How to find the percentage of NAs in a data.frame?

问题

I am trying to find the percentage of NAs in columns as well as inside the whole dataframe:

The first method which I have commented gives me zero and the second method which is not commented gives me a matrix. Not sure what I am missing. Any hint is truly appreciated!

cp.2006<-read.csv(file="cp2006.csv",head=TRUE)

#countNAs <- function(x) { 
#  sum(is.na(x)) 
#} 
#total=0
#for (i in col(cp.2006)) {
#  total=countNAs(i)+total
#}
#print(total)
count<-apply(cp.2006, 1, function(x) sum(is.na(x)))
dims<-dim(cp.2006)
num<-dims[1]*dims[2]
NApercentage<-(count/num) * 100
print(NApercentage)

回答1:

x = data.frame(x = c(1, 2, NA, 3), y = c(NA, NA, 4, 5))

For the whole dataframe:

sum(is.na(x))/prod(dim(x))

mean(is.na(x))

For columns:

apply(x, 2, function(col)sum(is.na(col))/length(col))

colMeans(is.na(x))

回答2:

You could also use dplyr::summarize_all for the column-wise proportions.

x %>% summarize_all(funs(sum(is.na(.)) / length(.)))

Which will give

     x   y
1 0.25 0.5

回答3:

If you are interested to find percentage of complete cases.

Using Same Example mentioned here.

x = data.frame(x = c(1, 2, NA, 3), y = c(NA, NA, 4, 5))

Output :

   x  y
1  1 NA
2  2 NA
3 NA  4
4  3  5

Finding Complete cases:

complete.cases(x)

Output :

[1] FALSE FALSE FALSE  TRUE

Percentage of complete cases:

mean(complete.cases(x))

Output:

[1] 0.25

That means 25% of complete rows are available in data provided. i.e Only fourth row is complete rest all contains NA values.

Cheers!

回答4:

Updated version of dplyr which doesnt support funs anymore:

x%>% summarise_all(list(name = ~sum(is.na(.))/length(.)))

来源：https://stackoverflow.com/questions/23597140/how-to-find-the-percentage-of-nas-in-a-data-frame

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csv

dataframe