编写一个程序来引入排列

writing a program to introduce permutations

基本上我想编写一个程序,将我的数据顺序随机化 n 次,然后完成生存分析并在 n

上绘制输出

因此,让我们从 matching() 包中获取以下通用数据,并创建一个包含接受治疗和未接受治疗的人的数据集。 Link to package

set.seed(123)

library(Matching)
data(lalonde)

lalonde$age_cat <- with(lalonde, ifelse(age < 24, 1, 2))
attach(lalonde)

lalonde$ID <- 1:length(lalonde$age)


#The covariates we want to match on
X = cbind(age_cat, educ, black, hisp, married, nodegr, u74, u75, re75, re74)
#The covariates we want to obtain balance on
BalanceMat <- cbind(age_cat, educ, black, hisp, married, nodegr, u74, u75, re75, re74,
                    I(re74*re75))
genout <- GenMatch(Tr=treat, X=X, BalanceMatrix=BalanceMat, estimand="ATE", M=1,
                   pop.size=16, max.generations=10, wait.generations=1)
detach(lalonde)

# now lets pair the the non-treated collisions to the treated
# BUT lets pair WITHOUT REPLACEMENT

mout <- Match(Y=NULL, Tr=lalonde$treat, X=X,
              Weight.matrix=genout, M=2,
              replace=FALSE, ties=TRUE)

summary(mout)
# we see that for 130 treated observations, we have 260 non-treated
# this is because we set M=2
# and yes length(lalonde$age[lalonde$treat==0]) == 260 but just follow me please
# but this was done for a specific reason

# now lets create a table for our 130+260 collisions
treated <- lalonde[mout$index.treated,]
# now we only want one occurence of the treated variables
library(dplyr)
treat_clean <- treated %>%
  group_by(ID) %>%
  slice(1)

non.treated <- lalonde[mout$index.control,]

# finally we can combine to form one clear data.set
matched.data <- rbind(treat_clean, non.treated)

我们现在可以进行条件逻辑回归来确定与 re78(1987 年赚取的钱)和治疗相关的 OR。为此,我们需要生存包。 Link to package

library(survival)

如果居住者在 1978 年的收入超过 8125,则假设成功

matched.data$success <- with(matched.data, ifelse(re78 > 8125, 1, 0))

output <- clogit(success ~ treat, matched.data, method = 'efron')

summary(output)

所以我们看到治疗 (treat=1) 的 OR 是 1.495

我们可以将其保存为:

iteration.1 <- exp(output$coefficients[1])

现在我们从匹配包 (link) 中读取 replace = FALSE 注意,如果为 FALSE, 比赛的顺序通常很重要。比赛将在 与数据排序的顺序相同

所以我想做的是创建一个函数,该函数将 n

Is essenece 我想在分析中引入排列。 当我们说 n=5

时,如何做到这一点

可以使用sample引入排列

data(lalonde)
lalonde$age_cat <- with(lalonde, ifelse(age < 24, 1, 2))
lalonde$ID <- 1:length(lalonde$age)
n <- 5
res <- rep(NA, n)
for (i in 1:n) {
    lalonde <- lalonde[sample(1:nrow(lalonde)), ] # randomise order
    ## rest of code 
    res[i] <- exp(output$coefficients[1])
}

plot(1:n, res, main="Odds Ratios")

我是 replicate 的超级粉丝,因为这样的事情:

X <- cbind(...)                         # what you had before
BalanceMat <- cbind(...)                # ditto
lalonde$ID <- seq.int(nrow(lalonde))

results <- replicate(1000, {
    ## not certain if it's just $ID order that matters
    lalonde$ID <- sample(nrow(lalonde))
    ## lalonde <- lalonde[ sample(nrow(lalonde)), ]

    ## ...
    ## rest of your computation
    ## ...

    #### optionally return everything 
    ## output
    #### return just the minimum
    exp(output$coefficients[1])
})

#### if you returned output earlier, you'll need this, otherwise not
## coef <- exp(sapply(results, function(z) z$coefficients[1]))

## plot as needed

我不知道你的意思是 ID 的顺序重要还是整个数据库的顺序;相应地调整 replicate 循环的前几行。