R Standardizing numeric variables in dataframe while retaining factor variables - r

Question

I have a dataframe (dcc) loaded in R which I have narrowed down to complete cases.
str(dcc)
'data.frame': 41715 obs. of 9 variables:
$ XCoord : num 661382 661412 661442 661472 661502 ...
$ YCoord : num 648092 648092 648092 648092 648092 ...
$ OBJECTID : int 1 2 3 4 5 6 7 8 9 10 ...
$ POINTID : int 1 2 3 4 5 6 7 8 9 10 ...
$ GRID_CODE : int 0 0 0 0 0 0 0 0 0 0 ...
$ APPL_COST_DIST_RIV_COAST: num 21350 21674 22185 22748 23448 ...
$ APPL_DEM30 : int 785 793 792 769 765 777 784 789 781 751 ...
$ APPL_DEM30_SLOPE : num 19.7 13.3 18.6 23.2 21 ...
$ APPL_SITE_NONSITE : Factor w/ 2 levels "0","1": 1 1 1 1 1 1 1 1 1 1 ...
I want to standardize the numeric and integer variables by subtracting the mean and dividing by the standard deviation. When I apply the following code, I inadvertently drop the factor variable APPL_SITE_NONSITE from the dataframe:
ind <- sapply(dcc, is.numeric)
dcc.s<-sapply(dcc[,ind], function(x) (x-mean(x))/sd(x))
dcc.s<-data.frame(dcc.s)
If I'm not mistaken, that happens because ind=FALSE for that variable. It seems like I need some combination of a for loop and if/else statement to standardize the numeric variables and leave the factor variable alone. I have tried a number of permutations, but keep getting errors. For example, the following code:
dcc.s <- for (i in 1:ncol(dcc)){ sapply(dcc[,i],
if (is.numeric(dcc[,i])==TRUE) {
function(x) (x-mean(x))/sd(x) }
else {dcc[,i]})
}
returns the error:
Error in match.fun(FUN) :
c("'if (is.numeric(dcc[, i]) == TRUE) {' is not a function, character or symbol", "' function(x) (x - mean(x))/sd(x)' is not a function, character or symbol", "'} else {' is not a function, character or symbol", "' dcc[, i]' is not a function, character or symbol", "'}' is not a function, character or symbol")
Perhaps this is a simple formatting error or misplaced bracket, but I'm thoroughly stuck. I am open to other approaches if there is an more elegant way to do this. Any help would be much appreciated.

Answer 1

You need to use rapply instead of sapply
set.seed(1)
> df=data.frame(A=rnorm(10),b=1:10,C=as.factor(rep(1:2,5)))
> str(df)
'data.frame': 10 obs. of 3 variables:
$ A: num -0.626 0.184 -0.836 1.595 0.33 ...
$ b: int 1 2 3 4 5 6 7 8 9 10
$ C: Factor w/ 2 levels "1","2": 1 2 1 2 1 2 1 2 1 2
The code you need to use:
> D=rapply(df,scale,c("numeric","integer"),how="replace")
> D
A b C
1 -0.97190653 -1.4863011 1
2 0.06589991 -1.1560120 2
3 -1.23987805 -0.8257228 1
4 1.87433300 -0.4954337 2
5 0.25276523 -0.1651446 1
6 -1.22045645 0.1651446 2
7 0.45507643 0.4954337 1
8 0.77649606 0.8257228 2
9 0.56826358 1.1560120 1
10 -0.56059319 1.4863011 2
> str(D)
'data.frame': 10 obs. of 3 variables:
$ A: num [1:10, 1] -0.9719 0.0659 -1.2399 1.8743 0.2528 ...
..- attr(*, "scaled:center")= num 0.132
..- attr(*, "scaled:scale")= num 0.781
$ b: num [1:10, 1] -1.486 -1.156 -0.826 -0.495 -0.165 ...
..- attr(*, "scaled:center")= num 5.5
..- attr(*, "scaled:scale")= num 3.03
$ C: Factor w/ 2 levels "1","2": 1 2 1 2 1 2 1 2 1 2
>

Answer 2

Here is a dplyr and scale solution.
For dplyr < 1.0.0
require(dplyr)
df %>% mutate_if(is.numeric, scale)
# a runif(20) rnorm(20)
#1 y 0.5783877 -0.004177104
#2 n -0.2344854 -0.866626472
#3 m 1.5629961 1.526857969
#4 h 0.9648646 -1.557975547
#5 u -0.7212756 0.533400304
#6 u 1.4753675 -0.072289864
#7 b 0.5346870 -0.464299111
#8 l -0.4287559 0.426600473
#9 m -1.2050841 -0.880135405
#10 h -0.6150410 -0.040636433
#11 r 1.3768249 -0.719785950
#12 a -1.3929511 0.083010969
#13 a -0.4422665 0.385574213
#14 l -0.7719473 -0.934716525
#15 m 1.4483803 0.131974911
#16 k 0.6291919 2.598581195
#17 k -1.0356817 -1.018890381
#18 s -1.0960083 1.560216350
#19 y -0.8826702 -0.367821579
#20 v 0.2554671 -0.318862011
For dplyr >= 1.0.0
df %>% mutate(across(where(is.numeric), scale))
Note that scale(x) will do the same as (x - mean(x)) / sd(x); if you want to scale based on different metrics (e.g. a robust/modified Z score based on the median and MAD) you can do that using sweep.
Sample data
set.seed(2017);
df <- cbind.data.frame(a = factor(sample(letters, 20, replace = T)), runif(20), rnorm(20));

Answer 3

ind <- sapply(dcc, is.numeric)
dcc.s <- as.data.frame(lapply(dcc[,ind], function(x) (x-mean(x))/sd(x)))
dcc.s <- cbind(dcc, dcc.s)
If you don't need the "old" dataframe you can also do
ind <- sapply(dcc, is.numeric)
dcc[,ind] <- vapply(dcc[,ind], function(x) (x-mean(x))/sd(x))