Divide only one column of nested data.frames by an integer value

Question

I wish to divide the third column of a dataframe by 5. These dataframes are nested and look like this:

[[44]]
    Ethnicity      Variant  Sum
 1:       ASW     ACCEPTOR    1
 2:       ASW          CDS   68
 3:       ASW   CGA_CNVWIN 1000
 4:       ASW     CGA_MIRB    0
 5:       ASW       DELETE    0
 6:       ASW      DISRUPT    0
 7:       ASW        DONOR    0
 8:       ASW   FRAMESHIFT    0
 9:       ASW       INSERT    1
10:       ASW       INTRON   54

I have used three commands each of which is successful but has off-target effects.

lapply(ASWldtSUM,function(x)(x/5))

returns

[[44]]
    Ethnicity Variant   Sum
 1:        NA      NA   0.2
 2:        NA      NA  13.6
 3:        NA      NA 200.0
 4:        NA      NA   0.0
 5:        NA      NA   0.0

which has the unfortunate effect of dividing ALL rows by 5, leading to issues when the class is not integer as in the $Sum column.

lapply(ASWldtSUM,function(x[,3])(x/5))

has the effect of returning only a single vector, which would work nicely if this were not a nested array of dataframes, but the statement

ASWdtSUM$NEWCOL<-lapply(ASWldtSUM,function(x[,3])(x/5))

Cannot simply be written because it is nested.

Using rapply as in the following statement:

rapply(ASWldtSUM,function(x) if (is.integer(x)) {(x/5)})

leads to a disordering of the results.

So, is there a simple way to either append a 4th column to each nested DataFrame, or to replace the third column of each DF (Sum) with that value divided by 5?

Answer 1

It is very simple, if ASWldtSUM is the name of the list containing the data frames, then you can do:

lapply(ASWldtSUM,FUN=function(x) { x[,3]=x[,3]/5; return(x) })

Basically you are replacing the (entire) third column with the division of the (entire) third colum by five.

In practice:

> ASWldtSUM1=data.frame(Ethnicity=rep('ASW',10),Variant=c("ACCEPTOR","CDS","CGA_CNVWIN","CGA_MIRB","DELETE","DISRUPT","DONOR","FRAMESHIFT","INSERT","INTRON"), Sum=c(1,68,1000,0,0,0,0,0,1,54))
> #created a first data.frame (equal to your example)
> ASWldtSUM2=data.frame(Ethnicity=rep('ASW',10),Variant=c("ACCEPTOR","CDS","CGA_CNVWIN","CGA_MIRB","DELETE","DISRUPT","DONOR","FRAMESHIFT","INSERT","INTRON"), Sum=c(1,2,3,4,5,6,7,8,9,10))
> #created a second data.frame (with different values for the third column)
> ASWldtSUM=list(ASWldtSUM1,ASWldtSUM2)
> #created a list of data frames
> lapply(ASWldtSUM,FUN=function(x) { x[,3]=x[,3]/5; return(x) })
> #apply the function to divide third column to each nested data.frame
[[1]]
   Ethnicity    Variant   Sum
1        ASW   ACCEPTOR   0.2
2        ASW        CDS  13.6
3        ASW CGA_CNVWIN 200.0
4        ASW   CGA_MIRB   0.0
5        ASW     DELETE   0.0
6        ASW    DISRUPT   0.0
7        ASW      DONOR   0.0
8        ASW FRAMESHIFT   0.0
9        ASW     INSERT   0.2
10       ASW     INTRON  10.8

[[2]]
   Ethnicity    Variant Sum
1        ASW   ACCEPTOR 0.2
2        ASW        CDS 0.4
3        ASW CGA_CNVWIN 0.6
4        ASW   CGA_MIRB 0.8
5        ASW     DELETE 1.0
6        ASW    DISRUPT 1.2
7        ASW      DONOR 1.4
8        ASW FRAMESHIFT 1.6
9        ASW     INSERT 1.8
10       ASW     INTRON 2.0
> #desired result

Answer 2

There are many ways of doing this. Here is one:

Create some sample data:

dat <- lapply(1:3, function(x)data.frame(a=sample(letters, 4), b=sample(LETTERS, 4), z=rnorm(4)))

dat
[[1]]
  a b          z
1 r M  0.3054329
2 v I -0.8051859
3 t Q -1.6082701
4 u D -0.2315290

[[2]]
  a b          z
1 j W -0.4692469
2 f S  0.3112689
3 a D  0.4208704
4 w Z  0.6903139

[[3]]
....

Next, use a small anonymous function inside lapply(). For better illustration, I multiply by 100 rather than divide by 5:

lapply(dat, function(x){x[3] <- x[3]*100; x})

[[1]]
  a b          z
1 r M   30.54329
2 v I  -80.51859
3 t Q -160.82701
4 u D  -23.15290

[[2]]
  a b         z
1 j W -46.92469
2 f S  31.12689
3 a D  42.08704
4 w Z  69.03139

[[3]]
....