一个线性数组的R线性模型（lm）预测函数

Question

I have an lm model in R that I have trained and serialized. Inside a function, where I pass as input the model and a feature vector (one single array), I have:

CREATE OR REPLACE FUNCTION lm_predict(
    feat_vec float[],
    model bytea
)
RETURNS float
AS
$$
    #R-code goes here.
    mdl <- unserialize(model)
    # class(feat_vec) outputs "array"
    y_hat <- predict.lm(mdl, newdata = as.data.frame.list(feat_vec))
    return (y_hat)
$$ LANGUAGE 'plr';

This returns the wrong y_hat !! I know this because this other solution works (the inputs to this function are still the model (in a bytearray) and one feat_vec (array)):

CREATE OR REPLACE FUNCTION lm_predict(
    feat_vec float[],
    model bytea
)
RETURNS float
AS
$$
    #R-code goes here.
    mdl <- unserialize(model)
    coef = mdl$coefficients
    y_hat = coef[1] + as.numeric(coef[-1]%*%feat_vec)
    return (y_hat)
$$ LANGUAGE 'plr';

What am I doing wrong?? It is the same unserialized model, the first option should give me the right answer as well...

Answer 1

The problem seems to be the use of newdata = as.data.frame.list(feat_vec) . As discussed in your previous question , this returns ugly column names. While when you call predict , newdata must have column names consistent with covariates names in your model formula. You should get some warning message when you call predict .

## example data
set.seed(0)
x1 <- runif(20)
x2 <- rnorm(20)
y <- 0.3 * x1 + 0.7 * x2 + rnorm(20, sd = 0.1)

## linear model
model <- lm(y ~ x1 + x2)

## new data
feat_vec <- c(0.4, 0.6)
newdat <- as.data.frame.list(feat_vec)
#  X0.4 X0.6
#1  0.4  0.6

## prediction
y_hat <- predict.lm(model, newdata = newdat)
#Warning message:
#'newdata' had 1 row but variables found have 20 rows 

What you need is

newdat <- as.data.frame.list(feat_vec,
                             col.names = attr(model$terms, "term.labels"))
#   x1  x2
#1 0.4 0.6

y_hat <- predict.lm(model, newdata = newdat)
#        1 
#0.5192413 

This is the same as what you can compute manually:

coef = model$coefficients
unname(coef[1] + sum(coef[-1] * feat_vec))
#[1] 0.5192413