我对 R 非常陌生。当值设置较低时，我的脚本工作正常。我需要计算概率并且需要至少 1,000,000 个样本。我的第一个循环返回此错误:
Error in new.matrix[i, ] <- as.vector(table(million.rep[, i])) : number of items to replace is not a multiple of replacement length
我的脚本:
actual.prob <- c(.14, .14, .16, .13, .19, .24)
number.in.bag <- c(8,6,20,9,10,5)
million.rep <- replicate(1000000, sample(1:6, 58, replace= T, actual.prob))
new.matrix <- matrix(nrow= 1000000, ncol=6)
# Note to other readers... the above line throws the error.
for(i in 1:1000000){
new.matrix[i,] <- as.vector(table(million.rep [, i]))
}
y <- c()
for(i in 1:1000000){
y[i] <- sum(new.matrix[i,] == number.in.bag)
}
y
sum(y == 6)
最佳答案
table 计算所有出现的元素。如果缺少任何 1:6 ，则输出的长度将小于 6，例如:
set.seed(1)
## first call; table output is length == 5
table(sample(1:6, size=10, replace=TRUE))
1 2 3 4 6
1 2 1 3 3
## second call; table output is length == 4
table(sample(1:6, size=10, replace=TRUE))
2 3 5 6
2 3 4 1
要避免此行为，您可以使用具有定义的级别的因子(factor(sample,levels=1:6)):
## third call using factors; table output is always length == 6
## please notice the "levels=1:6"
table(factor(sample(1:6, size=10, replace=TRUE), levels=1:6))
1 2 3 4 5 6
2 2 3 1 0 2
对于您的示例，您可以使用:
new.matrix[i,] <- as.vector(table(factor(million.rep [, i]), levels=1:6))
关于R错误: "number of items to replace is not a multiple of replacement length"，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21500585/