Question

I am using by to apply a function to a range columns of a data frame based on a factor. Everything works perfectly well if I use mean() as the function but if I use median() I get an error of the type “Error in median.default(x) : need numeric data” even if I don’t have NAs in the data frame.

The line that works using mean():

by(iris[,1:3], iris$Species, function(x) mean(x,na.rm=T))

> by(iris[,1:3], iris$Species, function(x) mean(x,na.rm=T))
iris$Species: setosa
Sepal.Length  Sepal.Width Petal.Length 
       5.006        3.428        1.462 
------------------------------------------------------------ 
iris$Species: versicolor
Sepal.Length  Sepal.Width Petal.Length 
       5.936        2.770        4.260 
------------------------------------------------------------ 
iris$Species: virginica
Sepal.Length  Sepal.Width Petal.Length 
       6.588        2.974        5.552 
Warning messages:
1: mean(<data.frame>) is deprecated.
 Use colMeans() or sapply(*, mean) instead. 
2: mean(<data.frame>) is deprecated.
 Use colMeans() or sapply(*, mean) instead. 
3: mean(<data.frame>) is deprecated.
 Use colMeans() or sapply(*, mean) instead. 

But if I use median() (note the na.rm=T option):

> by(iris[,1:3], iris$Species, function(x) median(x,na.rm=T))
Error in median.default(x, na.rm = T) : need numeric data

However if instead of choosing the range [,1:3] of columns I choose only one of the columns it works:

> by(iris[,1], iris$Species, function(x) median(x,na.rm=T))
iris$Species: setosa
[1] 5
------------------------------------------------------------ 
iris$Species: versicolor
[1] 5.9
------------------------------------------------------------ 
iris$Species: virginica
[1] 6.5

How can I achieve this behaviour while selecting a range of columns?

Answer 1

You are using a split-apply strategy when you use by. The objects being passed to the function are dataframes and you are getting the warning and error because of the non-existence of median.data.frame and the impending non-existence of mean.data.frame. It might work better if you used aggregate:

> aggregate(iris[,1:3], iris["Species"], function(x) mean(x,na.rm=T))
     Species Sepal.Length Sepal.Width Petal.Length
1     setosa        5.006       3.428        1.462
2 versicolor        5.936       2.770        4.260
3  virginica        6.588       2.974        5.552
> aggregate(iris[,1:3], iris["Species"], function(x) median(x,na.rm=T))
     Species Sepal.Length Sepal.Width Petal.Length
1     setosa          5.0         3.4         1.50
2 versicolor          5.9         2.8         4.35
3  virginica          6.5         3.0         5.55

aggregate works on the column vectors individually and then tabulates the results.