Creating Contingency Tables: "xtabs" command in R

I am working with the R programming table. Suppose I have the following data:

set.seed(123)

v1 <- c("2010-2011","2011-2012", "2012-2013", "2013-2014", "2014-2015") 
v2 <- c("A", "B", "C", "D", "E")
v3 <- c("Z", "Y", "X", "W" )

data_1 = data.frame(var_1 = rnorm(871, 10,10), var_2 = rnorm(871, 5,5))

data_1$dates <- as.factor(sample(v1, 871, replace=TRUE, prob=c(0.5, 0.2, 0.1, 0.1, 0.1)))

data_1$types <- as.factor(sample(v2, 871, replace=TRUE, prob=c(0.3, 0.2, 0.1, 0.1, 0.1)))

data_1$types2 <- as.factor(sample(v3, 871, replace=TRUE, prob=c(0.3, 0.5, 0.1, 0.1)))


data_2 = data.frame(var_1 = rnorm(412, 10,10), var_2 = rnorm(412, 5,5))

data_2$dates <- as.factor(sample(v1, 412, replace=TRUE, prob=c(0.5, 0.2, 0.1, 0.1, 0.1)))

data_2$types <- as.factor(sample(v2, 412, replace=TRUE, prob=c(0.3, 0.2, 0.1, 0.1, 0.1)))

data_2$types2 <- as.factor(sample(v3, 412, replace=TRUE, prob=c(0.3, 0.5, 0.1, 0.1)))

data_3 = data.frame(var_1 = rnorm(332, 10,10), var_2 = rnorm(332, 5,5))

data_3$dates <- as.factor(sample(v1, 332, replace=TRUE, prob=c(0.5, 0.2, 0.1, 0.1, 0.1)))

data_3$types <- as.factor(sample(v2, 332, replace=TRUE, prob=c(0.3, 0.2, 0.1, 0.1, 0.1)))

data_3$types2 <- as.factor(sample(v3, 332, replace=TRUE, prob=c(0.3, 0.5, 0.1, 0.1)))

data_1 <- data.frame(name="data_1", data_1)
data_2 <- data.frame(name="data_2", data_2)
data_3 <- data.frame(name="data_3", data_3)
my_data <- rbind(data_1, data_2, data_3)

I am trying to make a contingency table with this data ("my_data")

summary <- xtabs(~name+types+types, my_data)
ftable(summary, row.vars=1, col.vars=2:3)

However, this returns the following error:

Error in ftable.default(summary, row.vars = 1, col.vars = 2:3) : 
  incorrect specification for 'col.vars'

Does anyone know why this error is coming? Can someone please help me fix this?

Thanks!

Did you mean types+types2?

Otherwise, you have only 2 variables in summary, you can't select 3 of them.

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