如何计算标准偏差，而不考虑由R中其他两个向量的水平确定的行？

我用总参数计算同一行业和同一地区的公司的标准差

sd_lsales我们可以试试

df1$sd_lsales <- unsplit(lapply(split(df1[-1], df1$Region), function(x) {
                  sd1 <- sapply(unique(x$Sector), function(y) {
                  i1 <- x$Sector!=y
                  sd(x$lsales[i1])

                    })
                 sd1[x$Sector]
               }

             ), df1$Region)

df1$sd_lsales你能试试这个吗

library(dplyr)

sd_lsales <- aggregate(lsales, by=list(region, sector), function(x) sqrt((sum((x-mean(x))^2) - (x-mean(x))^2)/n()-1)   )

库（dplyr）
sd_lsales Hardik，很抱歉我的回复太晚了，我在您的代码中使用了dplyr包，答案是：n（）中的错误：不应直接调用此函数。不确定是什么原因造成的…@Hardik:现在尝试了长度（x）。产生与akrun不同的NAN和标准偏差。如果您从他的解决方案中复制粘贴df1，并将其与代码的结果进行比较。我再次检查了akrun代码的结果，结果是正确的。
df1 <- structure(list(Region = c("Vienna", "Vienna", "Vienna", "Berlin", 
"Berlin", "Berlin", "Berlin"), Sector = c("Food", "Other Manufacturing", 
 "Food", "Food", "Manufacturing", "Manufacturing", "IT"), lsales = c(363000000L, 
5930000L, 150000000L, 39200000L, 203900000L, 298000000L, 30339200L
 )), .Names = c("Region", "Sector", "lsales"), class = "data.frame", row.names = c("1", 
"2", "3", "505", "506", "507", "508"))

library(dplyr)

sd_lsales <- aggregate(lsales, by=list(region, sector), function(x) sqrt((sum((x-mean(x))^2) - (x-mean(x))^2)/n()-1)   )