Average neighbours inside a vector

Question

My data :

data <- c(1,5,11,15,24,31,32,65)

There are 2 neighbours: 31 and 32. I wish to remove them and keep only the mean value (e.g. 31.5), in such a way data would be :

data <- c(1,5,11,15,24,31.5,65)

It seems simple, but I wish to do it automatically, and sometimes with vectors containing more neighbours. For instance :

data_2 <- c(1,5,11,15,24,31,32,65,99,100,101,140)

Answer 1

Here is another idea that creates an id via cumsum(c(TRUE, diff(a) > 1)), where 1 shows the gap threshold, i.e.

#our group variable
grp <- cumsum(c(TRUE, diff(a) > 1))

#keep only groups with length 1 (i.e. with no neighbor)
i1 <- a[!!!ave(a, grp, FUN = function(i) length(i) > 1)] 

#Find the mean of the groups with more than 1 rows,
i2 <- unname(tapply(a, grp, function(i)mean(i[length(i) > 1])))

#Concatenate the above 2 (eliminating NAs from i2) to get final result
c(i1, i2[!is.na(i2)])
#[1]  1.0  5.0 11.0 15.0 24.0 65.0 31.5

You can also wrap it in a function. I left the gap as a parameter so you can adjust,

get_vec <- function(x, gap) {
    grp <- cumsum(c(TRUE, diff(x) > gap))
    i1 <- x[!!!ave(x, grp, FUN = function(i) length(i) > 1)]
    i2 <- unname(tapply(x, grp, function(i) mean(i[length(i) > 1])))
    return(c(i1, i2[!is.na(i2)]))
}

get_vec(a, 1)
#[1]  1.0  5.0 11.0 15.0 24.0 65.0 31.5

get_vec(a_2, 1)
#[1]   1.0   5.0  11.0  15.0  24.0  65.0 140.0  31.5 100.0

DATA:

a <- c(1,5,11,15,24,31,32,65)
a_2 <- c(1, 5, 11, 15, 24, 31, 32, 65, 99, 100, 101, 140)

Answer 2

Here is my solution, which uses run-length encoding to identify groups:

foo <- function(x) {
  y <- x - seq_along(x) #normalize to zero differences in groups
  ind <- rle(y) #run-length encoding
  ind$values <- ind$lengths != 1 #to find groups
  ind$values[ind$values] <- cumsum(ind$values[ind$values]) #group ids
  ind <- inverse.rle(ind)
  xnew <- x
  xnew[ind != 0] <- ave(x, ind, FUN = mean)[ind != 0] #calculate means
  xnew[!(duplicated(ind) & ind != 0)] #remove duplicates from groups
}

foo(data)
#[1]  1.0  5.0 11.0 15.0 24.0 31.5 65.0
foo(data_2)
#[1]   1.0   5.0  11.0  15.0  24.0  31.5  65.0 100.0 140.0
data_3 <- c(1, 2, 4, 1, 2)
foo(data_3)
#[1] 1.5 4.0 1.5

I assume that you don't need an extremely efficient solution. If you do, I'd recommend a simple C++ for loop in Rcpp.

Answer 3

I have a data.table based solution, same could be translated into dplyr I guess:

library(data.table)
df <- data.table(data2 = c(1,5,11,15,24,31,32,65,99,100,101,140))
df[,neighbours := ifelse(c(0,diff(data_2)) == 1,1,0)]
df[,neighbours := c(neighbours[1:(.N-1)],1),by = rleid(neighbours)]
df[,neigh_seq := rleid(neighbours)]

unique(df[,ifelse(neighbours == 1,mean(data2),data2),by = neigh_seq])

   neigh_seq    V1
1:         1   1.0
2:         1   5.0
3:         1  11.0
4:         1  15.0
5:         1  24.0
6:         2  31.5
7:         3  65.0
8:         4 100.0
9:         5 140.0

What it does : first line set neigbours to 1 if the difference with following number is 1

 1:     1          0
 2:     5          0
 3:    11          0
 4:    15          0
 5:    24          0
 6:    31          0
 7:    32          1
 8:    65          0
 9:    99          0
10:   100          1
11:   101          1
12:   140          0

I wanr to group so that neighbour variable is 1 for all neigbours. I need to add 1 to each end of each groups:

df[,neighbours := c(neighbours[1:(.N-1)],1),by = rleid(neighbours)]
    data2 neighbours
 1:     1          0
 2:     5          0
 3:    11          0
 4:    15          0
 5:    24          0
 6:    31          1
 7:    32          1
 8:    65          0
 9:    99          1
10:   100          1
11:   101          1
12:   140          0

Then after I just do a grouping on changing neighbour value, and set the value to mean if they are neihbours

df[,ifelse(neighbours == 1,mean(data2),data2),by = rleid(neighbours)]
    rleid    V1
 1:     1   1.0
 2:     1   5.0
 3:     1  11.0
 4:     1  15.0
 5:     1  24.0
 6:     2  31.5
 7:     2  31.5
 8:     3  65.0
 9:     4 100.0
10:     4 100.0
11:     4 100.0
12:     5 140.0

and take the unique values. And voila.