假设我将模型拟合为已标准化为 z 分数形式的数据：
#load data
data(cars)
#standardize variables so that they have mean 0 and st. dev 1
st.cars <- scale(cars)
st.cars <- as.data.frame(st.cars)
mod <- lm(speed ~ dist,data=st.cars)
我想使用这个模型进行预测：
predict(mod,st.cars)，但是，我想获得数据集中因变量原始尺度的预测。
有没有办法将预测重新缩放到原始比例？
该scale函数存储用于缩放属性中数据的scale和center值。这些可用于将缩放数据的预测转换回原始数据规模。
# Scale cars data:
scars <- scale(cars)
# Save scaled attibutes:
scaleList <- list(scale = attr(scars, "scaled:scale"),
center = attr(scars, "scaled:center"))
# scars is a matrix, make it a data frame like cars for modeling:
scars <- as.data.frame(scars)
smod <- lm(speed ~ dist, data = scars)
# Predictions on scaled data:
sp <- predict(smod, scars)
# Fit the same model to the original cars data:
omod <- lm(speed ~ dist, data = cars)
op <- predict(omod, cars)
# Convert scaled prediction to original data scale:
usp <- sp * scaleList$scale["speed"] + scaleList$center["speed"]
# Compare predictions:
all.equal(op, usp)
如果要使用模型通过smod模型对象预测新数据，则需要使用scaleList对象中的适当值缩放 newdata 值（不要scale直接在 newdata 上调用函数）。
我已经建立在 skaluzny 的答案之上，如果您想要一种更直观的方法来执行此操作，而不保存比例属性，而是使用默认情况下 scale() 函数的功能的知识（您实际上只需要这个答案的最后几行）。
尺度函数居中（减去平均值），然后尺度（除以数据的标准差）：
sdist <- scale(cars$dist)
head(sdist)
[,1]
[1,] -1.5902596
[2,] -1.2798136
[3,] -1.5126481
[4,] -0.8141446
[5,] -1.0469791
[6,] -1.2798136
sdist2<-(cars$dist-mean(cars$dist))/sd(cars$dist)
head(sdist2)
[1] -1.5902596 -1.2798136 -1.5126481 -0.8141446 -1.0469791 -1.2798136
# Note this only is oriented the other way because scale() function outputs a matrix:
sdist2<-as.matrix(sdist2)
head(sdist2)
# The output now looks identical
[,1]
[1,] -1.5902596
[2,] -1.2798136
[3,] -1.5126481
[4,] -0.8141446
[5,] -1.0469791
[6,] -1.2798136
因此，我们实际上可以使用原始数据的均值和标准差，而不是将事物存储为列表。
# Scale cars data:
scars <- scale(cars)
# Save scaled attibutes:
scaleList <- list(scale = attr(scars, "scaled:scale"),
center = attr(scars, "scaled:center"))
scaleList
$`scale`
speed dist
5.287644 25.769377
$center
speed dist
15.40 42.98
> sapply(cars,mean) # note that these values are the same as the `center` values above
speed dist
15.40 42.98
> sapply(cars,sd) # note that these values are the same as the `scale` values above
speed dist
5.287644 25.769377
所以现在我们可以检查如果我们只使用而不是缩放属性，预测值是否都mean()相同sd()：
# scars is a matrix, make it a data frame like cars for modeling:
scars <- as.data.frame(scars)
smod <- lm(speed ~ dist, data = scars)
# Predictions on scaled data:
sp <- predict(smod, scars)
# Fit the same model to the original cars data:
omod <- lm(speed ~ dist, data = cars)
op <- predict(omod, cars)
# Now the original answer was to use these stored attributes to modify the predictions:
usp1 <- sp * scaleList$scale["speed"] + scaleList$center["speed"]
# We can also simply use the standard deviation and mean from the original dataset:
usp2 <- sp * sd(cars$speed) + mean(cars$speed)
identical(usp1,usp2)
[1] TRUE
all.equal(op, usp1, usp2)
[1] TRUE
如果您这样做，这可能会更快/更有效，因为不需要创建额外的数据帧/对象：
Mod <- lm(scale(speed) ~ scale(dist), data = cars) # add scale() function directly to model
Unscaled_Pred <- predict(Mod, cars) * sd(cars$speed) + mean(cars$speed)
all.equal(op, Unscaled_Pred)
[1] TRUE # predictions are the same as the model that was never scaled