CODEWITHSUNDEEP
rfunctionstandardized
Addison
Addison

Reputation: 143

Create r function that standardizes multiple variables and creates new columns

I have a dataset in which I work with mean-centered and standardized versions of many of the variables. In my r code I have a large list of the scale() functions that I run for all of the variables but I am wondering if there is a way to write a simple function that will optimize this process.

For example: instead of having a huge list like this...

df$Z.ROW1 <- scale(df$ROW1, scale=T)
df$Z.ROW2 <- scale(df$ROW2, scale=T)
df$Z.ROW3 <- scale(df$ROW3, scale=T)
.....

Is there a way to write a function that will create new vectors and append them to the end of the data frame based on the variables I specify to be standardized?

I found this example online: 

set.seed(212)
df = matrix(rnorm(15), 5, 5))
colnames(df) <- c("ROW1", "ROW2", "ROW3", "ROW4", "ROW5")
df

          ROW1       ROW2       ROW3       ROW4       ROW5
[1,] -0.2391731  0.1544909  0.1503488 -0.2391731  0.1544909
[2,]  0.6769356  1.0368712  0.5096765  0.6769356  1.0368712
[3,] -2.4403360 -0.7796077 -0.7733148 -2.4403360 -0.7796077
[4,]  1.2408845  0.6212641  1.8756660  1.2408845  0.6212641
[5,] -0.3265144  0.2994313  0.7883057 -0.3265144  0.2994313

center.scale <- function(z) {
  scale(z, scale = T)
}
center.scale(df[,c("ROW1", "ROW2")])

         ROW1      ROW2
[1,] -0.01534097 -0.1657064
[2,]  0.63734894  1.1398052
[3,] -1.58357932 -1.5477370
[4,]  1.03913941  0.5249004
[5,] -0.07756806  0.0487378

Which gets close but it doesn't solve the issue of creating new vectors and appending them to the end of my existing dataset. Ideally, I would like it so that the only thing I need to change is the variable names in the center.scale() function. Thanks!

Upvotes: 3

Views: 1107

Answers (3)

Ista
Ista

Reputation: 10437

Here is a version that doesn't hard code scale options, and lets you select a subset of the original columns. It returns a data.frame since that will be useful in more situations, though you can easily modify it to return a matrix if you wish.

add_scaled <- function(data, vars = colnames(data), ...) {
    data.frame(data,
               setNames(data.frame(scale(data[, vars, drop = FALSE],
                                         ...)),
                        paste("Z", vars, sep = ".")))
}

By default it returns a data.frame with all columns standardized and appended.

df = matrix(rnorm(15), 5, 3)
colnames(df) <- paste0("Col", 1:ncol(df))
df
##            Col1       Col2       Col3
## [1,]  1.9659082 -1.2254071  0.1477912
## [2,]  0.2666273 -0.9123931  1.4747579
## [3,]  1.0813351  2.4138457 -1.5569830
## [4,]  0.9618084  1.3076966 -0.8646893
## [5,] -2.0246095  0.3043559 -1.3617747

add_scaled(df)
##         Col1       Col2       Col3     Z.Col1      Z.Col2     Z.Col3
## 1  1.9659082 -1.2254071  0.1477912  1.0040228 -1.05411792  0.4625295
## 2  0.2666273 -0.9123931  1.4747579 -0.1216110 -0.84828629  1.5207917
## 3  1.0813351  2.4138457 -1.5569830  0.4180659  1.33898111 -0.8970361
## 4  0.9618084  1.3076966 -0.8646893  0.3388893  0.61159985 -0.3449285
## 5 -2.0246095  0.3043559 -1.3617747 -1.6393669 -0.04817676 -0.7413566

If only some columns should be standardize you may select them.

add_scaled(df, vars = c("Col1", "Col3"))
##         Col1       Col2       Col3     Z.Col1     Z.Col3
## 1  1.9659082 -1.2254071  0.1477912  1.0040228  0.4625295
## 2  0.2666273 -0.9123931  1.4747579 -0.1216110  1.5207917
## 3  1.0813351  2.4138457 -1.5569830  0.4180659 -0.8970361
## 4  0.9618084  1.3076966 -0.8646893  0.3388893 -0.3449285
## 5 -2.0246095  0.3043559 -1.3617747 -1.6393669 -0.7413566

Finally, you can pass arguments to scale, so you don't lose any flexibility.

add_scaled(df, vars = "Col1", center = FALSE, scale = TRUE)
##         Col1       Col2       Col3     Z.Col1
## 1  1.9659082 -1.2254071  0.1477912  1.2353890
## 2  0.2666273 -0.9123931  1.4747579  0.1675502
## 3  1.0813351  2.4138457 -1.5569830  0.6795177
## 4  0.9618084  1.3076966 -0.8646893  0.6044064
## 5 -2.0246095  0.3043559 -1.3617747 -1.2722773

add_scaled(df, vars = "Col1", center = TRUE, scale = FALSE)
##         Col1       Col2       Col3     Z.Col1
## 1  1.9659082 -1.2254071  0.1477912  1.5156943
## 2  0.2666273 -0.9123931  1.4747579 -0.1835866
## 3  1.0813351  2.4138457 -1.5569830  0.6311212
## 4  0.9618084  1.3076966 -0.8646893  0.5115945
## 5 -2.0246095  0.3043559 -1.3617747 -2.4748234

Upvotes: 3

C. Braun
C. Braun

Reputation: 5201

If I understood your question correctly, you can cbind the output of scale to the original data, as @Dason suggests.

Example:

> df <- data.frame(ROW1 = c(1,2,1,1), ROW2 = c(1,2,3,4), ROW3 = c(5,8,6,5))
> df
  ROW1 ROW2 ROW3
1    1    1    5
2    2    2    8
3    1    3    6
4    1    4    5
> df <- cbind(df, scale(df, scale = T))
> names(df)[4:6] <- paste0('Z.', names(df)[4:6])
> df
  ROW1 ROW2 ROW3 Z.ROW1     Z.ROW2     Z.ROW3
1    1    1    5   -0.5 -1.1618950 -0.7071068
2    2    2    8    1.5 -0.3872983  1.4142136
3    1    3    6   -0.5  0.3872983  0.0000000
4    1    4    5   -0.5  1.1618950 -0.7071068

Upvotes: 1

Jake Kaupp
Jake Kaupp

Reputation: 8072

Like @Dason said, you just need to modify your function to cbind in your original data, and name the new columns accordingly. 

center.scale <- function(z) {

   x <- scale(z, scale = T)

   colnames(x) <- paste0("scale_", colnames(x))

   cbind(z, x)

}

center.scale(df[,c("ROW1", "ROW2")])

Results in:

          ROW1       ROW2  scale_ROW1 scale_ROW2
[1,] -0.2391731  0.1544909 -0.01534097 -0.1657064
[2,]  0.6769356  1.0368712  0.63734894  1.1398052
[3,] -2.4403360 -0.7796077 -1.58357932 -1.5477370
[4,]  1.2408845  0.6212641  1.03913941  0.5249004
[5,] -0.3265144  0.2994313 -0.07756806  0.0487378

Upvotes: 2

Related Questions