财务-使用plyr和RPORTFOIO的随机投资组合-需要拆分列
我试图将实际投资组合的表现与假设的随机投资组合的表现进行比较 这是我正在使用的数据集的一个示例。它显示了两个月的数据、投资组合中经理的姓名以及这些经理的回报、分配和属性财务-使用plyr和RPORTFOIO的随机投资组合-需要拆分列,r,finance,plyr,reshape2,R,Finance,Plyr,Reshape2,我试图将实际投资组合的表现与假设的随机投资组合的表现进行比较 这是我正在使用的数据集的一个示例。它显示了两个月的数据、投资组合中经理的姓名以及这些经理的回报、分配和属性 "date" "manager" "return" "allocation" "attribution" 2005-01-31 "manager01" -0.00763241754291056 0.146 6.94549996404861e-05 2005-01-31 "manager02" 0.0292205518315147
"date" "manager" "return" "allocation" "attribution"
2005-01-31 "manager01" -0.00763241754291056 0.146 6.94549996404861e-05
2005-01-31 "manager02" 0.0292205518315147 0.048 4.09087725641205e-05
2005-01-31 "manager03" -0.0354047394153526 0.049 -8.85118485383814e-05
2005-01-31 "manager04" 0.0424244772606645 0.124 -0.000148485670412326
2005-01-31 "manager05" -0.0574606103881735 0.134 0.000206858197397425
2005-01-31 "manager06" 0.0465278163188542 0.098 -0.000265208553017469
2005-01-31 "manager07" 0.157063203979822 0.142 -0.000219888485571751
2005-01-31 "manager08" -0.0594342759491509 0.071 2.97171379745754e-05
2005-01-31 "manager09" -0.0199466865109495 0.093 6.18347281839434e-05
2005-01-31 "manager10" 0.118839410130508 0.095 0.000190143056208813
2005-02-28 "manager01" 0.0403671815817711 0.119 -0.000460185870032191
2005-02-28 "manager02" 0.0246109773791459 0.064 -3.93775638066334e-05
2005-02-28 "manager03" 0.00868489880733732 0.065 -4.08190243944854e-05
2005-02-28 "manager04" -0.082332291530606 0.105 2.46996874591818e-05
2005-02-28 "manager05" -0.0903959999837099 0.114 -0.000117514799978823
2005-02-28 "manager06" 0.0514735666329574 0.081 -6.17682799595489e-05
2005-02-28 "manager07" -0.00914374153663751 0.164 -8.41224221370651e-05
2005-02-28 "manager08" -0.0367283709786134 0.083 -4.77468822721974e-05
2005-02-28 "manager09" -0.04752320926613 0.079 -3.8018567412904e-05
2005-02-28 "manager10" -0.0657464361573664 0.126 -0.000309008249939622
为了将数据输入R,将数据复制到剪贴板,然后
mydata<-read.table("clipboard",header=TRUE)
在rlongonly
功能中:
- m的值是我想要创建的随机投资组合的数量
- n的值是该时间段的分配数
- k的值是非零分配的数量
- x.u的值是分配的上限
mydata.new2<-ddply(mydata,.(date),mutate,new.attr=t(rlongonly(m=2,n=length(date),k=10,x.u=.15))*return)
dim(mydata.new)
如何拆分“new.attr”列,以便将数据熔化并绘制成图形?首先我要重新生成数据,您必须使用dput(mydata)并在下次发布结果 然后生成mydata.new2向量
library(plyr)
library(rportfolios)
mydata.new2<-ddply(mydata,
.(date),
mutate,
new.attr=t(rlongonly(m=2,n=length(date),k=10,x.u=.15))*return)
因为mydata.new2 data.frame中嵌套了data.framenew.attr。这是由于使用了mutate。这里最好使用转换,因为您不需要迭代地进行转换
因此:
library(reshape2)
drop<-names(mydata.new2) %in% c("manager","return","allocation")
melt(mydata.new2[!drop],id="date")
> Error in rbind(deparse.level, ...) :
> numbers of columns of arguments do not match
library(plyr)
library(rportfolios)
mydata.new2<-ddply(mydata,
.(date),
mutate,
new.attr=t(rlongonly(m=2,n=length(date),k=10,x.u=.15))*return)
mydata.new2[,-c(1,2)] <- numcolwise(round_any)(mydata.new2,0.0001)
head(mydata.new2)
date manager return allocation attribution new.attr.1 new.attr.2
1 2005-01-31 manager01 -0.0076 0.146 1e-04 -0.0009 -0.0007
2 2005-01-31 manager02 0.0292 0.048 0e+00 0.0032 0.0040
3 2005-01-31 manager03 -0.0354 0.049 -1e-04 -0.0024 -0.0049
4 2005-01-31 manager04 0.0424 0.124 -1e-04 0.0029 0.0025
5 2005-01-31 manager05 -0.0575 0.134 2e-04 -0.0047 -0.0042
6 2005-01-31 manager06 0.0465 0.098 -3e-04 0.0051 0.0039
library(reshape2)
drop<-names(mydata.new2) %in% c("manager","return","allocation")
melt(mydata.new2[!drop],id="date")
numbers of columns of arguments do not match
mydata.new2<-ddply(mydata,
.(date),
transform,
new.attr=t(rlongonly(m=2,n=length(date),k=10,x.u=.15))*return)
head(melt(mydata.new2[!drop],id="date"))
date variable value
1 2005-01-31 attribution 6.945500e-05
2 2005-01-31 attribution 4.090877e-05
3 2005-01-31 attribution -8.851185e-05
4 2005-01-31 attribution -1.484857e-04
5 2005-01-31 attribution 2.068582e-04
6 2005-01-31 attribution -2.652086e-04