How to Compare Two Factors with Different Levels

How can I compare two factors with different levels?

Convert to character then compare:

# data
A <- factor(1:5)
B <- factor(c(1:3,6,6))

str(A)
# Factor w/ 5 levels "1","2","3","4",..: 1 2 3 4 5
str(B)
# Factor w/ 4 levels "1","2","3","6": 1 2 3 4 4

mean(A == B)

Error in Ops.factor(A, B) : level sets of factors are different

mean(as.character(A) == as.character(B))
# [1] 0.6

Or another approach would be

mean(levels(A)[A] == levels(B)[B])

which is 2 times slower on a 1e8 dataset.

Compare the levels of two factors

If factor1 and factor2 are your two factors, just look at levels(factor1) and levels(factor2).

Same number of levels:

length(levels(factor1)) == length(levels(factor2))

Values in one and not the other:

setdiff(levels(factor1), levels(factor2))
setdiff(levels(factor2), levels(factor1))

R - Using If statement to compare factors with different levels

You could convert them to characters for the comparison. However, if you want to compare all of the rows you'll probably want to use ifelse:

ifelse(as.character(z$x) == as.character(z$y), 1, 0)

How to compare two R data frames to find missing factor-level?

Just take the set difference between the levels of the two factors.

F1 = factor(c('A', 'B', 'C'))
F2 = factor(c('B', 'C'))

setdiff(levels(F1), levels(F2))
 [1] "A"

Compare factor levels in R

I think you are looking for table function:

> table(a1, a2)
       a2
a1      [1,3] (3,4]
  [1,2]     4     0
  (2,3]     2     0
  (3,4]     0     3

Take difference between two levels of factor variable while retaining other factor variables in R

One option with dplyr would be

library(dplyr)

my.df %>% 
  group_by(Gene, Population) %>% 
  summarize(Coverage = Coverage[Color == "Blue"] - Coverage[Color == "Green"])

# A tibble: 4 x 3
# Groups:   Gene [?]
#   Gene  Population Coverage
#   <fct> <fct>         <dbl>
# 1 A_1   PopA       -0.00600
# 2 A_1   PopB       -0.420  
# 3 A_2   PopA       -0.01   
# 4 A_2   PopB        0.100 

Data

my.df <- 
  structure(list(Gene = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("A_1", "A_2"), class = "factor"), 
                 Population = structure(c(1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c("PopA", "PopB"), class = "factor"), 
                 Color = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c("Blue", "Green"), class = "factor"), 
                 Coverage = c(0.016, 0.022, 0.1322, 0.552, 0.13, 0.14, 1, 0.9)), class = "data.frame", row.names = c(NA, -8L))

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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