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我有一個按組劃分的數據框,每個組具有相同數量的觀察值。我已經為每個組隨機分配了1或0的值。對於給定值為1的組中的所有觀察值,我希望變量ysp用一定數量的隨機1和0填充。對於分配了0的組,我希望該變量ysp充滿所有0。
這是我到目前為止的代碼:
rm(列表= ls(全部= TRUE))
set.seed(1984)
ngroup <-50#組數
obs <-50#每組的觀察數
pgroup <-0.5#(1- p)具有至少1個非零obs的組的概率(僅在答案是整數的情況下有效)
p <-0.5#一旦選擇了我希望至少有一個非零obs的組數,我希望這些組中1%的p%。
constantdata <-data.frame(id = 1:ngroup)
假人<-c(0,1)
假人[樣本(1:nrow(constantdata),nrow(constantdata),FALSE)] <-rep(假人,c(pgroup * ngroup,(1-pgroup)* ngroup))
constantdata [“ probgr”] <-假人
fulldata <-constantdata [rep(1:ngroup,each = obs),]
fulldata $ ys <-rnorm(ngroup * obs)
#這就是我嘗試的方法
if(fulldata $ probgr = 1){
fulldata $ ysp [fulldata $ ys>分位數(fulldata $ ys,1-p)] <-1
fulldata $ ysp [fulldata $ ys <=分位數(fulldata $ ys,1-p)] <-0
}別的{
fulldata $ ysp = 0}
當然,這是行不通的。
我希望變量ysp擁有50%(pgroup%)隨機組全為0,另50%(1-pgroup%)組具有p%1s和0s的隨機分配。 
在編寫if(fulldata $ probgr = 1)的地方,您可能要表示if(fulldata $ probgr == 1)(相等性測試,而不是賦值)。另外,if不是向量運算。一種獲得所需內容的方法是,將ysp中的所有內容都設置為0,然後隨機更改probgr == 1的內容,如下所示:
fulldata $ ysp = 0
fulldata $ ysp [fulldata $ probgr == 1] =樣本(0:1,sum(fulldata $ probgr == 1),replace = TRUE)
1個
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