Date Server CPU
1/1/2012 WebA 30
1/1/2012 WebA 25
1/1/2012 WEbB 30
1/2/2012 WebA 45
1/2/2012 WebA 50
1/2/2012 WebA 60
dput(x)
structure(list(Date = structure(c(1L, 1L, 1L, 2L, 2L, 2L), .Label = c("1/1/2012",
"1/2/2012"), class = "factor"), Server = structure(c(1L, 1L,
2L, 1L, 1L, 1L), .Label = c("WebA", "WEbB"), class = "factor"),
CPU = c(30L, 25L, 30L, 45L, 50L, 60L)), .Names = c("Date",
"Server", "CPU"), class = "data.frame", row.names = c(NA, -6L
))
我应该为给定服务器的每个数据提供一个数据点。我需要删除重复的日期。当我选择日期时,CPU应该是最高的。例如,对于Date 1/2/2012,我的新行将是1/2/2012 WebA 60
。
如何删除R中的重复日期?
我可以这样做:
x[!duplicated(x[1:2]),]
来自CPU的,如何检查最高?
答案 0 :(得分:4)
使用aggregate
的解决方案:
aggregate(df$CPU, by=list(df$Date, df$Server), max)
# Group.1 Group.2 x
# 1 1/1/2012 WebA 30
# 2 1/2/2012 WebA 60
# 3 1/1/2012 WEbB 30
使用data.table
require(data.table)
dt <- data.table(df)
setkey(dt, "Date", "Server")
dt[, list(CPU.max = max(CPU)), by="Date,Server"]
# Date Server CPU.max
# 1: 1/1/2012 WebA 30
# 2: 1/1/2012 WEbB 30
# 3: 1/2/2012 WebA 60
编辑:根据OP的评论要求提供更多列:
df <- structure(list(Date = structure(c(1L, 1L, 1L, 2L, 2L, 2L),
.Label = c("1/1/2012", "1/2/2012"), class = "factor"),
Server = structure(c(1L, 1L, 2L, 1L, 1L, 1L),
.Label = c("WebA", "WEbB"), class = "factor"),
CPU = c(30L, 25L, 30L, 45L, 50L, 60L),
val1 = c(5L, 2L, 6L, 3L, 1L, 4L),
val2 = c(5L, 3L, 6L, 4L, 1L, 2L),
val3 = c(1L, 2L, 4L, 3L, 6L, 5L)),
.Names = c("Date", "Server", "CPU", "val1", "val2", "val3"),
row.names = c(NA, -6L), class = "data.frame")
> df
# Date Server CPU val1 val2 val3
# 1 1/1/2012 WebA 30 5 5 1
# 2 1/1/2012 WebA 25 2 3 2
# 3 1/1/2012 WEbB 30 6 6 4
# 4 1/2/2012 WebA 45 3 4 3
# 5 1/2/2012 WebA 50 1 1 6
# 6 1/2/2012 WebA 60 4 2 5
使用aggregate
的解决方案:使用聚合与公式(如下所示)通常更好,因为1)它保留了列名,2)它干净且易于理解3)它允许更容易合并到恢复其他列(由于(1))(这是你的问题,如果我做对了)。
df.agg <- aggregate(data = df, CPU ~ Date + Server, max)
merge(df.agg, df)
# Date Server CPU val1 val2 val3
# 1 1/1/2012 WebA 30 5 5 1
# 2 1/1/2012 WEbB 30 6 6 4
# 3 1/2/2012 WebA 60 4 2 5
使用data.table
的解决方案:
dt <- data.table(df, key=c("Date", "Server"))
# .SD holds the data.frame of the current group that is processed
dt[, .SD[which.max(CPU)], by=c("Date", "Server")]
# Date Server CPU val1 val2 val3
# 1: 1/1/2012 WebA 30 5 5 1
# 2: 1/1/2012 WEbB 30 6 6 4
# 3: 1/2/2012 WebA 60 4 2 5
答案 1 :(得分:2)
只是为了好玩,另一个有plyr
:
library(plyr)
ddply(x, c("Date", "Server"), summarize, cpumax=max(CPU))