RTrain3k - 1 year ago 89
R Question

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

I am using

`plyr::ddply`
to run a regression model

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

by the factor
`resp.id`
. 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
`NaN`
.

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?

Thanks.

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.

``````library(plyr)

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.

``````set.seed(0);
x1 <- rnorm(3); x2 <- rnorm(3); x3 <- rnorm(3)
y <- rnorm(3)

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

coef(summary(fit))
#             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
``````
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