RTrain3k RTrain3k - 8 months ago 53
R Question

Fitting a linear regression model by group gives NaN p-values

I am using

to run a regression model

model <- rating ~ A + B + C + D + E + F

by the factor
. I can create a data frame of the betas by each factor with:

indiv.betas <- ddply(data.coded, "resp.id",
function(df) coef(lm(model, data=df)))

I am now trying to extract the p-values for the variables by the factor using:

indiv.pvalues <- ddply(data.coded, "resp.id",
function(df) coef(summary(lm(model, data=df)))[, "Pr(>|t|)"])

Unfortunately, it just gives me a data frame with

Although, if I run a model across the entire data set, I can extract p-values from this one model as a data frame successfully with:

pvalue <- as.data.frame(coef(summary(lm(model, data=df)))[, "Pr(>|t|)"])

How can I create a data frame of the p-values by the factor?



When you fit a single model

rating ~ A + B + C + D + E + F

you get meaningful, non-NA result. While when you fit the same model for each subset / factor level by resp.id, you get NaN result. I am 100% sure that for some factor level, you don't have enough data to fit the above model. It would be a good idea, to first check how many data there are for each group. You can use:

N <- with(data.coded, tapply(rating, resp.id, FUN = length))

Your model has 7 coefficients (1 for intercept and 1 each for A, B, ..., F). So which(N < 7) will tell you which factor levels are producing NaN.

In this part, I will show that I am not able to reproduce your problem with iris dataset.


model <- Sepal.Length ~ Sepal.Width + Petal.Length + Petal.Width

ddply(iris, "Species", function(df) coefficients(lm(model, data=df)))

#     Species (Intercept) Sepal.Width Petal.Length Petal.Width
#1     setosa    2.351890   0.6548350    0.2375602   0.2521257
#2 versicolor    1.895540   0.3868576    0.9083370  -0.6792238
#3  virginica    0.699883   0.3303370    0.9455356  -0.1697527

ddply(iris, "Species", function(df) coef(summary(lm(model, data=df)))[, 4])

#     Species  (Intercept)  Sepal.Width Petal.Length Petal.Width
#1     setosa 3.034183e-07 6.834434e-09 2.593594e-01    0.470987
#2 versicolor 5.112246e-04 6.488965e-02 1.666695e-06    0.125599
#3  virginica 1.961563e-01 6.439972e-02 1.074269e-13    0.395875

In this part, I will show why NaN could appear when there are more coefficients than data.

x1 <- rnorm(3); x2 <- rnorm(3); x3 <- rnorm(3)
y <- rnorm(3)

fit <- lm(y ~ x1 + x2 + x3)  ## 3 data, 4 coefficients

#             Estimate Std. Error t value Pr(>|t|)
#(Intercept) 0.4217653        NaN     NaN      NaN
#x1          0.4124869        NaN     NaN      NaN
#x2          1.1489330        NaN     NaN      NaN