在R中递归地应用lapply

Question

Out of curiosity, I was testing whether recursive lapply gives me the same result as applying the function manually. I found out that lapply behaves erratically. So, here's what I did:

Example 1:

m<-c(2,3,4)
n<-c(5,6,3)
o<-c(1,1,1.5)
dc<-data.frame(m,n,o)

Now, let's analyze the fun part:

lapply(dc,mode)

gives:

lapply(dc,mode)
$m
[1] "numeric"

$n
[1] "numeric"

$o
[1] "numeric"

Let's compare above result running mode individually on say "m".

  mode(dc$m)

I got:

"numeric"

Ditto for others. This is all good because we have got atomic vectors.

Now, let's analyze another example:

Example 2:

a<-c(2,3,4,5,5,3)
b<-c(0,1,1,0,1,0)
b<-factor(b,levels = c(0,1),labels = c("F","M"))
c<-c("Hello","Hi")
datacheck<-data.frame(a,b,c)

Now, I would apply "str" function to a, b and c individually.

str(datacheck$b)
 Factor w/ 2 levels "F","M": 1 2 2 1 2 1
str(datacheck$c)
 Factor w/ 2 levels "Hello","Hi": 1 2 1 2 1 2
str(datacheck$a)
 num [1:6] 2 3 4 5 5 3

This is all good and expected because b and c are factors. "a" is just an array of numbers.

Now, when I run lapply, I get:

 lapply(datacheck,str)
 num [1:6] 2 3 4 5 5 3
 Factor w/ 2 levels "F","M": 1 2 2 1 2 1
 Factor w/ 2 levels "Hello","Hi": 1 2 1 2 1 2
$a
NULL

$b
NULL

$c
NULL

My question is: why are $a, $b and $c NULL and not numeric, what we found when we ran str() command independently? I looked around on SO and also read ?lapply, but I couldn't find an answer.

I'd appreciate your thoughts.

Answer 1

We need to use class

lapply(datacheck, class)

This returns a list , but if we need a vector

sapply(datacheck, class)
#       a         b         c 
#"numeric"  "factor"  "factor" 

---

If we need to get the str as a character output, we can do with capture.output as str just prints the output.

lapply(datacheck, function(x) trimws(capture.output(str(x))))
#$a
#[1] "num [1:6] 2 3 4 5 5 3"

#$b
#[1] "Factor w/ 2 levels \"F\",\"M\": 1 2 2 1 2 1"

#$c
#[1] "Factor w/ 2 levels \"Hello\",\"Hi\": 1 2 1 2 1 2"

By checking the

class(str(datacheck$a))
num [1:6] 2 3 4 5 5 3
#[1] "NULL"

we get a NULL as output which is why the lapply shows NULL

lapply(datacheck, str)

---

By checking the source code of str

 methods(str)
 #[1] str.data.frame* str.Date*       str.default*    str.dendrogram* str.logLik*     str.POSIXt*    

getAnywhere(str.default)
...
...

 cat(ss, sep = "\n") #just prints the output
 return(invisible())
 ...
 ...

Answer 2

The reason why lapply(datacheck,str) returns a list of NULL is explained in help(str) :

Value

str does not return anything, for efficiency reasons. The obvious side effect is output to the terminal.

So, the difference is what you see printed in the console window and what the function actually returns. Using lapply does make this visible.