## backtransform `scale()` for plotting

I have a explanatory variable that is centered using `scale()`

that is used to predict a response variable:

d <- data.frame( x=runif(100), y=rnorm(100) ) d <- within(d, s.x <- scale(x)) m1 <- lm(y~s.x, data=d)

I'd like to plot the predicted values, but using the original scale of `x`

rather than the centered scale. Is there a way to sort of backtransform or reverse scale `s.x`

?

Thanks!

Take a look at:

attributes(d$s.x)

You can use the attributes to unscale:

d$s.x * attr(d$s.x, 'scaled:scale') + attr(d$s.x, 'scaled:center')

For example:

> x <- 1:10 > s.x <- scale(x) > s.x [,1] [1,] -1.4863011 [2,] -1.1560120 [3,] -0.8257228 [4,] -0.4954337 [5,] -0.1651446 [6,] 0.1651446 [7,] 0.4954337 [8,] 0.8257228 [9,] 1.1560120 [10,] 1.4863011 attr(,"scaled:center") [1] 5.5 attr(,"scaled:scale") [1] 3.02765 > s.x * attr(s.x, 'scaled:scale') + attr(s.x, 'scaled:center') [,1] [1,] 1 [2,] 2 [3,] 3 [4,] 4 [5,] 5 [6,] 6 [7,] 7 [8,] 8 [9,] 9 [10,] 10 attr(,"scaled:center") [1] 5.5 attr(,"scaled:scale") [1] 3.02765

**How to un-transform exponential plot data to get back to original ,** However, my fitted values are no longer on the same scale as my original data. I'm trying to come up with what to do to adjust them back to the Multiple regression is used by many practitioners. In this post we have shown how to scale continuous predictors and transform back the regression coefficients to original scale. Scaled coefficients would help us to better interpret the results. The question of when to standardize the data is a different issue.

For a data frame or matrix:

set.seed(1) x = matrix(sample(1:12), ncol= 3) xs = scale(x, center = TRUE, scale = TRUE) x.orig = t(apply(xs, 1, function(r)r*attr(xs,'scaled:scale') + attr(xs, 'scaled:center'))) print(x) [,1] [,2] [,3] [1,] 4 2 3 [2,] 5 7 1 [3,] 6 10 11 [4,] 9 12 8 print(x.orig) [,1] [,2] [,3] [1,] 4 2 3 [2,] 5 7 1 [3,] 6 10 11 [4,] 9 12 8

Be careful when using functions like `identical()`

:

print(x - x.orig) [,1] [,2] [,3] [1,] 0 0 0.000000e+00 [2,] 0 0 8.881784e-16 [3,] 0 0 0.000000e+00 [4,] 0 0 0.000000e+00 identical(x, x.orig) # FALSE

**Statistical Methods in Biology: Design and Analysis of Experiments ,** backtransform `scale()` for plotting I have a explanatory variable that is centered using scale() that is used to predict a response variable: Convert data back from normalized values [duplicate] Ask Question backtransform `scale()` for plotting (6 answers) Closed 2 years ago. I have used scale to

I felt like this should be a proper function, here was my attempt at it:

#' Reverse a scale #' #' Computes x = sz+c, which is the inverse of z = (x - c)/s #' provided by the \code{scale} function. #' #' @param z a numeric matrix(like) object #' @param center either NULL or a numeric vector of length equal to the number of columns of z #' @param scale either NULL or a a numeric vector of length equal to the number of columns of z #' #' @seealso \code{\link{scale}} #' mtcs <- scale(mtcars) #' #' all.equal( #' unscale(mtcs), #' as.matrix(mtcars), #' check.attributes=FALSE #' ) #' #' @export unscale <- function(z, center = attr(z, "scaled:center"), scale = attr(z, "scaled:scale")) { if(!is.null(scale)) z <- sweep(z, 2, scale, `*`) if(!is.null(center)) z <- sweep(z, 2, center, `+`) structure(z, "scaled:center" = NULL, "scaled:scale" = NULL, "unscaled:center" = center, "unscaled:scale" = scale ) }

**The Design and Statistical Analysis of Animal Experiments,** However, we often want to interpret these differences on the scale on which the of predictions on the back-transformed scale for qualitative variables or plots of If you used a log10(x) transformation, then the back-transformation is 10**(x). But if you used ln(x), then go w/ e**(x). Re: the published back-transformed values, such mean values won't match the

**Using Back transformation,** The researcher decided to back-transform the predicted means onto the original scale before plotting them. The plot of back-transformed geometric means, with For the log transformation, you would back-transform by raising 10 to the power of your number. For example, the log transformed data above has a mean of 1.044 and a 95% confidence interval of ±0.344 log-transformed fish. The back-transformed mean would be 10 1.044 =11.1 fish.

I came across this problem and I think I found a simpler solution using linear algebra.

# create matrix like object a <- rnorm(1000,5,2) b <- rnorm(1000,7,5) df <- cbind(a,b) # get center and scaling values mean <- apply(df, 2, mean) sd <- apply(df, 2, sd) # scale data s.df <- scale(df, center = mean, scale = sd) #unscale data with linear algebra us.df <- t((t(s.df) * sd) + mean)

**unscale: Backtransform 'scale'-ed object in ggfortify: Data ,** I plot a nice graph and get complaints. A common I can back-transform the mean(log(value)) and find that it is nothing like the mean of the It is a real value and my sample size is insufficient to properly estimate the variability in the data. For example, with hierarchical forecasting the forecasts need to be aggregated, and medians do not aggregate but means do. It is easy enough to derive the mean forecast using a Taylor series expansion. Suppose represents the back-transformation function, is the mean on the transformed scale and is the variance on the transformed scale.

**Statistics: An Introduction Using R,** Loading required package: ggplot2 Sepal.Length Sepal.Width Petal.Length Petal.Width 1 5.1 3.5 1.4 0.2 2 4.9 3.0 1.4 0.2 3 4.7 3.2 1.3 0.2 4 4.6 3.1 1.5 0.2 5 5.0 An other possibility is the function scale_x_log10() and scale_y_log10(), which transform, respectively, the x and y axis scales into a log scale: base 10. Note that, the scale functions transform the data. If you fit anything to the data it would probably change the fitted values.

**OECD Series on Testing and Assessment Current Approaches in the ,** in plot, 72, 76 type = "n" for blank plots, 59, 64, 152, 190 proportion data, 272 with split, 309 type = "response", model output on backtransformed scale Ancova The back-transformed axis reveals a few key properties of the transformation (i.e., symmetry around 0.5, compressed scale in the middle of the distribution, and spread scale at the ends of the distribution), but obscures the fact that the confidence intervals are not symmetric around the mean.

Therefore, in plotting the model together with the data, the back transformed the MLEs of the regression parameters relate to the model on the original scale,