[R] Help on predict.lm

[Berend Hasselman]

On 27-03-2012, at 19:24, Nederjaard wrote:

> Hello, 
> 
> I'm new here, but will try to be as specific and complete as possible. I'm
> trying to use “lm“ to first estimate parameter values from a set of
> calibration measurements, and then later to use those estimates to calculate
> another set of values with “predict.lm”.
> 
> First I have a calibration dataset of absorbance values measured from
> standard solutions with known concentration of Bromide:
> 
>> stds
>      abs conc
> 1 -0.0021    0
> 2  0.1003  200
> 3  0.2395  500
> 4  0.3293  800
> 
> On this small calibration series, I perform a linear regression to find the
> parameter estimates of the relationship between absorbance (abs) and
> concentration (conc):
> 
>> linear1 <- lm(abs~conc, data=stds)
>> summary(linear1)
> 
> Call:
> lm(formula = abs ~ conc, data = stds)
> 
> Residuals:
>        1         2         3         4 
> -0.012600  0.006467  0.020667 -0.014533 
> 
> Coefficients:
>             Estimate Std. Error t value Pr(>|t|)   
> (Intercept) 1.050e-02  1.629e-02   0.645  0.58527   
> conc        4.167e-04  3.378e-05  12.333  0.00651 **
> ---
> Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 
> 
> Residual standard error: 0.02048 on 2 degrees of freedom
> Multiple R-squared: 0.987,      Adjusted R-squared: 0.9805 
> F-statistic: 152.1 on 1 and 2 DF,  p-value: 0.00651 
> 
> 
> 
> 
> 
> Now I come with another dataset, which contains measured absorbance values
> of Bromide in solution:
> 
>> brom
>    hours     abs
> 1    -1.0  0.0633
> 2     1.0  0.2686
> 3     5.0  0.2446
> 4    18.0  0.2274
> 5    29.0  0.2091
> 6    42.0  0.1961
> 7    53.0  0.1310
> 8    76.0  0.1504
> 9    91.0  0.1317
> 10   95.5  0.1169
> 11  101.0  0.0977
> 12  115.0  0.1023
> 13  123.5  0.0879
> 14  138.5  0.0724
> 15  147.5  0.0564
> 16  163.0  0.0495
> 17  171.0  0.0325
> 18  189.0  0.0182
> 19  211.0  0.0047
> 20  212.5      NA
> 21  815.5 -0.2112
> 22  816.5 -0.1896
> 23  817.5 -0.0783
> 24  818.5  0.2963
> 25  819.5  0.1448
> 26  839.5  0.0936
> 27  864.0  0.0560
> 28  888.0  0.0310
> 29  960.5  0.0056
> 30 1009.0 -0.0163
> 
> The values in column brom$abs, measured on 30 subsequent points in time need
> to be calculated to Bromide concentrations, using the previously established
> relationship “linear1”.  
> At first, I thought it could be done by:
> 
>> predict.lm(linear1, brom$abs)
> Error in eval(predvars, data, env) : 
>  numeric 'envir' arg not of length one
> 
> But, R gives the above error message. Then, after some searching around on
> different fora and R-communities (including this one), I learned that the
> “newdata” in “predict.lm” actually needs to be coerced into a separate
> dataframe. Thus:
> 
>> mabs <- data.frame(Abs = brom$abs)
>> predict.lm(linear1, mabs)
> Error in eval(expr, envir, enclos) : object 'conc' not found
> 

There is no column with name "conc" in your dataframe mabs.

You regressed abs on conc. For prediction you need data for conc and not abs.
So provide data for conc. Or change the regression around: lm(conc ~ abs, data=stds) if that makes any sense.

What you did with mabs wouldn't have worked anyway because Abs is not the same as abs.
And it wasn't necessary.

Berend


> Again, R gives an error...probably because I made an error, but I truly fail
> to see where. I hope somebody can explain to me clearly what I'm doing wrong
> and what I should do to instead.
> Any help is greatly appreciated, thanks !
> 
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