Fitting different splines in R (Cubic, Natural, Smoothing)










1














So I'm trying to fit a cubic, natural, and smoothing spline to the Auto dataset from the ISLR package. I'm having some trouble and am getting some warning/error messages which makes me think there is something wrong with my data or a matrix that I created.



What is really confusing is how this basic command throws an error.



natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(25, 50, 75)), data = Auto)



Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call
(arg 1)




There are additional errors/warnings in my code but the thing is: I had essentially copied the code from somewhere and I also ran it, which it worked for the Carseats dataset and modified it to change the variables to match the Auto dataset. This is why it is confusing me. I'm not understanding why I get errors for the Auto dataset but not the Carseats dataset. Does anyone have some insight?










share|improve this question




























    1














    So I'm trying to fit a cubic, natural, and smoothing spline to the Auto dataset from the ISLR package. I'm having some trouble and am getting some warning/error messages which makes me think there is something wrong with my data or a matrix that I created.



    What is really confusing is how this basic command throws an error.



    natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(25, 50, 75)), data = Auto)



    Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call
    (arg 1)




    There are additional errors/warnings in my code but the thing is: I had essentially copied the code from somewhere and I also ran it, which it worked for the Carseats dataset and modified it to change the variables to match the Auto dataset. This is why it is confusing me. I'm not understanding why I get errors for the Auto dataset but not the Carseats dataset. Does anyone have some insight?










    share|improve this question


























      1












      1








      1







      So I'm trying to fit a cubic, natural, and smoothing spline to the Auto dataset from the ISLR package. I'm having some trouble and am getting some warning/error messages which makes me think there is something wrong with my data or a matrix that I created.



      What is really confusing is how this basic command throws an error.



      natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(25, 50, 75)), data = Auto)



      Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call
      (arg 1)




      There are additional errors/warnings in my code but the thing is: I had essentially copied the code from somewhere and I also ran it, which it worked for the Carseats dataset and modified it to change the variables to match the Auto dataset. This is why it is confusing me. I'm not understanding why I get errors for the Auto dataset but not the Carseats dataset. Does anyone have some insight?










      share|improve this question















      So I'm trying to fit a cubic, natural, and smoothing spline to the Auto dataset from the ISLR package. I'm having some trouble and am getting some warning/error messages which makes me think there is something wrong with my data or a matrix that I created.



      What is really confusing is how this basic command throws an error.



      natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(25, 50, 75)), data = Auto)



      Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call
      (arg 1)




      There are additional errors/warnings in my code but the thing is: I had essentially copied the code from somewhere and I also ran it, which it worked for the Carseats dataset and modified it to change the variables to match the Auto dataset. This is why it is confusing me. I'm not understanding why I get errors for the Auto dataset but not the Carseats dataset. Does anyone have some insight?







      r spline






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      share|improve this question













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      share|improve this question








      edited Nov 11 '18 at 19:07









      jogo

      9,83692135




      9,83692135










      asked Nov 11 '18 at 18:52









      Leon

      82




      82






















          1 Answer
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          0














          The problem that you have is that you are defining the knots outside the range of the predictor variable. Here is a basic code that will work (I just defined knots that are within the range of the variable mpg).



          x <- ISLR::Auto

          natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(10,20,30,40)), data = x)
          summary(natural.splines.fit)


          I believe that you are trying to place the knots for the 25th, 50th, and 75th percentile, so I recommend first getting the values corresponding to those locations and then fitting the model.
          Here is how I did it



          target_quantiles <- unname(quantile(x$mpg, probs = c(0.25,0.5,0.75)))

          natural.splines.fit2 <- lm(horsepower ~ ns(mpg, knots = target_quantiles), data = x)
          summary(natural.splines.fit2)





          share|improve this answer




















          • Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
            – Leon
            Nov 11 '18 at 19:24










          • Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
            – Harro Cyranka
            Nov 11 '18 at 19:27










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          1 Answer
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          1 Answer
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          active

          oldest

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          active

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          0














          The problem that you have is that you are defining the knots outside the range of the predictor variable. Here is a basic code that will work (I just defined knots that are within the range of the variable mpg).



          x <- ISLR::Auto

          natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(10,20,30,40)), data = x)
          summary(natural.splines.fit)


          I believe that you are trying to place the knots for the 25th, 50th, and 75th percentile, so I recommend first getting the values corresponding to those locations and then fitting the model.
          Here is how I did it



          target_quantiles <- unname(quantile(x$mpg, probs = c(0.25,0.5,0.75)))

          natural.splines.fit2 <- lm(horsepower ~ ns(mpg, knots = target_quantiles), data = x)
          summary(natural.splines.fit2)





          share|improve this answer




















          • Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
            – Leon
            Nov 11 '18 at 19:24










          • Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
            – Harro Cyranka
            Nov 11 '18 at 19:27















          0














          The problem that you have is that you are defining the knots outside the range of the predictor variable. Here is a basic code that will work (I just defined knots that are within the range of the variable mpg).



          x <- ISLR::Auto

          natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(10,20,30,40)), data = x)
          summary(natural.splines.fit)


          I believe that you are trying to place the knots for the 25th, 50th, and 75th percentile, so I recommend first getting the values corresponding to those locations and then fitting the model.
          Here is how I did it



          target_quantiles <- unname(quantile(x$mpg, probs = c(0.25,0.5,0.75)))

          natural.splines.fit2 <- lm(horsepower ~ ns(mpg, knots = target_quantiles), data = x)
          summary(natural.splines.fit2)





          share|improve this answer




















          • Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
            – Leon
            Nov 11 '18 at 19:24










          • Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
            – Harro Cyranka
            Nov 11 '18 at 19:27













          0












          0








          0






          The problem that you have is that you are defining the knots outside the range of the predictor variable. Here is a basic code that will work (I just defined knots that are within the range of the variable mpg).



          x <- ISLR::Auto

          natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(10,20,30,40)), data = x)
          summary(natural.splines.fit)


          I believe that you are trying to place the knots for the 25th, 50th, and 75th percentile, so I recommend first getting the values corresponding to those locations and then fitting the model.
          Here is how I did it



          target_quantiles <- unname(quantile(x$mpg, probs = c(0.25,0.5,0.75)))

          natural.splines.fit2 <- lm(horsepower ~ ns(mpg, knots = target_quantiles), data = x)
          summary(natural.splines.fit2)





          share|improve this answer












          The problem that you have is that you are defining the knots outside the range of the predictor variable. Here is a basic code that will work (I just defined knots that are within the range of the variable mpg).



          x <- ISLR::Auto

          natural.splines.fit <- lm(horsepower ~ ns(mpg, knots = c(10,20,30,40)), data = x)
          summary(natural.splines.fit)


          I believe that you are trying to place the knots for the 25th, 50th, and 75th percentile, so I recommend first getting the values corresponding to those locations and then fitting the model.
          Here is how I did it



          target_quantiles <- unname(quantile(x$mpg, probs = c(0.25,0.5,0.75)))

          natural.splines.fit2 <- lm(horsepower ~ ns(mpg, knots = target_quantiles), data = x)
          summary(natural.splines.fit2)






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 11 '18 at 19:01









          Harro Cyranka

          1,1601513




          1,1601513











          • Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
            – Leon
            Nov 11 '18 at 19:24










          • Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
            – Harro Cyranka
            Nov 11 '18 at 19:27
















          • Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
            – Leon
            Nov 11 '18 at 19:24










          • Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
            – Harro Cyranka
            Nov 11 '18 at 19:27















          Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
          – Leon
          Nov 11 '18 at 19:24




          Oh, so the values I put in the "knots" command is the actual value? I thought they represented quantiles. Thanks for clearing that up.
          – Leon
          Nov 11 '18 at 19:24












          Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
          – Harro Cyranka
          Nov 11 '18 at 19:27




          Yes. You have to define the breakpoints as values of the variable. Please, if that solved your issue, would you mind upvoting and checkmarking the answer?
          – Harro Cyranka
          Nov 11 '18 at 19:27

















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