Pamela Krone-Davis
2012-Jul-13 16:00 UTC
[R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect
Hi,
I have been using lm in R to do a linear regression and find the slope
coefficients and value for R-squared. The R-squared value reported by R
(R^2 = 0.9558) is very different than the R-squared value when I use the
same equation in Exce (R^2 = 0.328). I manually computed R-squared and the
Excel value is correct. I show my code for the determination of R^2 in R.
When I do not set 0 as the intercept, the R^2 value is the same in R and
Excel. In both cases the slope coefficient from R and from Excel are
identical.
k is a data frame with two columns.
M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different
R^2 values in R and Excel
M2 = lm(k[,1]~k[,2])
sumM1 = summary(M1)
sumM2 = summary(M2) ## get same value as Excel when intercept is not
set to 0
Below is what R returns for sumM1:
lm(formula = k[, 1] ~ k[, 2] + 0)
Residuals:
Min 1Q Median 3Q Max
-0.057199 -0.015857 0.003793 0.013737 0.056178
Coefficients:
Estimate Std. Error t value Pr(>|t|)
k[, 2] 1.05022 0.04266 24.62 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.02411 on 28 degrees of freedom
Multiple R-squared: 0.9558, Adjusted R-squared: 0.9543
F-statistic: 606.2 on 1 and 28 DF, p-value: < 2.2e-16
Way manual determination was performed. The value returned coincides with
the value from Excel:
#### trying to figure out why the R^2 for R and Excel are so different.
sqerr = (k[,1] - predict(M1))^2
sqtot = (k[,1] - mean(k[,1]) ^2
R2 = 1 - sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as
excel value
I am very puzzled by this. How does R compute the value for R^2 in this
case? Did i write the lm incorrectly?
Thanks
Pam
PS In case you are interested, the data I am using for hte two columns is
below.
k[, 1]
1]
[1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441
0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484
0.06358752 0.07945473
[15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827
0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638
0.08419016 0.11995240
[29] 0.14425678
> k[,2]
[1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10
0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12
--
Pam Krone-Davis
Project Research Assistant and Grant Manager
PO Box 22122
Carmel, CA 93922
(831)582-3684 (o)
(831)324-0391 (h)
[[alternative HTML version deleted]]
William Dunlap
2012-Jul-13 17:03 UTC
[R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect
What does Excel give for the following data, where the by-hand formula you gave is obviously wrong? > x <- c(1, 2, 3) > y <- c(13.1, 11.9, 11.0) > M1 <- lm(y~x+0) > sqerr <- (y- predict(M1)) ^ 2 > sqtot <- (y - mean(y)) ^ 2 > 1 - sum(sqerr)/sum(sqtot) [1] -37.38707 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com> -----Original Message----- > From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On > Behalf Of Pamela Krone-Davis > Sent: Friday, July 13, 2012 9:01 AM > To: r-help at r-project.org > Subject: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is > incorrect > > Hi, > > I have been using lm in R to do a linear regression and find the slope > coefficients and value for R-squared. The R-squared value reported by R > (R^2 = 0.9558) is very different than the R-squared value when I use the > same equation in Exce (R^2 = 0.328). I manually computed R-squared and the > Excel value is correct. I show my code for the determination of R^2 in R. > When I do not set 0 as the intercept, the R^2 value is the same in R and > Excel. In both cases the slope coefficient from R and from Excel are > identical. > > k is a data frame with two columns. > > M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different > R^2 values in R and Excel > M2 = lm(k[,1]~k[,2]) > sumM1 = summary(M1) > sumM2 = summary(M2) ## get same value as Excel when intercept is not > set to 0 > > Below is what R returns for sumM1: > > lm(formula = k[, 1] ~ k[, 2] + 0) > > Residuals: > Min 1Q Median 3Q Max > -0.057199 -0.015857 0.003793 0.013737 0.056178 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > k[, 2] 1.05022 0.04266 24.62 <2e-16 *** > --- > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.02411 on 28 degrees of freedom > Multiple R-squared: 0.9558, Adjusted R-squared: 0.9543 > F-statistic: 606.2 on 1 and 28 DF, p-value: < 2.2e-16 > > Way manual determination was performed. The value returned coincides with > the value from Excel: > > #### trying to figure out why the R^2 for R and Excel are so different. > sqerr = (k[,1] - predict(M1))^2 > sqtot = (k[,1] - mean(k[,1]) ^2 > > R2 = 1 - sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as > excel value > > I am very puzzled by this. How does R compute the value for R^2 in this > case? Did i write the lm incorrectly? > > Thanks > Pam > > PS In case you are interested, the data I am using for hte two columns is > below. > > k[, 1] > 1] > [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441 > 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484 > 0.06358752 0.07945473 > [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827 > 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638 > 0.08419016 0.11995240 > [29] 0.14425678 > > > k[,2] > [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10 > 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12 > > > -- > Pam Krone-Davis > Project Research Assistant and Grant Manager > PO Box 22122 > Carmel, CA 93922 > (831)582-3684 (o) > (831)324-0391 (h) > > [[alternative HTML version deleted]]
John Sorkin
2012-Jul-13 17:22 UTC
[R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect
Pamela R squared with a non-zero, and with a zero intercept can be very different as the regression line that you get with and without a zero intercept can be very different. Have you plotted your data plot(k[,2],k[,1]) to see if a zero intercept is reasonable for your data? Have you drawn the regression lines that you get from your models and compared the lines to the plots of your data? John>>> Pamela Krone-Davis <pkrone-davis at csumb.edu> 7/13/2012 12:00:36 PM >>>Hi, I have been using lm in R to do a linear regression and find the slope coefficients and value for R-squared. The R-squared value reported by R (R^2 = 0.9558) is very different than the R-squared value when I use the same equation in Exce (R^2 = 0.328). I manually computed R-squared and the Excel value is correct. I show my code for the determination of R^2 in R. When I do not set 0 as the intercept, the R^2 value is the same in R and Excel. In both cases the slope coefficient from R and from Excel are identical. k is a data frame with two columns. M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different R^2 values in R and Excel M2 = lm(k[,1]~k[,2]) sumM1 = summary(M1) sumM2 = summary(M2) ## get same value as Excel when intercept is not set to 0 Below is what R returns for sumM1: lm(formula = k[, 1] ~ k[, 2] + 0) Residuals: Min 1Q Median 3Q Max -0.057199 -0.015857 0.003793 0.013737 0.056178 Coefficients: Estimate Std. Error t value Pr(>|t|) k[, 2] 1.05022 0.04266 24.62 <2e-16 *** --- Signif. codes: 0 ?***? 0.001 ?**? 0.01 ?*? 0.05 ?.? 0.1 ? ? 1 Residual standard error: 0.02411 on 28 degrees of freedom Multiple R-squared: 0.9558, Adjusted R-squared: 0.9543 F-statistic: 606.2 on 1 and 28 DF, p-value: < 2.2e-16 Way manual determination was performed. The value returned coincides with the value from Excel: #### trying to figure out why the R^2 for R and Excel are so different. sqerr = (k[,1] - predict(M1))^2 sqtot = (k[,1] - mean(k[,1]) ^2 R2 = 1 - sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as excel value I am very puzzled by this. How does R compute the value for R^2 in this case? Did i write the lm incorrectly? Thanks Pam PS In case you are interested, the data I am using for hte two columns is below. k[, 1] 1] [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484 0.06358752 0.07945473 [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638 0.08419016 0.11995240 [29] 0.14425678> k[,2][1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12 -- Pam Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (h) [[alternative HTML version deleted]] Confidentiality Statement: This email message, including any attachments, is for th...{{dropped:6}}
William Dunlap
2012-Jul-13 17:38 UTC
[R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect
While excluding the intercept may make sense, your formula for r^2 assumes that there was an intercept (that is why mean(y) is in your expression for sqtot). Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Pamela Krone-Davis [mailto:pkrone-davis@csumb.edu] Sent: Friday, July 13, 2012 10:32 AM To: William Dunlap Subject: Re: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect Hi William, Thanks for getting back to me. When I use the values you provided, it would not make sense to set an intercept of 0. For my data, 0 does make sense as an intercept. When I do not set a 0 intercept using your data points, I get the same value for R-squared in R and in Excel and manually. Thanks Pam On Fri, Jul 13, 2012 at 10:03 AM, William Dunlap <wdunlap@tibco.com<mailto:wdunlap@tibco.com>> wrote: What does Excel give for the following data, where the by-hand formula you gave is obviously wrong? > x <- c(1, 2, 3) > y <- c(13.1, 11.9, 11.0) > M1 <- lm(y~x+0) > sqerr <- (y- predict(M1)) ^ 2 > sqtot <- (y - mean(y)) ^ 2 > 1 - sum(sqerr)/sum(sqtot) [1] -37.38707 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com<http://tibco.com>> -----Original Message----- > From: r-help-bounces@r-project.org<mailto:r-help-bounces@r-project.org> [mailto:r-help-bounces@r-project.org<mailto:r-help-bounces@r-project.org>] On > Behalf Of Pamela Krone-Davis > Sent: Friday, July 13, 2012 9:01 AM > To: r-help@r-project.org<mailto:r-help@r-project.org> > Subject: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is > incorrect > > Hi, > > I have been using lm in R to do a linear regression and find the slope > coefficients and value for R-squared. The R-squared value reported by R > (R^2 = 0.9558) is very different than the R-squared value when I use the > same equation in Exce (R^2 = 0.328). I manually computed R-squared and the > Excel value is correct. I show my code for the determination of R^2 in R. > When I do not set 0 as the intercept, the R^2 value is the same in R and > Excel. In both cases the slope coefficient from R and from Excel are > identical. > > k is a data frame with two columns. > > M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different > R^2 values in R and Excel > M2 = lm(k[,1]~k[,2]) > sumM1 = summary(M1) > sumM2 = summary(M2) ## get same value as Excel when intercept is not > set to 0 > > Below is what R returns for sumM1: > > lm(formula = k[, 1] ~ k[, 2] + 0) > > Residuals: > Min 1Q Median 3Q Max > -0.057199 -0.015857 0.003793 0.013737 0.056178 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > k[, 2] 1.05022 0.04266 24.62 <2e-16 *** > --- > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.02411 on 28 degrees of freedom > Multiple R-squared: 0.9558, Adjusted R-squared: 0.9543 > F-statistic: 606.2 on 1 and 28 DF, p-value: < 2.2e-16 > > Way manual determination was performed. The value returned coincides with > the value from Excel: > > #### trying to figure out why the R^2 for R and Excel are so different. > sqerr = (k[,1] - predict(M1))^2 > sqtot = (k[,1] - mean(k[,1]) ^2 > > R2 = 1 - sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as > excel value > > I am very puzzled by this. How does R compute the value for R^2 in this > case? Did i write the lm incorrectly? > > Thanks > Pam > > PS In case you are interested, the data I am using for hte two columns is > below. > > k[, 1] > 1] > [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441 > 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484 > 0.06358752 0.07945473 > [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827 > 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638 > 0.08419016 0.11995240 > [29] 0.14425678 > > > k[,2] > [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10 > 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12 > > > -- > Pam Krone-Davis > Project Research Assistant and Grant Manager > PO Box 22122 > Carmel, CA 93922 > (831)582-3684 (o) > (831)324-0391 (h) > > [[alternative HTML version deleted]]-- Pam Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (h) [[alternative HTML version deleted]]
William Dunlap
2012-Jul-13 20:05 UTC
[R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect
S+ and, I assume, R compute r^2 when there is no intercept as sum(fitted(M1)^2) / sum(y^2) where M1 is the fitted model and y the response. See, for example, http://web.ist.utl.pt/~ist11038/compute/errtheory/,regression/regrthroughorigin.pdf for the derivation of this formula. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com From: Pamela Krone-Davis [mailto:pkrone-davis@csumb.edu] Sent: Friday, July 13, 2012 11:05 AM To: William Dunlap Subject: Re: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect Thanks William, I have actually tried a couple of different formulas for determining R^2. The second formula does return a different value and assumes no intercept. The second formula is attached in the image. However, when I use the manual formula for R^2 that is shown, I get an R^2 value that matches the second formula when I don't set the intercept. ISo I think the formula I showed works for both cases. M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 M2 = lm(k[,1]~k[,2]) sqerrM2 = (k[,1] - predict(M2))^2 sqtotM2 = (k[,1] - mean(k[,1])) ^2 sqerrM1 = (k[,1] - predict(M1))^2 sqtotM1 = (k[,1] - mean(k[,1])) ^2 R2M1 = 1 - sum(sqerrM1)/sum(sqtotM1) ## get 0.328 same as excel value R2M2 = 1 - sum(sqerrM2)/sum(sqtotM2) ## same as Excel 0.408 and as R in this case> R2M1[1] 0.3284381> R2M2[1] 0.4083052 How does R compute the R-squared value? Thanks Pam On Fri, Jul 13, 2012 at 10:38 AM, William Dunlap <wdunlap@tibco.com<mailto:wdunlap@tibco.com>> wrote: While excluding the intercept may make sense, your formula for r^2 assumes that there was an intercept (that is why mean(y) is in your expression for sqtot). Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com<http://tibco.com> From: Pamela Krone-Davis [mailto:pkrone-davis@csumb.edu<mailto:pkrone-davis@csumb.edu>] Sent: Friday, July 13, 2012 10:32 AM To: William Dunlap Subject: Re: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is incorrect Hi William, Thanks for getting back to me. When I use the values you provided, it would not make sense to set an intercept of 0. For my data, 0 does make sense as an intercept. When I do not set a 0 intercept using your data points, I get the same value for R-squared in R and in Excel and manually. Thanks Pam On Fri, Jul 13, 2012 at 10:03 AM, William Dunlap <wdunlap@tibco.com<mailto:wdunlap@tibco.com>> wrote: What does Excel give for the following data, where the by-hand formula you gave is obviously wrong? > x <- c(1, 2, 3) > y <- c(13.1, 11.9, 11.0) > M1 <- lm(y~x+0) > sqerr <- (y- predict(M1)) ^ 2 > sqtot <- (y - mean(y)) ^ 2 > 1 - sum(sqerr)/sum(sqtot) [1] -37.38707 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com<http://tibco.com>> -----Original Message----- > From: r-help-bounces@r-project.org<mailto:r-help-bounces@r-project.org> [mailto:r-help-bounces@r-project.org<mailto:r-help-bounces@r-project.org>] On > Behalf Of Pamela Krone-Davis > Sent: Friday, July 13, 2012 9:01 AM > To: r-help@r-project.org<mailto:r-help@r-project.org> > Subject: [R] R-squared with Intercept set to 0 (zero) for linear regression in R is > incorrect > > Hi, > > I have been using lm in R to do a linear regression and find the slope > coefficients and value for R-squared. The R-squared value reported by R > (R^2 = 0.9558) is very different than the R-squared value when I use the > same equation in Exce (R^2 = 0.328). I manually computed R-squared and the > Excel value is correct. I show my code for the determination of R^2 in R. > When I do not set 0 as the intercept, the R^2 value is the same in R and > Excel. In both cases the slope coefficient from R and from Excel are > identical. > > k is a data frame with two columns. > > M1 = lm(k[,1]~k[,2] + 0) ## set intercept to 0 and get different > R^2 values in R and Excel > M2 = lm(k[,1]~k[,2]) > sumM1 = summary(M1) > sumM2 = summary(M2) ## get same value as Excel when intercept is not > set to 0 > > Below is what R returns for sumM1: > > lm(formula = k[, 1] ~ k[, 2] + 0) > > Residuals: > Min 1Q Median 3Q Max > -0.057199 -0.015857 0.003793 0.013737 0.056178 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > k[, 2] 1.05022 0.04266 24.62 <2e-16 *** > --- > Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 > > Residual standard error: 0.02411 on 28 degrees of freedom > Multiple R-squared: 0.9558, Adjusted R-squared: 0.9543 > F-statistic: 606.2 on 1 and 28 DF, p-value: < 2.2e-16 > > Way manual determination was performed. The value returned coincides with > the value from Excel: > > #### trying to figure out why the R^2 for R and Excel are so different. > sqerr = (k[,1] - predict(M1))^2 > sqtot = (k[,1] - mean(k[,1]) ^2 > > R2 = 1 - sum(sqerr)/sum(sqtot) ## for 1D get 0.328 same as > excel value > > I am very puzzled by this. How does R compute the value for R^2 in this > case? Did i write the lm incorrectly? > > Thanks > Pam > > PS In case you are interested, the data I am using for hte two columns is > below. > > k[, 1] > 1] > [1] 0.17170228 0.10881539 0.11843669 0.11619201 0.08441067 0.09424441 > 0.04782264 0.09526496 0.11596476 0.10323453 0.06487894 0.08916484 > 0.06358752 0.07945473 > [15] 0.11213532 0.06531185 0.11503484 0.13679548 0.13762677 0.13126827 > 0.12350649 0.12842441 0.13075654 0.15026602 0.14536351 0.07841638 > 0.08419016 0.11995240 > [29] 0.14425678 > > > k[,2] > [1] 0.11 0.10 0.11 0.10 0.10 0.09 0.10 0.09 0.09 0.11 0.09 0.10 0.09 0.10 > 0.09 0.10 0.10 0.10 0.11 0.10 0.11 0.11 0.12 0.13 0.15 0.10 0.09 0.11 0.12 > > > -- > Pam Krone-Davis > Project Research Assistant and Grant Manager > PO Box 22122 > Carmel, CA 93922 > (831)582-3684<tel:%28831%29582-3684> (o) > (831)324-0391<tel:%28831%29324-0391> (h) > > [[alternative HTML version deleted]]-- Pam Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684<tel:%28831%29582-3684> (o) (831)324-0391<tel:%28831%29324-0391> (h) -- Pam Krone-Davis Project Research Assistant and Grant Manager PO Box 22122 Carmel, CA 93922 (831)582-3684 (o) (831)324-0391 (h) [[alternative HTML version deleted]]