jon*_*nas 1 r function summary dataframe dplyr
我想从一个比例来引导置信区间data.frame.我想在我的一个列中获得变量的结果.我已设法为矢量执行引导程序,但不知道如何data.frame从此处将其扩展到a .一个简化示例,将阈值设置为10并查看数据中小于10的比例.
矢量解决方案
library(boot)
vec <- abs(rnorm(1000)*10) #generate example vector
data_to_tb <- vec
tb <- function(data) {
sum(data < 10, na.rm = FALSE)/length(data) #function for generating the proportion
}
tb(data_to_tb)
boot.out <- boot(data = data_to_tb, function(u,i) tb(u[i]), R = 999)
quantile(boot.out$t, c(.025,.975))
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从这里开始,我想对data.frame包含两列的内容做同样的事情.data.frame如果可能的话,我希望以" (x,样本,比例,CI)列的形式返回结果" :
x n proportion CI
A xx xx xx
B xx xx xx
C xx xx xx
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如果dplyr可以使用包装会更好.以下是我的数据的简化示例:
例:
dataframe <- data.frame(x = sample(c("A","B","C"),100,replace = TRUE), vec =abs(rnorm(100)*10))
head(dataframe)
## x vec
## 1 B 0.06735163
## 2 C 0.48612358
## 3 B 2.34190635
## 4 C 0.36393262
## 5 A 7.99762969
## 6 B 1.43293330
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您可以使用group_by,并summarise从dplyr以达到预期的效果.请参阅下面的代码.
# load required package
require(dplyr)
# function to calculate the confidence interval
CIfun <- function(v, probs = c(.025, .975)) {
quantile(boot(data = v, function(u,i) tb(u[i]), R = 999)$t, probs)
}
# using summarise from dplyr
dataframe %>% group_by(x) %>%
summarise(n = n(),
proportion = tb(vec),
`2.5%` = CIfun(vec, .025),
`97.5%`= CIfun(vec, .975))
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