Imlerith - 3 months ago 41
R Question

# Normalize data in R data.frame column

Suppose I have the following data:

a <- data.frame(var1=letters,var2=runif(26))

Suppose I want to scale every value in
var2
such that the sum of the
var2
column is equal to 1 (basically turn the var2 column into a probability distribution)

I have tried the following:

a\$var2 <- lapply(a\$var2,function(x) (x-min(a\$var2))/(max(a\$var2)-min(a\$var2)))

this not only gives an overall sum greater than 1 but also turns the
var2
column into a list on which I can't do operations like
sum

Is there any valid way of turning this column into a probability distribution?

Suppose you have a vector x with non-negative values and no NA, you can normalize it by

x / sum(x)

which is a proper probability mass function.

The transform you take:

(x - min(x)) / (max(x) - min(x))

only rescales x onto [0, 1], but does not ensure "summation to 1".

Regarding you code

There is no need to use lapply here:

lapply(a\$var2, function(x) (x-min(a\$var2)) / (max(a\$var2) - min(a\$var2)))

Just use vectorized operation

a\$var2 <- with(a, (var2 - min(var2)) / (max(var2) - min(var2)))

As you said, lapply gives you a list, and that is what "l" in "lapply" refers to. You can use unlist to collapse that list into a vector; or, you can use sapply, where "s" implies "simplification (when possible)".

Source (Stackoverflow)