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R - Pass column names as Variable with names contain I()
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!How to sort a dataframe by multiple column(s)Drop data frame columns by nameHow to interpret output from three-piece linear regression in RWhich linear model summary row corresponds to which term in formula?Pass offset arguments into lm functionGet p value and R squared values for Simple linear model by group in RHow to interpret and quantify coefficients of slope and intercept in segmented regressionHow to conduct linear hypothesis test on regression coefficients with a clustered covariance matrix?Coefficients of all Dummy Factor variables in GLMFast pairwise simple linear regression between variables in a data frame
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I'm performing the polynomial regression and testing the linear combination of the coefficient. But I'm running to some problems that when I tried to test the linear combination of the coefficient.
LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
summary(LnModel_1)
It output the values below:
Call:
lm(formula = PROF ~ UI_1 + UI_2 + I(UI_1^2) + UI_1:UI_2 + I(UI_2^2))
Residuals:
Min 1Q Median 3Q Max
-3.4492 -0.5405 0.1096 0.4226 1.7346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.66274 0.06444 72.354 < 2e-16 ***
UI_1 0.25665 0.07009 3.662 0.000278 ***
UI_2 0.25569 0.09221 2.773 0.005775 **
I(UI_1^2) -0.15168 0.04490 -3.378 0.000789 ***
I(UI_2^2) -0.08418 0.05162 -1.631 0.103643
UI_1:UI_2 -0.02849 0.05453 -0.522 0.601621
Then I use names(coef()) to extract the coefficient names
names(coef(LnModel_1))
output:
[1] "(Intercept)" "UI_1" "UI_2" "I(UI_1^2)"
"I(UI_2^2)""UI_1:UI_2"
For some reasons, when I use glht(), it give me an error on UI_2^2
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*I(UI_2^2) =0
") )
Output:
Error: multcomp:::expression2coef::walkCode::eval: within ‘UI_2^2’, the term
‘UI_2’ must not denote an effect. Apart from that, the term must evaluate to
a real valued constant
Don't know why it would give me this error message. How to input the I(UI_2^2) coefficient to the glht()
Thank you very much
r non-linear-regression coefficients hypothesis-test
asked Mar 9 at 6:01
cloudcloud
83
add a comment |
I'm performing the polynomial regression and testing the linear combination of the coefficient. But I'm running to some problems that when I tried to test the linear combination of the coefficient.
LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
summary(LnModel_1)
It output the values below:
Call:
lm(formula = PROF ~ UI_1 + UI_2 + I(UI_1^2) + UI_1:UI_2 + I(UI_2^2))
Residuals:
Min 1Q Median 3Q Max
-3.4492 -0.5405 0.1096 0.4226 1.7346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.66274 0.06444 72.354 < 2e-16 ***
UI_1 0.25665 0.07009 3.662 0.000278 ***
UI_2 0.25569 0.09221 2.773 0.005775 **
I(UI_1^2) -0.15168 0.04490 -3.378 0.000789 ***
I(UI_2^2) -0.08418 0.05162 -1.631 0.103643
UI_1:UI_2 -0.02849 0.05453 -0.522 0.601621
Then I use names(coef()) to extract the coefficient names
names(coef(LnModel_1))
output:
[1] "(Intercept)" "UI_1" "UI_2" "I(UI_1^2)"
"I(UI_2^2)""UI_1:UI_2"
For some reasons, when I use glht(), it give me an error on UI_2^2
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*I(UI_2^2) =0
") )
Output:
Error: multcomp:::expression2coef::walkCode::eval: within ‘UI_2^2’, the term
‘UI_2’ must not denote an effect. Apart from that, the term must evaluate to
a real valued constant
Don't know why it would give me this error message. How to input the I(UI_2^2) coefficient to the glht()
Thank you very much
r non-linear-regression coefficients hypothesis-test
asked Mar 9 at 6:01
cloudcloud
83
2
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32
add a comment |
I'm performing the polynomial regression and testing the linear combination of the coefficient. But I'm running to some problems that when I tried to test the linear combination of the coefficient.
LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
summary(LnModel_1)
It output the values below:
Call:
lm(formula = PROF ~ UI_1 + UI_2 + I(UI_1^2) + UI_1:UI_2 + I(UI_2^2))
Residuals:
Min 1Q Median 3Q Max
-3.4492 -0.5405 0.1096 0.4226 1.7346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.66274 0.06444 72.354 < 2e-16 ***
UI_1 0.25665 0.07009 3.662 0.000278 ***
UI_2 0.25569 0.09221 2.773 0.005775 **
I(UI_1^2) -0.15168 0.04490 -3.378 0.000789 ***
I(UI_2^2) -0.08418 0.05162 -1.631 0.103643
UI_1:UI_2 -0.02849 0.05453 -0.522 0.601621
Then I use names(coef()) to extract the coefficient names
names(coef(LnModel_1))
output:
[1] "(Intercept)" "UI_1" "UI_2" "I(UI_1^2)"
"I(UI_2^2)""UI_1:UI_2"
For some reasons, when I use glht(), it give me an error on UI_2^2
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*I(UI_2^2) =0
") )
Output:
Error: multcomp:::expression2coef::walkCode::eval: within ‘UI_2^2’, the term
‘UI_2’ must not denote an effect. Apart from that, the term must evaluate to
a real valued constant
Don't know why it would give me this error message. How to input the I(UI_2^2) coefficient to the glht()
Thank you very much
r non-linear-regression coefficients hypothesis-test
asked Mar 9 at 6:01
cloudcloud
83
I'm performing the polynomial regression and testing the linear combination of the coefficient. But I'm running to some problems that when I tried to test the linear combination of the coefficient.
LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
summary(LnModel_1)
It output the values below:
Call:
lm(formula = PROF ~ UI_1 + UI_2 + I(UI_1^2) + UI_1:UI_2 + I(UI_2^2))
Residuals:
Min 1Q Median 3Q Max
-3.4492 -0.5405 0.1096 0.4226 1.7346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.66274 0.06444 72.354 < 2e-16 ***
UI_1 0.25665 0.07009 3.662 0.000278 ***
UI_2 0.25569 0.09221 2.773 0.005775 **
I(UI_1^2) -0.15168 0.04490 -3.378 0.000789 ***
I(UI_2^2) -0.08418 0.05162 -1.631 0.103643
UI_1:UI_2 -0.02849 0.05453 -0.522 0.601621
Then I use names(coef()) to extract the coefficient names
names(coef(LnModel_1))
output:
[1] "(Intercept)" "UI_1" "UI_2" "I(UI_1^2)"
"I(UI_2^2)""UI_1:UI_2"
For some reasons, when I use glht(), it give me an error on UI_2^2
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*I(UI_2^2) =0
") )
Output:
Error: multcomp:::expression2coef::walkCode::eval: within ‘UI_2^2’, the term
‘UI_2’ must not denote an effect. Apart from that, the term must evaluate to
a real valued constant
Don't know why it would give me this error message. How to input the I(UI_2^2) coefficient to the glht()
Thank you very much
r non-linear-regression coefficients hypothesis-test
r non-linear-regression coefficients hypothesis-test
asked Mar 9 at 6:01
cloudcloud
83
asked Mar 9 at 6:01
cloudcloud
83
edited Mar 9 at 8:41
cloud
asked Mar 9 at 6:01
cloudcloud
83
asked Mar 9 at 6:01
cloudcloud
83
asked Mar 9 at 6:01
cloudcloud
83
83
2
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32
add a comment |
2
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32
2
2
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32
add a comment |
1 Answer
1
active
oldest
votes
The issue seems to be that I(UI^2) can be interpreted as an expression in R in the same fashion you did here LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
Therefore, you should indicate R that you want to evaluate a string inside your string:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*`I(UI_2^2)` =0
") )
Check my example (since I cannot reproduce your problem):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*`I(UI_2^2)` =0") )
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
Thank you so much. It works.
– cloud
Mar 11 at 9:04
add a comment |
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oldest
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votes
The issue seems to be that I(UI^2) can be interpreted as an expression in R in the same fashion you did here LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
Therefore, you should indicate R that you want to evaluate a string inside your string:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*`I(UI_2^2)` =0
") )
Check my example (since I cannot reproduce your problem):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*`I(UI_2^2)` =0") )
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
Thank you so much. It works.
– cloud
Mar 11 at 9:04
add a comment |
The issue seems to be that I(UI^2) can be interpreted as an expression in R in the same fashion you did here LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
Therefore, you should indicate R that you want to evaluate a string inside your string:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*`I(UI_2^2)` =0
") )
Check my example (since I cannot reproduce your problem):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*`I(UI_2^2)` =0") )
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
Thank you so much. It works.
– cloud
Mar 11 at 9:04
add a comment |
The issue seems to be that I(UI^2) can be interpreted as an expression in R in the same fashion you did here LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
Therefore, you should indicate R that you want to evaluate a string inside your string:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*`I(UI_2^2)` =0
") )
Check my example (since I cannot reproduce your problem):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*`I(UI_2^2)` =0") )
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
The issue seems to be that I(UI^2) can be interpreted as an expression in R in the same fashion you did here LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
Therefore, you should indicate R that you want to evaluate a string inside your string:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*`I(UI_2^2)` =0
") )
Check my example (since I cannot reproduce your problem):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*`I(UI_2^2)` =0") )
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
answered Mar 9 at 9:52
LocoGrisLocoGris
2,7481828
2,7481828
Thank you so much. It works.
– cloud
Mar 11 at 9:04
add a comment |
Thank you so much. It works.
– cloud
Mar 11 at 9:04
Thank you so much. It works.
– cloud
Mar 11 at 9:04
Thank you so much. It works.
– cloud
Mar 11 at 9:04
add a comment |
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2
Suggest you use edit facilities to add Minimal, Complete, and Verifiable example
– 42-
Mar 9 at 6:32