使用二项分布执行统计分析
Perform statistical analysis using binomial distribution
我正在尝试使用二项分布来测试 "random" 模型是否只是随机响应 "virginica" 50% 的时间,"setosa" 25% 的时间和 "versicolor" 最后 25% 的时间来查看我的逻辑回归模型是否更准确,反之亦然。这可以做到吗?这是我的尝试...
library(datasets)
iris$dummy_virginica_iris <- 0
iris$dummy_virginica_iris[iris$Species == 'virginica'] <- 1
iris$dummy_virginica_iris
# Logistic regression model.
glm <- glm(dummy_virginica_iris ~ Petal.Width + Sepal.Width,
data = iris,
family = 'binomial')
summary(glm)
# Classifer.
glm.pred <- predict(glm, type="response")
virginica <- ifelse(glm.pred > .5, TRUE, FALSE)
table(iris$Species, virginica)
# Table of predictions.
table(virginica, iris$dummy_virginica_iris)
# Binomial distribution??
rbinom(160, 1, 0.5)
您可以使用 sample
来执行此操作:
set.seed(1)
rando <- sample(c('virginica', 'setosa', 'versicolor'), # vector of possible responses
prob = c(1/2, 1/4, 1/4), # probabilities of those responses
size = length(virginica), # number of responses desired
replace = TRUE) # specify sampling with replacement
table(rando, iris$dummy_virginica_iris)
rando 0 1
setosa 27 8
versicolor 21 18
virginica 52 24
rando_virginica <- ifelse(rando == 'virginica', TRUE, FALSE)
table(rando_virginica, iris$dummy_virginica_iris)
rando_virginica 0 1
FALSE 48 26
TRUE 52 24
我正在尝试使用二项分布来测试 "random" 模型是否只是随机响应 "virginica" 50% 的时间,"setosa" 25% 的时间和 "versicolor" 最后 25% 的时间来查看我的逻辑回归模型是否更准确,反之亦然。这可以做到吗?这是我的尝试...
library(datasets)
iris$dummy_virginica_iris <- 0
iris$dummy_virginica_iris[iris$Species == 'virginica'] <- 1
iris$dummy_virginica_iris
# Logistic regression model.
glm <- glm(dummy_virginica_iris ~ Petal.Width + Sepal.Width,
data = iris,
family = 'binomial')
summary(glm)
# Classifer.
glm.pred <- predict(glm, type="response")
virginica <- ifelse(glm.pred > .5, TRUE, FALSE)
table(iris$Species, virginica)
# Table of predictions.
table(virginica, iris$dummy_virginica_iris)
# Binomial distribution??
rbinom(160, 1, 0.5)
您可以使用 sample
来执行此操作:
set.seed(1)
rando <- sample(c('virginica', 'setosa', 'versicolor'), # vector of possible responses
prob = c(1/2, 1/4, 1/4), # probabilities of those responses
size = length(virginica), # number of responses desired
replace = TRUE) # specify sampling with replacement
table(rando, iris$dummy_virginica_iris)
rando 0 1
setosa 27 8
versicolor 21 18
virginica 52 24
rando_virginica <- ifelse(rando == 'virginica', TRUE, FALSE)
table(rando_virginica, iris$dummy_virginica_iris)
rando_virginica 0 1
FALSE 48 26
TRUE 52 24