Scaling a numeric matrix in R with values 0 to 1

Question

Here is an excerpt of numeric matrix that I have

 [1,]   30 -33.129487   3894754.1 -39.701738 -38.356477 -34.220534
 [2,]   29 -44.289487  -8217525.9 -44.801         


        

Answer 1

Not the prettiest but this just got the job done, since I needed to do this in a dataframe.

column_zero_one_range_scale  <- function(
input_df,
columns_to_scale #columns in input_df to scale, must be numeric
){

input_df_replace <- input_df

columncount <- length(columns_to_scale)
for(i in 1:columncount){

columnnum <- columns_to_scale[i]

if(class(input_df[,columnnum]) !='numeric' & class(input_df[,columnnum])!='integer')
  {print(paste('Column name ',colnames(input_df)[columnnum],' not an integer or numeric, will skip',sep='')) }


if(class(input_df[,columnnum]) %in% c('numeric','integer'))
{
  vec <- input_df[,columnnum]
  rangevec <- max(vec,na.rm=T)-min(vec,na.rm=T)
  vec1 <- vec - min(vec,na.rm=T)
  vec2 <- vec1/rangevec
}
input_df_replace[,columnnum] <- vec2
colnames(input_df_replace)[columnnum] <- paste(colnames(input_df)[columnnum],'_scaled')

}

return(input_df_replace)


}

Answer 2

And if you were still to use scale:

maxs <- apply(a, 2, max)
mins <- apply(a, 2, min)
scale(a, center = mins, scale = maxs - mins)

Answer 3

Install the clusterSim package and run the following command:

normX = data.Normalization(x,type="n4");

Answer 4

Try the following, which seems simple enough:

## Data to make a minimal reproducible example
m <- matrix(rnorm(9), ncol=3)

## Rescale each column to range between 0 and 1
apply(m, MARGIN = 2, FUN = function(X) (X - min(X))/diff(range(X)))
#           [,1]      [,2]      [,3]
# [1,] 0.0000000 0.0000000 0.5220198
# [2,] 0.6239273 1.0000000 0.0000000
# [3,] 1.0000000 0.9253893 1.0000000

Answer 5

scales package has a function called rescale:

set.seed(2020)

x <- runif(5, 100, 150)
scales::rescale(x)
#1.0000000 0.5053362 0.9443995 0.6671695 0.0000000