R help - Dec 2013 - Predicting response from fitted linear model with incomplete new sample data

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
>
Thanks
Chris

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