fit1 <- eulerr(c("A" = 1, "B" = 0.4, "C" = 3, "A&B" = 0.2))
# Same result as above
fit2 <- eulerr(c("A" = 1, "B" = 0.4, "C" = 3,
"A&B" = 0.2, "A&C" = 0, "B&C" = 0,
"A&B&C" = 0) )
# Using the matrix method
mat <- cbind(A = sample(c(TRUE, TRUE, FALSE), size = 50, replace = TRUE),
B = sample(c(TRUE, FALSE), size = 50, replace = TRUE))
fit3 <- eulerr(mat)
# Using grouping via the 'by' argument
dat <- data.frame(
A = sample(c(TRUE, FALSE), size = 100, replace = TRUE),
B = sample(c(TRUE, TRUE, FALSE), size = 100, replace = TRUE),
gender = sample(c("Men", "Women"), size = 100, replace = TRUE),
nation = sample(c("Sweden", "Denmark"), size = 100, replace = TRUE)
)
fit4 <- eulerr(dat[, 1:2], by = dat[, 3:4])
# A set with no perfect solution
rel <- c("a" = 3491, "b" = 3409, "c" = 3503,
"a&b" = 120, "a&c" = 114, "b&c" = 132, "a&b&c" = 126)
# Use the cost function from eulerAPE (the default)
fit5 <- eulerr(rel, cost = "eulerAPE")
# Use the stress function from venneuler
fit6 <- eulerr(rel, cost = "venneuler")
par(mfrow = c(1, 2))
plot(fit5); plot(fit6)
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