Reputation: 560
This is home work.
I am new to R.
I have two data frames each containing two columns of data. I have to find a function that normalize the first data frame to a mean of 0 and a variance of 1 - for both columns. Then I want to apply that function on the second data frame.
I have tried this:
my_scale_test <- function(x,y) {
apply(y, 2, function(x,y) {
(y - mean(x ))/sd(x)
})
}
where x is the first data frame and y is the data frame to be normalized.
Can some one help me?
Edit:
I have now tried this aswell, but not working either:
scale_func <- function(x,y) {
xmean <- mean(x)
xstd <- sd(x)
yout <- y
for (i in 1:length(x[1,]))
yout[,i] <- yout[,i] - xmean[i]
for (i in 1:length(x[1,]))
yout[,i] <- yout[,i]/xsd[i]
invisible(yout)
}
Edit 2: I found this working function for MatLab (which i tried to translate in edit 1):
function [ Xout ] = scale( Xbase, Xin )
Xmean = mean(Xbase);
Xstd = std(Xbase);
Xout = Xin;
for i=1:length(Xbase(1,:))
Xout(:,i) = Xout(:,i) - Xmean(i);
end
for i=1:length(Xbase(1,:))
Xout(:,i) = Xout(:,i)/Xstd(i);
end
end
Can someone help me translate it?
Upvotes: 0
Views: 96
Reputation: 180
Since you are new to R, let's try something really basic.
my_scale_test <- function(x, y) {
y.nrow <- nrow(y)
x.mean <- data.frame(t(apply(x, 2, mean)))
x.sd <- data.frame(t(apply(x, 2, sd)))
# To let x.mean and x.sd have the same dimension as y, let's repeat the rows.
x.mean <- x.mean[rep(1, y.nrow), ]
x.sd <- x.sd[rep(1, y.nrow), ]
(y - x.mean)/x.sd
}
To test, try
set.seed(1)
x <- data.frame(matrix(rnorm(10), nrow = 5))
y <- x
result <- my_scale_test(x, y)
apply(result, 2, mean)
apply(result, 2, sd)
Upvotes: 2