使用R重塑和打印linux dstat CSV文件

我正试图使用

reformae2

来格式化这90个观测数据，以便以合理的方式绘制它，但我不知道如何将所有

foo.5

列转换为一个我可以用ggplot2轻松绘制的东西

> library(RCurl)
> x <- getURL("https://gist.githubusercontent.com/hamiltont/27d9f93ad53d8978bccd/raw/ec6118805b9ab8606a303dc81450a6f760864cf9/dstat.csv")
> dstat = read.csv(file="stats", skip=6, header=TRUE)
> names(dstat)
  [1] "usr"    "sys"    "idl"    "wai"    "hiq"   
  [6] "siq"    "usr.1"  "sys.1"  "idl.1"  "wai.1" 
 [11] "hiq.1"  "siq.1"  "usr.2"  "sys.2"  "idl.2" 
 [16] "wai.2"  "hiq.2"  "siq.2"  "usr.3"  "sys.3" 
 [21] "idl.3"  "wai.3"  "hiq.3"  "siq.3"  "usr.4" 
 [26] "sys.4"  "idl.4"  "wai.4"  "hiq.4"  "siq.4" 
 [31] "usr.5"  "sys.5"  "idl.5"  "wai.5"  "hiq.5" 
 [36] "siq.5"  "usr.6"  "sys.6"  "idl.6"  "wai.6" 
 [41] "hiq.6"  "siq.6"  "usr.7"  "sys.7"  "idl.7" 
 [46] "wai.7"  "hiq.7"  "siq.7"  "usr.8"  "sys.8" 
 [51] "idl.8"  "wai.8"  "hiq.8"  "siq.8"  "usr.9" 
 [56] "sys.9"  "idl.9"  "wai.9"  "hiq.9"  "siq.9" 
 [61] "usr.10" "sys.10" "idl.10" "wai.10" "hiq.10"
 [66] "siq.10" "usr.11" "sys.11" "idl.11" "wai.11"
 [71] "hiq.11" "siq.11" "usr.12" "sys.12" "idl.12"
 [76] "wai.12" "hiq.12" "siq.12" "usr.13" "sys.13"
 [81] "idl.13" "wai.13" "hiq.13" "siq.13" "usr.14"
 [86] "sys.14" "idl.14" "wai.14" "hiq.14" "siq.14"
 [91] "usr.15" "sys.15" "idl.15" "wai.15" "hiq.15"
 [96] "siq.15" "usr.16" "sys.16" "idl.16" "wai.16"
[101] "hiq.16" "siq.16" "usr.17" "sys.17" "idl.17"
[106] "wai.17" "hiq.17" "siq.17" "usr.18" "sys.18"
[111] "idl.18" "wai.18" "hiq.18" "siq.18" "usr.19"
[116] "sys.19" "idl.19" "wai.19" "hiq.19" "siq.19"
[121] "usr.20" "sys.20" "idl.20" "wai.20" "hiq.20"
[126] "siq.20" "usr.21" "sys.21" "idl.21" "wai.21"
[131] "hiq.21" "siq.21" "usr.22" "sys.22" "idl.22"
[136] "wai.22" "hiq.22" "siq.22" "usr.23" "sys.23"
[141] "idl.23" "wai.23" "hiq.23" "siq.23" "read"  
[146] "writ"   "recv"   "send"   "recv.1" "send.1"
[151] "recv.2" "send.2" "in."    "out"    "int"   
[156] "csw"    "X20"    "X21"    "X23"    "X76"   
[161] "X77"    "X78"    "X79"    "X80"    "X81"   
[166] "X82"    "X83"    "X1m"    "X5m"    "X15m"  
[171] "used"   "buff"   "cach"   "free"   "run"   
[176] "blk"    "new"    "read.1" "writ.1" "used.1"
[181] "free.1" "epoch"  "X.aio"  "files"  "inodes"
[186] "msg"    "sem"    "shm"    "pos"    "lck"   
[191] "rea"    "wri"    "raw"    "tot"    "tcp"   
[196] "udp"    "raw.1"  "frg"    "lis"    "act"   
[201] "syn"    "tim"    "clo"    "raw.2"  "tot.1" 
[206] "tcp.1"  "udp.1"  "raw.3"  "frg.1"  "lis.1" 
[211] "act.1"  "syn.1"  "tim.1"  "clo.1"  "lis.2" 
[216] "act.2"  "dgm"    "str"    "lis.3"  "act.3" 
[221] "majpf"  "minpf"  "alloc"  "free.2" "util"  
[226] "call"   "retr"   "refr"   "call.1" "erca"  
[231] "erau"   "ercl"   "xdrc"  

如果我能说服我的数据框架如下所示：

observation | usr_type | value
1             usr        22
1             usr.1      32
1             usr.2      12
<snip>
2             usr        21
2             usr.1      20
2             usr.2      20

观测值| usr|U型|值
1 usr 22
1 usr.1 32
1 usr.2 12
2 usr 21
2 usr.1 20
2 usr.2 20

有几种方法可以做到这一点，但我建议使用我的“splitstackshape”软件包中的

Stacked

基本方法如下：

ggplot(dstat, aes(x=observation, y=usr, color=usr_type) +
  geom_line()

library(splitstackshape) ## Also loads "data.table"
DT <- as.data.table(dstat, keep.rownames = TRUE)
Stacked(DT, id.vars = "rn", var.stubs = "usr", 
        sep = "var.stubs", keep.all = FALSE)
#       rn .time_1   usr
#    1:  1         0.027
#    2:  1      .1 8.068
#    3:  1     .10 0.022
#    4:  1     .11 3.276
#    5:  1     .12 0.003
#   ---                 
# 2060:  9      .5 0.000
# 2061:  9      .6 0.000
# 2062:  9      .7 0.000
# 2063:  9      .8 0.000
# 2064:  9      .9 0.000

sep=“var.stubs”

参数基本上表示列名中没有真正的分隔符。对于您的数据集，这并不完全正确，但它使您不必将每个集合的第一列重命名为类似“usr.0”的名称。如果数据已经以该形式存在，则可以使用

作为

sep

库的值（Reforme2）
library(reshape2)   
dstat$obs<-1:nrow(dstat)
u<-names(dstat)[grep("usr", names(dstat))]
dstat1<-dstat[,c("obs",u)]
dstat2<-melt(dstat1, id=c("obs"))
dstat2<-dstat2[order(dstat2$obs),]

dstat$obs
dstat-tcmn--output node.csv#output dstat
mail-a node.csv-s“备份文件”容器。sui@xxx.com

RStudio&plotly

library(plotly)
ds = read.table("C:\\Users\\tank.sui\\Desktop\\node.csv",header=TRUE,sep=",",skip=6)
p <- plot_ly(x=strptime(date.time,"%m-%d %H:%M:%S"),y=idl,name="Test",data=ds)
p

library（plotly）
ds=read.table（“C:\\Users\\tank.sui\\Desktop\\node.csv”，header=TRUE，sep=“，”，skip=6）
这看起来很完美，但我不能让它像你在我的数据上显示的那样工作。当使用
sep=“var.stubs”
时，它在names（x）@Hamy中返回一个错误
error，我认为您链接到的文件是您的数据。您使用的是什么版本的“splitstackshape”和“data.table”？什么操作系统？是的，这就是为什么我也感到困惑的原因：-P
splitstackshape 1.2.0版
，
data.table 1.9.2版
，以及
OSX10.9.5版
@Hamy，请同时更新“splitstackshape”和“data.table”。CRAN上最新的“splitstackshape”是1.4.2，“data.table”是1.9.4，我相信。这就解决了它。我必须从源代码（例如，
install.packages（“splitstackshape”，type=“source”）
）编译这两个包，因为没有可用的二进制文件。RStudio似乎默认使用二进制文件进行软件包更新，因此我没有得到新的东西（它确实警告了我，但我错过了消息
有一个二进制版本可用（将安装），但源版本更晚：
dstat -tcmn --output node.csv  #output dstat

mail -a node.csv -s "Backup File" tank.sui@xxx.com < /dev/null  #send as attachment

library(plotly)
ds = read.table("C:\\Users\\tank.sui\\Desktop\\node.csv",header=TRUE,sep=",",skip=6)
p <- plot_ly(x=strptime(date.time,"%m-%d %H:%M:%S"),y=idl,name="Test",data=ds)
p