Count Trailing and Leading NA for each vector - r

Question

I have some vectors in the following format
v1 <- c(NA,NA,NA,10,10,10,10)
v2 <- c(NA,NA, 3, 3, 3,NA,NA)
v3 <- c( 5, 5, NA,NA,NA,NA,NA)
For each vector I want to calculate how many leading NAs and trailing NAs.
For v1, LeadNA = 3, TrailNA = 0
For v2, LeadNA = 2, TrailNA = 2
For v3, LeadNA = 0, TrailNA = 5

Answer 1

1) Cumsum - An option would be to create a logical vector with cumsum on the presence of non-NA elements and get the sum (base R - No packages used)
f1 <- function(vec, trail = FALSE) {
if(trail) {
vec <- rev(vec)
}
sum(!cumsum(!is.na(vec)))
}
f1(v1)
#[1] 3
f1(v1, TRUE)
#[1] 0
sapply(mget(paste0("v", 1:3)), f1)
# v1 v2 v3
# 3 2 0
sapply(mget(paste0("v", 1:3)), f1, TRUE)
# v1 v2 v3
# 0 2 5
2 rle - Another base R option is rle (No packages are used)
with(rle(is.na(v2)), lengths[values & seq_along(values) %in% c(1, length(values))])

Answer 2

This turns out to be similar to #Bulat's solution
count_nas <- function(x) {
nas <- is.na(x)
if (sum(nas) == length(x)) {
warning('all elements were NA')
return(c(start_na = NA_integer_, end_na = NA_integer_))
}
c(start_na = which.min(nas) - 1,
end_na = which.min(rev(nas)) - 1)
}
count_nas(v1)
#start_na end_na
# 3 0
sapply(list(v1,v2,v3), count_nas)
# [,1] [,2] [,3]
#start_na 3 2 0
#end_na 0 2 5
As far as performance, this is the fastest method with #akrun's methods being in the ballpark.
v_test3 <- sample(10000)
v_test3[c(1:3, 9998:10000)] <- NA_integer_
Unit: microseconds
expr min lq mean median uq max neval
akrun_cumsum 175.7 182.15 193.580 186.55 200.80 354.7 100
akrun_rle 168.6 199.25 210.635 209.25 221.00 289.3 100
g_grothen_zoo 1848.5 1904.45 2008.994 1941.40 2001.35 4799.6 100
g_grothen_reduce 12467.3 12888.10 14174.157 13445.15 15054.35 28241.6 100
www_rleid 5357.2 5439.40 5741.471 5517.15 5947.15 8470.4 100
bulat_and_cole 63.5 66.45 73.681 71.25 75.75 96.9 100
Code for reproducibility:
library(microbenchmark)
library(zoo)
library(data.table)
v_test3 <- sample(10000)
v_test3[c(1:3, 9998:10000)] <- NA_integer_
count_nas <- function(x) {
nas <- is.na(x)
if (sum(nas) == length(x)) {
warning('all elements were NA')
return(c(start_na = NA_integer_, end_na = NA_integer_))
}
c(start_na = which.min(nas) - 1,
end_na = which.min(rev(nas)) - 1)
}
countNA <- function(x) {
len <- function(fromLast = FALSE) length(na.locf(x, fromLast = fromLast))
if (all(is.na(x))) c(left = NA, right = NA)
else length(x) - c(left = len(), right = len(TRUE))
}
f1 <- function(vec, trail = FALSE) {
if(trail) {
vec <- rev(vec)
}
sum(!cumsum(!is.na(vec)))
}
count_fun <- function(x){
y <- rleid(x)
z <- split(x, y)[c(1, length(unique(y)))]
ans <- sapply(z, function(x) sum(is.na(x)))
return(unname(ans))
}
countNA2 <- function(x) {
f <- function(x) sum(Reduce(all, is.na(x), acc = TRUE))
if (all(is.na(x))) c(left = NA, right = NA)
else c(left = f(x), right = f(rev(x)))
}
microbenchmark(
akrun_cumsum = {
f1(v_test3, TRUE)
f1(v_test3, FALSE)
}
,
akrun_rle = {
with(rle(is.na(v_test3)), lengths[values & seq_along(values) %in% c(1, length(values))])
}
,
g_grothen_zoo = {
countNA(v_test3)
}
,
g_grothen_reduce = {
countNA2(v_test3)
}
,
www_rleid = {
count_fun(v_test3)
}
,
bulat_and_cole = {
count_nas(v_test3)
}
)

Answer 3

Wrapper over which.max:
leading.nas <- function(x) {
if (length(x) == 0) {
0L
}
else {
which.min(!is.na(x)) - 1
}
}

Answer 4

A function returns two numbers. The first is the count of leading NA. The second is the count of trailing NA. It requires the rleid function from the data.table package.
library(data.table)
count_fun <- function(x){
y <- rleid(x)
z <- split(x, y)[c(1, length(unique(y)))]
ans <- sapply(z, function(x) sum(is.na(x)))
return(unname(ans))
}
count_fun(v1)
# [1] 3 0
count_fun(v2)
# [1] 2 2
count_fun(v3)
# [1] 0 5

Answer 5

1) na.locf Remove the leading NAs using na.locf and determine the difference in length between the original and reduced vector. Do the same for the trailing NAs. It is not clear what should be returned if the input vector is empty or all NAs so we return NA for both left and right.
library(zoo)
countNA <- function(x) {
len <- function(fromLast = FALSE) length(na.locf(x, fromLast = fromLast))
if (all(is.na(x))) c(left = NA, right = NA)
else length(x) - c(left = len(), right = len(TRUE))
}
countNA(v1)
## left right
## 3 0
countNA(v2)
## left right
## 2 2
countNA(v3)
## left right
## 0 5
It would also be possible to use na.fill to perform this calculation.
2) Reduce A second approach is to use Reduce. It gives the same answer. No packages are used.
countNA2 <- function(x) {
f <- function(x) sum(Reduce(all, is.na(x), acc = TRUE))
if (all(is.na(x))) c(left = NA, right = NA)
else c(left = f(x), right = f(rev(x)))
}