在R中使用is.na和Sapply函数

Question

Can anyone tell me what the line of code written below do?

sapply(X, function(x) sum(is.na(x))) / nrow(airports) * 100

What is understood is that it will drop NA s when it applies the sum function but keeps them in the matrix.

Any help is appreciated.

Thank you

Answer 1

Enough comments, time for an answer:

sapply(X,      # apply to each item of X (each column, if X is a data frame)
  function(x)  # this function:
    sum(is.na(x))  # count the NAs
) / nrow(airports) * 100  # then divide the result by the number of rows in the the airports object
  # and multiply by 100

In words, it counts the number of missing values in each column of X , then divides the result by the number of rows in airports and multiplies by 100. Calculating the percentage of missing values in each column, assuming X has the same number of rows as airports .

It's strange to mix and match the columns of X with the nrow(airports) , I would expect those to be the same (that is, either sapply(airports, ...) / nrow(airports) or sapply(X, ...) / nrow(X) .

As I mentioned in comments, nothing is being "dropped". If you wanted to do a sum ignoring the NA values, you do sum(foo, na.rm = TRUE) . Instead, here, *what is being summed is is.na(x) , that is we are summing whether or not each value is missing: counting missing values. sum(is.na(foo)) is the idiomatic way to count the number of NA values in foo .

In this case, where the goal is a percent not a count, we can simplify by using mean() instead of sum() / n :

# slightly simpler, consistent object
sapply(airports, function(x) mean(is.na(x))) * 100

We could also use is.na() on the entire data so we don't need the "anonymous function":

# rearrange for more simplicity
sapply(is.na(airports), mean) * 100