The difference of na.rm and na.omit in R

Question

I've just started with R and I've executed these statements:

library(datasets)
head(airquality)
s <- split(airquality,airquality$Month)
sapply(s, function(x) {colMeans(x[,c("Ozone", "Solar.R", "Wind")], na.rm = TRUE)})
lapply(s, function(x) {colMeans(na.omit(x[,c("Ozone", "Solar.R", "Wind")])) }) 

For the sapply, it returns the following:

             5         6          7          8         9
Ozone    23.61538  29.44444  59.115385  59.961538  31.44828
Solar.R 181.29630 190.16667 216.483871 171.857143 167.43333
Wind     11.62258  10.26667   8.941935   8.793548  10.18000

And for lapply, it returns the following:

$`5`
    Ozone   Solar.R      Wind 
 24.12500 182.04167  11.50417 

$`6`
    Ozone   Solar.R      Wind 
 29.44444 184.22222  12.17778 

$`7`
     Ozone    Solar.R       Wind 
 59.115385 216.423077   8.523077 

$`8`
    Ozone   Solar.R      Wind 
 60.00000 173.08696   8.86087 

$`9`
    Ozone   Solar.R      Wind 
 31.44828 168.20690  10.07586 

Now, my question would be, why are the returned values similar, but not the same? Isn't na.rm = TRUE and na.omit supposed to be doing the exact same thing? Omit the missing values and calculate the mean only for the values that we have? And in that case, shouldn't I have had the same values in both result sets?

Thank you so much for any input!

Answer 1

They are not supposed to give the same result. Consider this example:

exdf<-data.frame(a=c(1,NA,5),b=c(3,2,2))
#   a b
#1  1 3
#2 NA 2
#3  5 2
colMeans(exdf,na.rm=TRUE)
#       a        b 
#3.000000 2.333333
colMeans(na.omit(exdf))
#  a   b 
#3.0 2.5

Why is this? In the first case, the mean of column b is calculated through (3+2+2)/3. In the second case, the second row is removed in its entirety (also the value of b which is not-NA and therefore considered in the first case) by na.omit and so the b mean is just (3+2)/2.