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
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