Question

假设我有以下df

df <- structure(list(var1 = c(1, 0, 1, 0, 0 , 1 ), var2 = c(0, 
0, 0, 1, 1, 0), var99 = c(0, 1, 1, 1, 1, 0), value = c(154, 
120, 100, 180, 200, 460)), .Names = c("var1", "var2", "var99", "value" ), row.names = c(NA, -6L), class = "data.frame")

我想要实现这个输出数据：

structure(list(var = c("var1", "var2", "var99"), mean = c(238, 
190, 150)), .Names = c("var", "mean"), row.names = c(NA, -3L), class = 
"data.frame")

这是：获得列的平均值＆＃39;值＆＃39;对于每隔一列：var1，var2，...，var99。只会计算带有1的行以计算平均值。

我用for循环完成了它：

l <- vector("list", 3)
for (i in 1:3)
l[[i]] <- mean(df$value[df[,i]==1], na.rm = T)
i <- i+1

有人可以建议我在可能的情况下用Base R省略循环吗？

Answer 1

sapply(df[, -4], weighted.mean, x=df[, 4])

或者

colSums(sweep(df[, -4], 1, df[, 4], `*`)) / colSums(df[, -4])

Answer 2

或者：

sapply(subset(df, select = -value), function(x) mean(df$value[x == 1]))

根据每个其他列

2 个答案: