Scaling data in R gives spurious Error “length of

Question

I am trying to scale a data.frame in the range of 0 and 1 using the following code:

for(i in 1:nrow(data))
{
  x <- data[i, ]
  data[i, ] <- scale(x, min(x), max(x)-min(x))
}

Data:
 x1   x2  x3  x4  x5  x6  x7  x8  x9  x10  x11  x12  x13  x14  x15  x16  x17 
 15   6   6   0   9   3   1   4   5    1    1   13    0    0   20    5   28
 2  24  14   7   0  15   7   0  11   3    3    4   15    7    0   30    0  344
 3  10   5   2   0   6   2   0   5   0    0    2    7    1    0   11    0  399
 4   9   4   2   0   5   2   0   4   0    0    2    6    1    0   10    0   28
 5   6   2   1   0   3   1   0   2   0    0    1    3    1    0    6    0   82
 6   9   4   2   0   5   2   0   4   0    0    2    6    1    0   10    0   42

But I am getting the following error message:

Error in scale.default(x, min(x), max(x) - min(x)) (from #4) : 
  length of 'center' must equal the number of columns of 'x'

Answer 1

there is also another way to scale the data by creating a function

 data_norm<- function(x) {((x-min(x))/(max(x)-min(x)))}
 variables_norm<- as.data.frame(lapply(data[1:17], data_norm)) 
 summary(variables_norm)

Answer 2

The center and scale arguments to scale have to have length equal to the number of columns in x. It looks like data is a data.frame, so your x has as many columns as your data.frame does and hence the conflict. You can get past this snag three ways:

• drop the row into an atomic vector before passing to scale (which will treat it as a single column): scale(as.numeric(x), ...)
• convert data into a matrix, which drops row extractions into atomic vectors automatically.
• use @agstudy's apply suggestion, which would work whether it's a data.frame or a matrix and is arguably the "right" way to do this in R.

Answer 3

Using this data , your example works for me:

data <- matrix(sample(1:1000,17*6), ncol=17,nrow=6)
for(i in 1:nrow(data)){
  x <- data[i, ]
  data[i, ] <- scale(x, min(x), max(x)-min(x))
}

Here another option using scale , without a loop. You need just to provide a scale and a center with same columns that your matrix.

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

EDIT how to access the result.

The scale returns a matrix with 2 attributes. To get a data.frame, you need just to coerce the scale result to a data.frame.

dat.scale <- scale(data, center = mins, scale = maxs - mins)
dat.sacle <- as.data.frame(dat.scale)