidyverse fan and would ideally like something i can use in my dplyr chain. i suspect that i can work with this solution in my chain but i sure wouldn't mind knowing if there's a "tidier" method. – Dan Strobridge Dec 21 '18 at 19:28
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    `d%>%mutate(m = map_dbl(1:n(), ~min(abs(.x - which(!is.na(x))))))` – Onyambu Dec 21 '18 at 19:35
3

Here is a way using data.table

library(data.table)
setDT(d)
d[, out := pmin(cumsum(is.na(x)), rev(cumsum(is.na(x)))), by = rleid(is.na(x))]
d
#     x y out
# 1:  0 0   0
# 2: NA 1   1
# 3:  0 0   0
# 4: NA 1   1
# 5: NA 2   2
# 6: NA 1   1
# 7:  0 0   0
# 8: NA 1   1
# 9: NA 2   2
#10: NA 3   3
#11: NA 2   2
#12: NA 1   1
#13:  0 0   0

For each group of NAs we calculation the parallel minimum of cumsum(is.na(x)) and its reverse. That works because the values in the groups of all non-NAs will be 0. Call setDF(d) if you want to continue with a data.frame.

Instead of calculating cumsum(is.na(x)) twice, we could also do

d[, out := {
  tmp <- cumsum(is.na(x))
  pmin(tmp, rev(tmp))
  }, by = rleid(is.na(x))]

This might give a performance gain, but I didn't test.

Using dplyr syntax this would read

library(dplyr)
d %>% 
  group_by(grp = data.table::rleid(is.na(x))) %>% 
  mutate(out = pmin(cumsum(is.na(x)), rev(cumsum(is.na(x))))) %>% 
  ungroup()
# A tibble: 13 x 4
#       x     y   grp   out
#   <dbl> <dbl> <int> <int>
# 1     0     0     1     0
# 2    NA     1     2     1
# 3     0     0     3     0
# 4    NA     1     4     1
# 5    NA     2     4     2
# 6    NA     1     4     1
# 7     0     0     5     0
# 8    NA     1     6     1
# 9    NA     2     6     2
#10    NA     3     6     3
#11    NA     2     6     2
#12    NA     1     6     1
#13     0     0     7     0

The same idea in base R

rle_x <- rle(is.na(d$x))
grp <- rep(seq_along(rle_x$lengths), times = rle_x$lengths)

transform(d, out = ave(is.na(x), grp, FUN = function(i) pmin(cumsum(i), rev(cumsum(i)))))
markus
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0

Here a solution using vapply

d$y <- 0
d$y[is.na(d$x)] <- vapply(which(diff(cumsum(is.na(d$x))) != 0), 
                          function (k) min(abs(which(diff(cumsum(is.na(d$x))) == 0) - k)), 
                          numeric(1))
d
    x y
1   0 0
2  NA 1
3   0 0
4  NA 1
5  NA 2
6  NA 1
7   0 0
8  NA 1
9  NA 2
10 NA 3
11 NA 2
12 NA 1
13  0 0

with 

d <- structure(list(x = c(0, NA, 0, NA, NA, NA, 0, NA, NA, NA, NA, NA, 0)), 
               class = "data.frame", row.names = c(NA, -13L))
niko
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