# Help on predict.lm

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## Help on predict.lm

 This post was updated on . 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 Again, R gives an error...because I made an error somewhere, 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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## Re: Help on predict.lm

 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 ! > > -- > View this message in context: http://r.789695.n4.nabble.com/Help-on-predict-lm-tp4509586p4509586.html> Sent from the R help mailing list archive at Nabble.com. > [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Help on predict.lm

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## Re: Help on predict.lm

 FORTUNE!!! -- Bert On Tue, Mar 27, 2012 at 11:44 AM, Peter Ehlers <[hidden email]> wrote: > > R tries hard to keep you from committing scientific abuse. > As stated, your problem seems to me akin to > > 1. Given that a man's age can be modelled as a function >   of the grayness of his hair, > 2. predict a man's age from the temperature in Barcelona. > ... Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.