Transform NA values based on first registration and nearest values

Question

I already made a similar question but now I want just to restrict the new values of NA.

I have some data like this:

Date 1   Date 2    Date 3    Date 4    Date 5   Date 6
A  NA       0.1       0.2       NA        0.3    0.2
B  0.1      NA        NA        0.3       0.2    0.1
C  NA       NA        NA        NA        0.3    NA
D  0.1      0.2       0.3       NA        0.1    NA
E  NA       NA        0.1       0.2       0.1    0.3

I would like to change the NA values of my data based on the first date a value is registered. So for example for A, the first registration is Date 2. Then I want that before that registration the values of NA in A are 0, and after the first registration the values of NA become the mean of the nearest values (mean of date 3 and 5).

In case the last value is an NA, transform it into the last registered value (as in C and D). In the case of E all NA values will become 0.

Get something like this:

Date 1   Date 2    Date 3    Date 4    Date 5   Date 6 
A  0       0.1       0.2        0.25      0.3    0.2
B  0.1     0.2       0.2        0.3       0.2    0.1
C  0       0         0          0         0.3    0.3
D  0.1     0.2       0.3        0.2       0.1    0.1
E  0       0         0.1        0.2       0.1    0.3

Can you help me? I'm not sure how to do it in R.

Enginery

Answer 1

Here is a way using na.approx from the zoo package and apply with MARGIN = 1 (so this is probably not very efficient but get's the job done).

library(zoo)
df1 <- as.data.frame(t(apply(dat, 1, na.approx, method = "constant", f = .5, na.rm = FALSE)))

This results in

df1
#   V1  V2  V3   V4  V5
#A  NA 0.1 0.2 0.25 0.3
#B 0.1 0.2 0.2 0.30 0.2
#C  NA  NA  NA   NA 0.3
#E  NA  NA 0.1 0.20 0.1

Replace NAs and rename columns.

df1[is.na(df1)] <- 0
names(df1) <- names(dat)
df1
#  Date_1 Date_2 Date_3 Date_4 Date_5
#A    0.0    0.1    0.2   0.25    0.3
#B    0.1    0.2    0.2   0.30    0.2
#C    0.0    0.0    0.0   0.00    0.3
#E    0.0    0.0    0.1   0.20    0.1

---

explanation

Given a vector

x <- c(0.1, NA, NA, 0.3, 0.2)
na.approx(x)

returns x with linear interpolated values

#[1] 0.1000000 0.1666667 0.2333333 0.3000000 0.2000000

But OP asked for constant values so we need the argument method = "constant" from the approx function.

na.approx(x, method = "constant") 
# [1] 0.1 0.1 0.1 0.3 0.2

But this is still not what OP asked for because it carries the last observation forward while you want the mean for the closest non-NA values. Therefore we need the argument f (also from approx)

na.approx(x, method = "constant", f = .5)
# [1] 0.1 0.2 0.2 0.3 0.2 # looks good

From ?approx

f : for method = "constant" a number between 0 and 1 inclusive, indicating a compromise between left- and right-continuous step functions. If y0 and y1 are the values to the left and right of the point then the value is y0 if f == 0, y1 if f == 1, and y0*(1-f)+y1*f for intermediate values. In this way the result is right-continuous for f == 0 and left-continuous for f == 1, even for non-finite y values.

Lastly, if we don't want to replace the NAs at the beginning and end of each row we need na.rm = FALSE.

From ?na.approx

na.rm : logical. If the result of the (spline) interpolation still results in NAs, should these be removed?

data

dat <- structure(list(Date_1 = c(NA, 0.1, NA, NA), Date_2 = c(0.1, NA, 
NA, NA), Date_3 = c(0.2, NA, NA, 0.1), Date_4 = c(NA, 0.3, NA, 
0.2), Date_5 = c(0.3, 0.2, 0.3, 0.1)), .Names = c("Date_1", "Date_2", 
"Date_3", "Date_4", "Date_5"), class = "data.frame", row.names = c("A", 
"B", "C", "E"))

EDIT

If there are NAs in the last column we can replace these with the last non-NAs before we apply na.approx as shown above.

dat$Date_6[is.na(dat$Date_6)] <- dat[cbind(1:nrow(dat),
                                           max.col(!is.na(dat), ties.method = "last"))][is.na(dat$Date_6)]

Antimagnetic

Nice! That is the route I took but couldn't quite get the na.approx to work. Can you please explain the arguments?(method and f). FYI na.approx is from zoo package, not Base R – Tedious

@Tedious added some explaination. Best – Antimagnetic

Great Thanks! So f = .5 is not continuous? It's just to tell it to take the mean? i.e. We can not specify (f = .6)? – Tedious

@Tedious f could be a number between 0 (default) and 1. 0 means last observation carried forward, 1 would be next observation carried backwards, for any number in between is will result in y0 * (1 - f) + y1 * f, see ?approx – Antimagnetic

Great! I have one question, if the data has a NA in the last column, how do I transform that last NA as the last registered value? – Enginery

Ahh, got it. Thanks. – Tedious

Answer 2

This is another possible answer, using na.locf from the zoo package. Edit: apply is actually not required; This solution fills in the last observed value if this value is missing.

# create the dataframe
Date1 <- c(NA,.1,NA,NA)
Date2 <- c(.1, NA,NA,NA)
Date3 <- c(.2,NA,NA,.1)
Date4 <- c(NA,.3,NA,.2)
Date5 <- c(.3,.2,.3,.1)
Date6 <- c(.1,NA,NA,NA)
df <- as.data.frame(cbind(Date1,Date2,Date3,Date4,Date5,Date6))
rownames(df) <- c('A','B','C','D')

> df
  Date1 Date2 Date3 Date4 Date5 Date6
A    NA   0.1   0.2    NA   0.3   0.1
B   0.1    NA    NA   0.3   0.2    NA
C    NA    NA    NA    NA   0.3    NA
D    NA    NA   0.1   0.2   0.1    NA



# Load library
library(zoo)
df2 <- t(na.locf(t(df),na.rm = F)) # fill last observation carried forward
df3 <- t(na.locf(t(df),na.rm = F, fromLast = T)) # last obs carried backward

df4 <- (df2 + df3)/2 # mean of both dataframes

df4 <- t(na.locf(t(df4),na.rm = F)) # fill last observation carried forward
df4[is.na(df4)] <- 0 # NA values are 0

  Date1 Date2 Date3 Date4 Date5 Date6
A   0.0   0.1   0.2  0.25   0.3   0.1
B   0.1   0.2   0.2  0.30   0.2   0.2
C   0.0   0.0   0.0  0.00   0.3   0.3
D   0.0   0.0   0.1  0.20   0.1   0.1

Pronunciamento

Answer 3

Here's another option with base R + rollmean from zoo (clearly easy to rewrite in base R for this case with window size k = 2).

t(apply(df, 1, function(x) {
  means <- c(0, rollmean(na.omit(x), 2), tail(na.omit(x), 1))
  replace(x, is.na(x), means[1 + cumsum(!is.na(x))[is.na(x)]])
}))
#   Date1 Date2 Date3 Date4 Date5 Date6
# A   0.0   0.1   0.2  0.25   0.3   0.2
# B   0.1   0.2   0.2  0.30   0.2   0.1
# C   0.0   0.0   0.0  0.00   0.3   0.3
# D   0.1   0.2   0.3  0.20   0.1   0.1
# E   0.0   0.0   0.1  0.20   0.1   0.3

---

Explanation. Suppose that x is the first row of df:

#   Date1 Date2 Date3 Date4 Date5 Date6
# A    NA   0.1   0.2    NA   0.3   0.2

Then

means
# [1] 0.00 0.15 0.25 0.25 0.20

is a vector of 0, rolling means of two the following non-NA elements, and the last non-NA element. Then all we need to do is to replace those elements of x that are is.na(x). We will replace them by the elements of means at indices 1 + cumsum(!is.na(x))[is.na(x)]. That's the trickier part. Here

cumsum(!is.na(x))
# [1] 0 1 2 2 3 4

Meaning that the first element of x has seen 0 non-NA elements, while, say, the last one has seen 4 non-NA elements so far. Then

cumsum(!is.na(x))[is.na(x)]
# [1] 0 2

is about those NA elements in x that we want to replace. Notice that then

1 + cumsum(!is.na(x))[is.na(x)]
# [1] 1 3

corresponds to the elements of means that we want to use for replacement.

Placable

Answer 4

I am finding the function below too complicated but it works, so here it goes.

fun <- function(x){
  if(anyNA(x)){
    inx <- which(!is.na(x))
    if(inx[1] > 1) x[seq_len(inx[1] - 1)] <- 0
    prev <- inx[1]
    for(i in inx[-1]){
      if(i - prev > 1){
        m <- mean(c(x[i], x[prev]))
        while(prev < i){
          x[prev] <- m
          prev <- prev + 1
        }
      }
      prev <- i
    }
  }
  x
}

res <- t(apply(df1, 1, fun))
res <- as.data.frame(res)
res
#  Date.1 Date.2 Date.3 Date.4 Date.5
#A    0.0    0.1   0.25   0.25    0.3
#B    0.2    0.2   0.20   0.30    0.2
#C    0.0    0.0   0.00   0.00    0.3
#E    0.0    0.0   0.10   0.20    0.1

Data.

df1 <- read.table(text = "
Date.1   Date.2    Date.3    Date.4    Date.5
A  NA       0.1       0.2       NA        0.3
B  0.1      NA        NA        0.3       0.2
C  NA       NA        NA        NA        0.3
E  NA       NA        0.1       0.2       0.1                  
", header = TRUE)

Apocarpous