Scaling a Numeric Matrix in R with Values 0 to 1

Scaling a numeric matrix in R with values 0 to 1

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

scaling r dataframe to 0-1 with NA values

Here's the answer from the 2nd question you link:

function(x) {(x - min(x)) / (max(x) - min(x))}

We can modify this to work with NAs (using the built-in NA handling in min and max

stdize = function(x, ...) {(x - min(x, ...)) / (max(x, ...) - min(x, ...))}

Then you can call it and pass through na.rm = T.

x = rexp(100)
x[sample(1:100, size = 10)] <- NA
stdize(x)  # lots of NA
stdize(x, na.rm = T) # works!

Or, using the o data frame from your question:

o_std = lapply(o, stdize, na.rm = T)

The NAs will still be there at the end.

Range standardization (0 to 1) in R

s = sort(rexp(100))

range01 <- function(x){(x-min(x))/(max(x)-min(x))}

range01(s)

  [1] 0.000000000 0.003338782 0.007572326 0.012192201 0.016055006 0.017161145
  [7] 0.019949532 0.023839810 0.024421602 0.027197168 0.029889484 0.033039408
 [13] 0.033783376 0.038051265 0.045183382 0.049560233 0.056941611 0.057552543
 [19] 0.062674982 0.066001242 0.066420884 0.067689067 0.069247825 0.069432174
 [25] 0.070136067 0.076340460 0.078709590 0.080393512 0.085591881 0.087540132
 [31] 0.090517295 0.091026499 0.091251213 0.099218526 0.103236344 0.105724733
 [37] 0.107495340 0.113332392 0.116103438 0.124050331 0.125596034 0.126599323
 [43] 0.127154661 0.133392300 0.134258532 0.138253452 0.141933433 0.146748798
 [49] 0.147490227 0.149960293 0.153126478 0.154275371 0.167701855 0.170160948
 [55] 0.180313542 0.181834891 0.182554291 0.189188137 0.193807559 0.195903010
 [61] 0.208902645 0.211308713 0.232942314 0.236135220 0.251950116 0.260816843
 [67] 0.284090255 0.284150541 0.288498370 0.295515143 0.299408623 0.301264703
 [73] 0.306817872 0.307853369 0.324882091 0.353241217 0.366800517 0.389474449
 [79] 0.398838576 0.404266315 0.408936260 0.409198619 0.415165553 0.433960390
 [85] 0.440690262 0.458692639 0.464027428 0.474214070 0.517224262 0.538532221
 [91] 0.544911543 0.559945121 0.585390414 0.647030109 0.694095422 0.708385079
 [97] 0.736486707 0.787250428 0.870874773 1.000000000

Adding ... will allow you to pass through na.rm = T if you want to omit missing values from the calculation (they will still be present in the results):

range01 <- function(x, ...){(x - min(x, ...)) / (max(x, ...) - min(x, ...))}

How can i rescale every column in my data frame to a 0-100 scale? (in r)

Using scale, if dat is the name of your data frame:

## for one column
dat$a <- scale(dat$a, center = FALSE, scale = max(dat$a, na.rm = TRUE)/100)
## for every column of your data frame
dat <- data.frame(lapply(dat, function(x) scale(x, center = FALSE, scale = max(x, na.rm = TRUE)/100)))

For a simple case like this, you could also write your own function.

fn <- function(x) x * 100/max(x, na.rm = TRUE)
fn(c(0,1,0))
# [1]   0 100   0
## to one column
dat$a <- fn(dat$a)
## to all columns of your data frame
dat <- data.frame(lapply(dat, fn))

Standardize data columns in R

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:

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