我试图将我的数据集折叠成一个变量的均值和第75百分位数，但是我似乎找不到正确的方式来声明我想要第75百分位数。
six_month_agg <- summaryBy(pd ~ industry + region + date, FUN=c(mean, 0.75), data=six_month_pd)
anauzrmj1#
我们可以使用quantile
quantile
library(doBy) summaryBy(pd ~ industry + region + date, FUN= function(x) c(Mean = mean(x), Quantile = quantile(x, probs = 0.75)), data=six_month_pd)
使用可重现的示例
data(warpbreaks) out <- summaryBy(breaks ~ wool + tension, warpbreaks, FUN=function(x) c(Mean = mean(x), Quantile = quantile(x, probs = .75))) str(out) #'data.frame': 6 obs. of 4 variables: # $ wool : Factor w/ 2 levels "A","B": 1 1 1 2 2 2 # $ tension : Factor w/ 3 levels "L","M","H": 1 2 3 1 2 3 # $ breaks.Mean : num 44.6 24 24.6 28.2 28.8 ... # $ breaks.Quantile.75%: num 54 30 28 31 39 21
1条答案
按热度按时间anauzrmj1#
我们可以使用
quantile使用可重现的示例