R 上 Na 值（数据框和变量）的百分比

Question

I would like to calculate percentage of NA -values in a dataframe and for variables.

My dataframe has this:

mean(is.na(dataframe))
# 0.03354

How I read this result? Na 0,033%? I don't understand.

For the individual variables I did the following for the count of NA s

sapply(DATAFRAME, function(x) sum(is.na(x)))

Then, for the percentage of NA -values:

colMeans(is.na(VARIABLEX)) 

Doesn't work because I get the following error:

"x must be an array of at least two dimension"

Why does this error occur? Anyway, afterwards I tried the following:

mean(is.na(VariableX))
# 0.1188

Should I interpret this as having 0.11% NA -values?

Answer 1

I'd just divide the number of rows containing NAs by the total number of rows:

df <- data.frame(data = c(NA, NA, NA, NA, 2, 4, NA, 7, NA))

percent_NA <- NROW(df[is.na(df$data),])/NROW(df)

Which gives:

> percent_NA
[1] 0.6666667

Which means there are 66,67% NAs in my dataframe

Answer 2

I don't understand the issue you are trying to solve. It all works as expected.
First, a dataset since you haven't provided one.

set.seed(6180)  # make it reproducible
dat <- data.frame(x = sample(c(1:4, NA), 100, TRUE),
                  y = sample(c(1:5, NA), 100, TRUE))

Now the code for sums.

s <- sapply(dat, function(x) sum(is.na(x)))
s
# x  y 
#18 13
sum(s)
#[1] 31
sum(is.na(dat))
#[1] 31

colSums(is.na(dat))
# x  y 
#18 13

The same goes for means, be it mean or colMeans .
EDIT.
Here is the code to get the means of NA values per column/variable and a grand total.

sapply(dat, function(x) mean(is.na(x)))
#   x    y 
#0.18 0.13
colMeans(is.na(dat))   # Same result, faster
#   x    y 
#0.18 0.13
mean(is.na(dat))       # overall mean
#[1] 0.155