Range Standardization (0 to 1) in R

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, ...))}

Normalize variables from 0 to 1 with different range of variables?

I think you can do in one line:

sapply(df, function(x) (x - min(x, na.rm = T)) / (max(x, na.rm = T) - min(x, na.rm=T)))

      x1        x2
[1,] 0.0 0.0000000
[2,] 0.5 1.0000000
[3,] 1.0 0.4444444
[4,] 0.5 0.7777778
[5,] 0.0 0.2222222

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:

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

How to scale int values in R between 0 and 100

With treshold I mean, that all values < 50 keep their value.

I got it now with:

employ.data$value <- replace(employ.data$value, employ.data$value > 50, 100)

And my result:

> employ.data
employee value  startdate
1   John Doe     1 2010-11-01
2 Peter Gynn     3 2008-03-25
3 Jolie Hope   100 2007-03-14 

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.

Standardizing a vector in R so that values shift towards boundaries

You can use min-max standardization. Usually min max std. is used to scale values between 0 and 1. However, you can scale values to any range [a, b] by using the following equation:

X_Scaled = a + (x - min(x)) * (b-a) / (max(x) - min(x))

So in your case, let's break it down to two steps.

First: you want positive values to be centered around 0.75 and negative values centered around -0.75. So we can just filter for the values in your data.

data <- runif(100, -1, 1)

positive_vals <- data[data > 0]
negative_vals <- data[data < 0]

Second step: You want to control how much they move towards this value of 0.75. So you could define a range and a center. Say, a range of 0.05 and a center of 0.75 gives us a = 0.7 and b=0.8, right? We can do the same for the negative center.

range <- 0.05
upper_center <- 0.75
lower_center <- -0.75

b1 <- upper_center + range
a1 <- upper_center - range

b2 <- lower_center + range
a2 <- lower_center - range

Finally, we apply the min-max equation for both cases, taking care to preserve the original positions of the positive and negative values in the original array.

# normalize them using, say, min-max
positive_vals <- a1 + ((positive_vals - min(positive_vals)) * (b1 - a1)) / (max(positive_vals) - min(positive_vals))
negative_vals <- a2 + ((negative_vals - min(negative_vals)) * (b2 - a2)) / (max(negative_vals) - min(negative_vals))

new_data <- data
new_data[data > 0] <- positive_vals
new_data[data < 0] <- negative_vals

# Plot the results!
plot(data)
points(new_data, col = "red")

If you're not satisfied with moving values so close to 0.75, just increase the range. You can also move the centers by defining different values.

Using your data provided:

R Normalize Many Columns

Try scale like below

cbind(data, `colnames<-`(scale(data[normalize_these]), paste0(normalize_these, "NEW")))

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If you would like to use data.table, below might be an option

setDT(data)
data[, paste0(normalize_these, "NEW") := lapply(.SD, scale), .SDcols = normalize_these]

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