我正在尝试创建单独的数据框,其中包括主机名数据的Avg,Max和第95百分位数。
数据框看起来像这样:
Hostname Avg Max 95th Percentile
Web01 10 90 92
Web02 5 80 75
dput(d)
structure(list(Hostname = structure(c(8L, 8L, 9L, 5L, 6L, 7L,
1L, 2L, 3L, 4L), .Label = c("db01", "db02", "farm01", "farm02",
"tom01", "tom02", "tom03", "web01", "web03"), class = "factor"),
Date = structure(c(6L, 10L, 5L, 3L, 2L, 1L, 8L, 9L, 7L, 4L
), .Label = c("10/5/2015 1:15", "10/5/2015 1:30", "10/5/2015 2:15",
"10/5/2015 4:30", "10/5/2015 8:30", "10/5/2015 8:45", "10/6/2015 8:15",
"10/6/2015 8:30", "9/11/2015 5:00", "9/11/2015 6:00"), class = "factor"),
Cpubusy = c(31L, 20L, 30L, 20L, 18L, 20L, 41L, 21L, 29L,
24L), UsedPercentMemory = c(99L, 98L, 95L, 99L, 99L, 99L,
99L, 98L, 63L, 99L)), .Names = c("Hostname", "Date", "Cpubusy",
"UsedPercentMemory"), class = "data.frame", row.names = c(NA,
-10L))
在r中有一种简单的方法可以做到这一点,我试图避免循环。
我试过了:
dd %>% group_by(Hostname) %>% summarise_each(funs(mean, max))
我无法弄清楚第95百分位数。
答案 0 :(得分:1)
不确定这是否是有效的方法
library(dplyr)
library(lazyeval)
dd %>%
group_by(Hostname) %>%
summarise_(Mean = interp(~mean(var, na.rm=TRUE), var=as.name(m)),
Max=interp(~max(var, na.rm=TRUE), var=as.name(m)),
Quantile= interp(~quantile(var, prob=0.95), var=as.name(m)))