Lm Function in R Does Not Give Coefficients for All Factor Levels in Categorical Data

`lm` summary not display all factor levels

This issue is raised over and over again, but unfortunately no satisfying answer has been made which can be an appropriate duplicate target. Looks like I need to write one.

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Most people know this is related to "contrasts", but not everyone knows why it is needed, and how to understand its result. We have to look at model matrix in order to fully digest this.

Suppose we are interested in a model with two factors: ~ f + g (numerical covariates do not matter so I include none of them; the response does not appear in model matrix, so drop it, too). Consider the following reproducible example:

set.seed(0)

f <- sample(gl(3, 4, labels = letters[1:3]))
# [1] c a a b b a c b c b a c
#Levels: a b c

g <- sample(gl(3, 4, labels = LETTERS[1:3]))
# [1] A B A B C B C A C C A B
#Levels: A B C

We start with a model matrix with no contrasts at all:

X0 <- model.matrix(~ f + g, contrasts.arg = list(
                   f = contr.treatment(n = 3, contrasts = FALSE),
                   g = contr.treatment(n = 3, contrasts = FALSE)))

#   (Intercept) f1 f2 f3 g1 g2 g3
#1            1  0  0  1  1  0  0
#2            1  1  0  0  0  1  0
#3            1  1  0  0  1  0  0
#4            1  0  1  0  0  1  0
#5            1  0  1  0  0  0  1
#6            1  1  0  0  0  1  0
#7            1  0  0  1  0  0  1
#8            1  0  1  0  1  0  0
#9            1  0  0  1  0  0  1
#10           1  0  1  0  0  0  1
#11           1  1  0  0  1  0  0
#12           1  0  0  1  0  1  0

Note, we have:

unname( rowSums(X0[, c("f1", "f2", "f3")]) )
# [1] 1 1 1 1 1 1 1 1 1 1 1 1

unname( rowSums(X0[, c("g1", "g2", "g3")]) ) 
# [1] 1 1 1 1 1 1 1 1 1 1 1 1

So span{f1, f2, f3} = span{g1, g2, g3} = span{(Intercept)}. In this full specification, 2 columns are not identifiable. X0 will have column rank 1 + 3 + 3 - 2 = 5:

qr(X0)$rank
# [1] 5

So, if we fit a linear model with this X0, 2 coefficients out of 7 parameters will be NA:

y <- rnorm(12)  ## random `y` as a response
lm(y ~ X - 1)  ## drop intercept as `X` has intercept already

#X0(Intercept)           X0f1           X0f2           X0f3           X0g1  
#      0.32118        0.05039       -0.22184             NA       -0.92868  
#         X0g2           X0g3  
#     -0.48809             NA  

What this really implies, is that we have to add 2 linear constraints on 7 parameters, in order to get a full rank model. It does not really matter what these 2 constraints are, but there must be 2 linearly independent constrains. For example, we can do either of the following:

• drop any 2 columns from X0;
• add two sum-to-zero constrains on parameters, like we require coefficients for f1, f2 and f3 sum to 0, and the same for g1, g2 and g3.
• use regularization, for example, adding ridge penalty to f and g.

Note, these three ways end up with three different solutions:

• contrasts;
• constrained least squares;
• linear mixed models or penalized least squares.

The first two are still in the scope of fixed effect modelling. By "contrasts", we reduce the number of parameters until we get a full rank model matrix; while the other two does not reduce the number of parameters, but effectively reduces the effective degree of freedom.

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Now, you are certainly after the "contrasts" way. So, remember, we have to drop 2 columns. They can be

• one column from f and one column from g, giving to a model ~ f + g, with f and g contrasted;
• intercept, and one column from either f or g, giving to a model ~ f + g - 1.

Now you should be clear, that within the framework of dropping columns, there is no way you can get what you want, because you are expecting to drop only 1 column. The resulting model matrix will still be rank-deficient.

If you really want to have all coefficients there, use constrained least squares, or penalized regression / linear mixed models.

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Now, when we have interaction of factors, things are more complicated but the idea is still the same. But given that my answer is already long enough, I don't want to continue.

R's lm() function removes factor levels (aside from intercept estimator) without reporting error or warning

Yes your missingness is probably causing your results.

Consider the following reproducible example.

x1 <- sample(1:5, 1000, replace=T)  
x2 <- sample(1:3, 1000, replace=T)
y <- 2*x1 + 3*x2 + rnorm(1000)
#no missings (everything works fine)
lm(y~as.factor(x1) + as.factor(x2))
lm(y~ -1 + as.factor(x1) + as.factor(x2)) # no intercept
#with missings
x1[x2==2]<-NA #create specific missingness
table(x1,useNA = "always")
table(x2,useNA = "always")
table(x1,x2,useNA = "always") #you see the missing pattern
lm(y~ as.factor(x1) + as.factor(x2))
lm(y~ -1 + as.factor(x1) + as.factor(x2))

If you run the code, you will see that for factor x2 two categories are omitted. The first is the ref.cat. the second (x2)2 because of the missings.

Coefficients for all levels of a categorical factor unchanged in lmer after adding variables

This is apparently about statistics, not programming. Consider asking it on Cross Validated instead.

It seems that the answer to your question lies in the way you set up your example data. An additional control/predictor variable X2 only influences estimates of another, previously included predictor X1 they are (at least a bit) correlated. In reality, this is mostly true, as you hardly get r = .00 correlations in real-life data. But you set up your data in a way such that treatment is orthogonal to control_a and control_b. Therefore, including either control does not impact the coefficients of the treatment dummies.

Multiple Linear Regression using lm() in r output with categorical variable is incomplete?

i think that your problem is because when you adjust a linear regression with a categorical variable, one of the categories will be used as a "reference" and its value will be represented (be part) in the "intercept".

So, to see the coefficient for "four doors" you should change the base reference of your variable. You can do this with:

train_sample$num.of.doors = relevel(train_sample$num.of.doors, ref=2)

Keep in mind that with this change, "two doors" will now be the base reference.

Comparing all factor levels to the grand mean: can I tweak contrasts in linear model fitting to show all levels?

This answer shows you how to obtain the following coefficient table:

#               Estimate Std. Error     t value  Pr(>|t|)
#(Intercept) -0.02901982  0.4680937 -0.06199574 0.9544655
#A           -0.19238543  0.6619845 -0.29061922 0.7902750
#B            0.40884591  0.6619845  0.61760645 0.5805485
#C           -0.21646049  0.6619845 -0.32698723 0.7651640

Amazing, isn't it? It mimics what you see from summary(fit), i.e.,

#               Estimate Std. Error     t value  Pr(>|t|)
#(Intercept) -0.02901982  0.4680937 -0.06199574 0.9544655
#x1          -0.19238543  0.6619845 -0.29061922 0.7902750
#x2           0.40884591  0.6619845  0.61760645 0.5805485

But now we have all factor levels displayed.

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Why lm summary does not display all factor levels?

In 2016, I answered this Stack Overflow question: `lm` summary not display all factor levels and since then, it has become the target for marking duplicated questions on similar topics.

To recap, the basic idea is that in order to have a full-rank design matrix for least squares fitting, we must apply contrasts to a factor variable. Let's say that the factor has N levels, then no matter what type of contrasts we choose (see ?contrasts for a list), it reduces the raw N dummy variables to a new set of N - 1 variables. Therefore, only N - 1 coefficients are associated with an N-level factor.

However, we can transform the N - 1 coefficients back to the original N coefficients using the contrasts matrix. The transformation enables us to obtain a coefficient table for all factor levels. I will now demonstrate how to do this, based on OP's reproducible example:

set.seed(1)
y <- rnorm(6, 0, 1)
x <- factor(rep(LETTERS[1:3], each = 2))
fit <- lm(y ~ x, contrasts = list(x = contr.sum))

In this example, the sum-to-zero contrast is applied to factor x. To know more on how to control contrasts for model fitting, see my answer at How to set contrasts for my variable in regression analysis with R?.

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R code walk-through

For a factor variable of N levels subject to sum-to-zero contrasts, we can use the following function to get the N x (N - 1) transformation matrix that maps the (N - 1) coefficients estimated by lm back to the N coefficients for all levels.

ContrSumMat <- function (fctr, sparse = FALSE) {
  if (!is.factor(fctr)) stop("'fctr' is not a factor variable!")
  N <- nlevels(fctr)
  Cmat <- contr.sum(N, sparse = sparse)
  dimnames(Cmat) <- list(levels(fctr), seq_len(N - 1))
  Cmat
}

For the example 3-level factor x, this matrix is:

Cmat <- ContrSumMat(x)
#   1  2
#A  1  0
#B  0  1
#C -1 -1

The fitted model fit reports 3 - 1 = 2 coefficients for this factor. We can extract them as:

## coefficients After Contrasts
coef_ac <- coef(fit)[2:3]
#        x1         x2 
#-0.1923854  0.4088459 

Therefore, the level-specific coefficients are:

## coefficients Before Contrasts
coef_bc <- (Cmat %*% coef_ac)[, 1]
#         A          B          C 
#-0.1923854  0.4088459 -0.2164605 

## note that they sum to zero as expected
sum(coef_bc)
#[1] 0

Similarly, we can get the covariance matrix after contrasts:

var_ac <- vcov(fit)[2:3, 2:3]
#           x1         x2
#x1  0.4382235 -0.2191118
#x2 -0.2191118  0.4382235

and transform it to the one before contrasts:

var_bc <- Cmat %*% var_ac %*% t(Cmat)
#           A          B          C
#A  0.4382235 -0.2191118 -0.2191118
#B -0.2191118  0.4382235 -0.2191118
#C -0.2191118 -0.2191118  0.4382235

Interpretation:

• The model intercept coef(fit)[1] is the grand mean.

• The computed coef_bc is the deviation of each level's mean from the grand mean.

• The diagonal entries of var_bc gives the estimated variance of these deviations.

We can then proceed to compute t-statistics and p-values for these coefficients, as follows.

## standard error of point estimate `coef_bc`
std.error_bc <- sqrt(diag(var_bc))
#        A         B         C 
#0.6619845 0.6619845 0.6619845 

## t-statistics (Null Hypothesis: coef_bc = 0)
t.stats_bc <- coef_bc / std.error_bc
#         A          B          C 
#-0.2906192  0.6176065 -0.3269872 

## p-values of the t-statistics
p.value_bc <- 2 * pt(abs(t.stats_bc), df = fit$df.residual, lower.tail = FALSE)
#        A         B         C 
#0.7902750 0.5805485 0.7651640 

## construct a coefficient table that mimics `coef(summary(fit))`
stats.tab_bc <- cbind("Estimate" = coef_bc,
                      "Std. Error" = std.error_bc,
                      "t value" = t.stats_bc,
                      "Pr(>|t|)" = p.value_bc)
#    Estimate Std. Error    t value  Pr(>|t|)
#A -0.1923854  0.6619845 -0.2906192 0.7902750
#B  0.4088459  0.6619845  0.6176065 0.5805485
#C -0.2164605  0.6619845 -0.3269872 0.7651640

We can also augment it by including the result for the grand mean (i.e., the model intercept).

## extract statistics of the intercept
intercept.stats <- coef(summary(fit))[1, , drop = FALSE]
#               Estimate Std. Error     t value  Pr(>|t|)
#(Intercept) -0.02901982  0.4680937 -0.06199574 0.9544655

## augment the coefficient table
stats.tab <- rbind(intercept.stats, stats.tab_bc)
#               Estimate Std. Error     t value  Pr(>|t|)
#(Intercept) -0.02901982  0.4680937 -0.06199574 0.9544655
#A           -0.19238543  0.6619845 -0.29061922 0.7902750
#B            0.40884591  0.6619845  0.61760645 0.5805485
#C           -0.21646049  0.6619845 -0.32698723 0.7651640

We can also print this table with significance stars.

printCoefmat(stats.tab)
#            Estimate Std. Error t value Pr(>|t|)
#(Intercept) -0.02902    0.46809 -0.0620   0.9545
#A           -0.19239    0.66199 -0.2906   0.7903
#B            0.40885    0.66199  0.6176   0.5805
#C           -0.21646    0.66199 -0.3270   0.7652

Emm? Why are there no stars? Well, in this example all p-values are very large. The stars will show up if p-values are small. Here is a convincing demo:

fake.tab <- stats.tab
fake.tab[, 4] <- fake.tab[, 4] / 100
printCoefmat(fake.tab)
#            Estimate Std. Error t value Pr(>|t|)   
#(Intercept) -0.02902    0.46809 -0.0620 0.009545 **
#A           -0.19239    0.66199 -0.2906 0.007903 **
#B            0.40885    0.66199  0.6176 0.005805 **
#C           -0.21646    0.66199 -0.3270 0.007652 **
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Oh, this is so beautiful. For the meaning of these stars, see my answer at: Interpeting R significance codes for ANOVA table?

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Closing Remarks

It should be possible to write a function (or even an R package) to perform such table transformation. However, it might take great effort to make such function flexible enough, to handle:

• all type of contrasts (this is easy to do);

• complicated model terms, like interaction between a factor and other numeric/factor variables (this seems really involving!!).

So, I will stop here for the moment.

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Miscellaneous Replies

Are the model scores that I get from the lm's summary still accurate, even though it isn't displaying all levels of the factor?

Yes, they are. lm conducts accurate least squares fitting.

In addition, the transformation of coefficient table does not affect R-squares, degree of freedom, residuals, fitted values, F-statistics, ANOVA table, etc.

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