[R] Help on how to use predict

[David Winsemius]

Eduardo;

I think you would be more successful if you put your data in a  
dataframe, offered it to lm with column names only in the formula,   
and then used the newdata argument with predict with the column names  
matching the column names in the original data.

-- 
David.
On Aug 15, 2011, at 5:46 AM, Eduardo M. A. M.Mendes wrote:

> Dear R-Users
>
>
>
> My problem is quite simple: I need to use a fitted model to predict  
> the next
> point (that is, just one single point in a curve).
>
>
>
> The data was divided in two parts:  identification (x and y - class  
> matrix)
> and validation (xt and yt  - class matrix).  I don't use all values  
> in x and
> y but only the 10 nearest points (x[b,] and y[b,]) for each  
> regression (b is
> a vector with the indexes).
>
>
>
>> fit.a=lm(y[b] ~ x[b,],method="qr")
>
>> summary(fit.a)
>
>
>
> Call:
>
> lm(formula = y[b] ~ x[b, ], method = "qr")
>
>
>
> Residuals:
>
>         1          2          3          4          5           
> 6          7
>
>
> 6.939e-18 -2.393e-02 -3.912e-02  1.344e-02 -5.926e-02 -1.821e-02  
> -4.075e-02
>
>
>         8          9         10
>
> 4.075e-02  3.938e-02  8.769e-02
>
>
>
> Coefficients:
>
>             Estimate Std. Error t value Pr(>|t|)
>
> (Intercept)  1.852793   0.842459   2.199    0.115
>
> x[b, ]1     -0.086324   0.056841  -1.519    0.226
>
> x[b, ]2      0.001114   0.001666   0.668    0.552
>
> x[b, ]3      0.002501   0.004376   0.571    0.608
>
> x[b, ]4     -0.003589   0.009041  -0.397    0.718
>
> x[b, ]5     -0.276498   0.119545  -2.313    0.104
>
> x[b, ]6     -0.003010   0.003574  -0.842    0.462
>
>
>
> Residual standard error: 0.07893 on 3 degrees of freedom
>
> Multiple R-squared: 0.7932,       Adjusted R-squared: 0.3795
>
> F-statistic: 1.917 on 6 and 3 DF,  p-value: 0.3169
>
>
>
> Once the fiited model is found (it does not matter how bad it is - the
> variables are poorly correlated), I need to predict the next value
>
>
>
>> predict(fit.a,xt[j,])
>
> Error in eval(predvars, data, env) :
>
>  numeric 'envir' arg not of length one
>
>
>
> It seems that predict needs a dataframe class but even if I change  
> xt[b,] to
> as.data.frame(xt[b,]) the result is not what I expect.
>
>
>
>> predict(fit.a,as.data.frame(t(xt[j,])))
>
>        1         2         3         4         5         6         7
> 8
>
> 0.7834919 0.8243357 0.7780093 0.7810451 0.8084342 0.8057823 1.0304123
> 0.9729126
>
>        9        10
>
> 0.7708979 0.8464298
>
> Warning message:
>
> 'newdata' had 1 rows but variable(s) found have 10 rows
>
>
>
> My feeling is that I did not understand how to enter the formula in  
> lm in
> the first place.
>
>
>
> Many thanks
>
>
>
> Ed
>
>
>
>
> 	[[alternative HTML version deleted]]
>
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> and provide commented, minimal, self-contained, reproducible code.

David Winsemius, MD
Heritage Laboratories
West Hartford, CT