Standardize data columns in R

This is 3 years old. Still, I feel I have to add the following:

The most common normalization is the z-transformation, where you subtract the mean and divide by the standard deviation of your variable. The result will have mean=0 and sd=1.

For that, you don't need any package.

zVar <- (myVar - mean(myVar)) / sd(myVar)

That's it.

I have to assume you meant to say that you wanted a mean of 0 and a standard deviation of 1. If your data is in a dataframe and all the columns are numeric you can simply call the scale function on the data to do what you want.

dat <- data.frame(x = rnorm(10, 30, .2), y = runif(10, 3, 5))
scaled.dat <- scale(dat)

# check that we get mean of 0 and sd of 1
colMeans(scaled.dat)  # faster version of apply(scaled.dat, 2, mean)
apply(scaled.dat, 2, sd)

Using built in functions is classy. Like this cat:

Realizing that the question is old and one answer is accepted, I'll provide another answer for reference.

scale is limited by the fact that it scales all variables. The solution below allows to scale only specific variable names while preserving other variables unchanged (and the variable names could be dynamically generated):

library(dplyr)

set.seed(1234)
dat <- data.frame(x = rnorm(10, 30, .2), 
                  y = runif(10, 3, 5),
                  z = runif(10, 10, 20))
dat

dat2 <- dat %>% mutate_at(c("y", "z"), ~(scale(.) %>% as.vector))
dat2

which gives me this:

> dat
          x        y        z
1  29.75859 3.633225 14.56091
2  30.05549 3.605387 12.65187
3  30.21689 3.318092 13.04672
4  29.53086 3.079992 15.07307
5  30.08582 3.437599 11.81096
6  30.10121 4.621197 17.59671
7  29.88505 4.051395 12.01248
8  29.89067 4.829316 12.58810
9  29.88711 4.662690 19.92150
10 29.82199 3.091541 18.07352

and

> dat2 <- dat %>% mutate_at(c("y", "z"), ~(scale(.) %>% as.vector))
> dat2
          x          y           z
1  29.75859 -0.3004815 -0.06016029
2  30.05549 -0.3423437 -0.72529604
3  30.21689 -0.7743696 -0.58772361
4  29.53086 -1.1324181  0.11828039
5  30.08582 -0.5946582 -1.01827752
6  30.10121  1.1852038  0.99754666
7  29.88505  0.3283513 -0.94806607
8  29.89067  1.4981677 -0.74751378
9  29.88711  1.2475998  1.80753470
10 29.82199 -1.1150515  1.16367556

EDIT 1 (2016): Addressed Julian's comment: the output of scale is Nx1 matrix so ideally we should add an as.vector to convert the matrix type back into a vector type. Thanks Julian!

EDIT 2 (2019): Quoting Duccio A.'s comment: For the latest dplyr (version 0.8) you need to change dplyr::funcs with list, like dat %>% mutate_each_(list(~scale(.) %>% as.vector), vars=c("y","z"))

EDIT 3 (2020): Thanks to @mj_whales: the old solution is deprecated and now we need to use mutate_at.