As far as I can discern, your question makes no sense at all.
Suppose you *know* that y = 2 + 3*x1 + 4*x2.
Now what should you predict when x1 = 6 (with x2 "missing"/unknown)?
See fortune("magic").
On 19/12/13 07:18, Chris Wilkinson wrote:
I would like to predict a new response from a fitted linear model where the
new data is a single case with a missing value. My reading of the help on
predict() is inconclusive on whether this is possible.
Leaving out the missing value or setting it to NA both fail but differently,
see example code below.
y <- runif(50)
x1 <- rnorm(50)
x2 <- rnorm(50)
dat <- data.frame(y, x1, x2)
mod <- lm(y~.,data=dat)
summary(mod)
Call:
lm(formula = y ~ ., data = dat)
Residuals:
Min 1Q Median 3Q Max
-0.50467 -0.28997 0.01457 0.27970 0.47791
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.50098 0.04577 10.945 1.6e-14 ***
x1 -0.01762 0.04172 -0.422 0.675
x2 -0.02753 0.04920 -0.560 0.578
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.3177 on 47 degrees of freedom
Multiple R-squared: 0.009301, Adjusted R-squared: -0.03286
F-statistic: 0.2206 on 2 and 47 DF, p-value: 0.8028
predict(mod, newdata=data.frame(x1=0.1, x2=0.3)) #OK as expected
1
0.4909624
predict(mod, newdata=data.frame(x1=0.1)) # x2 missing
Error in model.frame.default(Terms, newdata, na.action = na.action, xlev =
object$xlevels) :
variable lengths differ (found for 'x2')
In addition: Warning message:
'newdata' had 1 row but variables found have 50 rows
predict(mod, newdata=data.frame(x1=0.1, x2=NA)) #x2=NA
Error: variable 'x2' was fitted with type "numeric" but type "logical" was
supplied
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