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DF <- data.frame(factor1=rep(1:4,1000), factor2 = rep(1:4,each=1000),base = rnorm(4000,0,1),dep=rnorm(4000,400,5))

DF$f1_1 = DF$factor1 == 1
DF$f1_2 = DF$factor1 == 2
DF$f1_3 = DF$factor1 == 3
DF$f1_4 = DF$factor1 == 4

DF$f2_1 = DF$factor2 == 1
DF$f2_2 = DF$factor2 == 2
DF$f2_3 = DF$factor2 == 3
DF$f2_4 = DF$factor2 == 4

I want to run the following regression:

Dep = (f1_1 + f1_2 + f1_3 + f1_4)*(f2_1 + f2_2 + f2_3 + f2_4)*(base+base^2+base^3+base^4+base^5)

Is there a smarter way to do it?

zx8754
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Leonhardt Guass
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2 Answers2

2

You should code factor1 and factor2 as real factor variables. Also, it is better to use poly for polynomials. Here is what we can do:

DF <- data.frame(factor1=rep(1:4,1000), factor2 = rep(1:4,each=1000),
                 base = rnorm(4000,0,1), dep = rnorm(4000,400,5))

DF$factor1 <- as.factor(DF$factor1)
DF$factor2 <- as.factor(DF$factor2)

fit <- lm(dep ~ factor1 * factor2 * poly(base, degree = 5))

By default, poly generates orthogonal basis for numerical stability. If you want ordinary polynomials like base + base ^ 2 + base ^ 3 + ..., use poly(base, degree = 5, raw = TRUE).

Be aware, you will get lots of parameters from this model, as you are fitting a fifth order polynomial for each pair of levels between factor1 and factor2.

Consider a small example.

set.seed(0)
f1 <- sample(gl(3, 20, labels = letters[1:3]))    ## randomized balanced factor
f2 <- sample(gl(3, 20, labels = LETTERS[1:3]))    ## randomized balanced factor
x <- runif(3 * 20)  ## numerical covariate
y <- rnorm(3 * 20)  ## toy response

fit <- lm(y ~ f1 * f2 * poly(x, 2))

#Call:
#lm(formula = y ~ f1 * f2 * poly(x, 2))
#
#Coefficients:
#        (Intercept)                  f1b                  f1c  
#            -0.5387               0.8776               0.1572  
#                f2B                  f2C          poly(x, 2)1  
#             0.5113               1.0139               5.8345  
#        poly(x, 2)2              f1b:f2B              f1c:f2B  
#             2.4373               1.0666               0.1372  
#            f1b:f2C              f1c:f2C      f1b:poly(x, 2)1  
#            -1.4951              -1.4601              -6.2338  
#    f1c:poly(x, 2)1      f1b:poly(x, 2)2      f1c:poly(x, 2)2  
#           -11.0760              -2.3668               1.9708  
#    f2B:poly(x, 2)1      f2C:poly(x, 2)1      f2B:poly(x, 2)2  
#            -3.7127              -5.8253               5.6227  
#    f2C:poly(x, 2)2  f1b:f2B:poly(x, 2)1  f1c:f2B:poly(x, 2)1  
#            -7.3582              20.9179              11.6270  
#f1b:f2C:poly(x, 2)1  f1c:f2C:poly(x, 2)1  f1b:f2B:poly(x, 2)2  
#             1.2897              11.2041              12.8096  
#f1c:f2B:poly(x, 2)2  f1b:f2C:poly(x, 2)2  f1c:f2C:poly(x, 2)2  
#            -9.8476              10.6664               4.5582

Note, even for 3 factor levels each and a 3rd order polynomial, we already end up with great number of coefficients.

Zheyuan Li
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0

Using I () forces the formula to treat +-×/ as arithmetic rather than model operators. Example: lm (y ~ I (x1 +x2))

PeterPancake
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