我试图模拟三个不同比例接受治疗的患者死亡的十年风险。我已经每年这样做十年了，结果证明它是一个很长的代码。我想要的是将其转换为每月十年，并且为了避免数百行代码，我想使用for循环。
set.seed(1234)
N <- 750000
id <- c(1:N)
###creates a sex variable for men and appends women
treated <- rep.int(0,125000)
treated <- append(treated, rep.int(1,125000))
treated <- append(treated, rep.int(0,100000))
treated <- append(treated, rep.int(1,150000))
treated <- append(treated, rep.int(0,75000))
treated <- append(treated, rep.int(1,175000))
groupname <- rep.int(1,250000)
groupname <- c(groupname, rep.int(2,250000))
groupname <- c(groupname, rep.int(3,250000))
data = data.frame(treated, id, groupname)
class(data$treated)
data$treated <- factor(data$treated, levels = c(0,1), labels = c("untreated","treated"))
data$groupname <- factor(data$groupname, levels = c(1,2,3), labels = c("group 1", "group 2", "group 3"))
然后我生成每个“wave”，就像这样十年（基本相同的代码，只为每个波分配一个新的列名）：
data$year_0 <- 1
data$year_1 <- ifelse(data$treated=="treated",rbinom(N, 1, 1-0.035/4), rbinom(N, 1, 1-0.05/4))
data$year_2 <- ifelse(data$treated=="treated",
ifelse(data$year_1 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_1 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_3 <- ifelse(data$treated=="treated",
ifelse(data$year_2 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_2 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_4 <- ifelse(data$treated=="treated",
ifelse(data$year_3 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_3 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_5 <- ifelse(data$treated=="treated",
ifelse(data$year_4 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_4 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_6 <- ifelse(data$treated=="treated",
ifelse(data$year_5 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_5 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_7 <- ifelse(data$treated=="treated",
ifelse(data$year_6 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_6 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_8 <- ifelse(data$treated=="treated",
ifelse(data$year_7 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_7 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_9 <- ifelse(data$treated=="treated",
ifelse(data$year_8 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_8 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
data$year_10 <- ifelse(data$treated=="treated",
ifelse(data$year_9 =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_9 =="0", 0, rbinom(N, 1, 1-0.05/4))
)
###converts to long format
data_long <- reshape(data, direction="long", varying= c(list(4:14)), sep = "_",
idvar="id", timevar=c("year"))
class(data_long$year)
data_long$year <- as.numeric(data_long$year)
data_long$year <- data_long$year -1
我想用for循环来做这个，所以我可以模拟120个月 我写了这段代码
for (i in 1:10){ n <- ifelse(data$treated=="treated",
ifelse(data$year_[(i-1)] =="0", 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data$year_[(i-1)] =="0",
0, rbinom(N, 1, 1-0.05/4))
)
data$year_[i] <- n
}
##1: I data$year_[i] <- n :
##error number of items to replace is not a multiple of replacement length
据我所知，此错误表示循环编码的方式返回不兼容的长度数据。通常我可以通过谷歌进行故障排除，但是因为代码在我不在for循环中时运行 我不明白问题出在哪里。 我认为错误可能是在[i]的解释中，而不是可用于命名列的字符串，但除了已经提到的字符之外，使用粘贴只会导致此警告。
##Fejl i `$<-.data.frame`(`*tmp*`, "year_", value = c(NA, NA, NA, NA, :
##replacement has 750001 rows, data has 750000
并且谷歌在这个问题上的结果似乎并没有将此视为一个问题。 所以现在的问题是，我不知道问题是什么。
答案 0 :(得分：0)
您可以将列year_i放在一个额外的矩阵中。然后，您可以使用cbind()按列扩展矩阵列：
set.seed(1234)
N <- 750000
data = data.frame(treated=rep(c(0,1,0,1,0,1), c(125000, 125000, 100000, 150000, 75000, 175000)), id=1:N,
groupname=rep(1:3, each=250000))
data$treated <- factor(data$treated, levels = c(0,1), labels = c("untreated","treated"))
data$groupname <- factor(data$groupname, levels = c(1,2,3), labels = c("group 1", "group 2", "group 3"))
Year <- matrix(1, N, 1) # data$year_0 <- 1
Year <- cbind(Year, ifelse(data$treated=="treated",rbinom(N, 1, 1-0.035/4), rbinom(N, 1, 1-0.05/4))) # data$year_1
for (i in 2:10) {
lastcol <- Year[,ncol(Year)]
Year <- cbind(Year,
ifelse(data$treated=="treated",
ifelse(lastcol==0, 0, rbinom(N, 1, 1-0.035/4)),
ifelse(lastcol==0, 0, rbinom(N, 1, 1-0.05/4)))
)
}
你可以通过预分配加快一点（但最重要的是采样）：
set.seed(1234)
K <- 10 # year_0 ... year_K
Year <- matrix(NA, N, K+1)
Year[,1] <- 1 # year_0
Year[,2] <- ifelse(data$treated=="treated", rbinom(N, 1, 1-0.035/4), rbinom(N, 1, 1-0.05/4)) # data$year_1
for (i in 3:(K+1)) Year[,i] <- ifelse(data$treated=="treated",
ifelse(Year[,i-1]==0, 0, rbinom(N, 1, 1-0.035/4)),
ifelse(Year[,i-1]==0, 0, rbinom(N, 1, 1-0.05/4)))
如果需要，可以将数据框和矩阵Year放在一起。如果是这样，最好将列名分配给矩阵：
colnames(Year) <- paste0("year_", 0:K)
答案 1 :(得分：0)
考虑使用括号引用[[...]]到列名称来传递带有paste0()的字符串，条件为第一年，然后是所有其他年份：
data$year_0 <- 1
for (i in 1:10){
if (i == 1){
n <- ifelse(data$treated=="treated", rbinom(N, 1, 1-0.035/4), rbinom(N, 1, 1-0.05/4))
}
else {
n <- ifelse(data$treated=="treated",
ifelse(data[[paste0("year_", i-1)]] == 0, 0, rbinom(N, 1, 1-0.035/4)),
ifelse(data[[paste0("year_", i-1)]] == 0, 0, rbinom(N, 1, 1-0.05/4))
)
}
data[[paste0("year_", i)]] <- n
}