[英]zoo series and aggregate returns in R
I want to have overall returns for data series over the whole of zoo series time perid, which I have in both a prices or daily returns;我想在整个动物园系列时间段内获得数据系列的总体回报,我在价格或每日回报中都有;
eg例如
GOLD PA PL SLV
2001-05-22 0.000000000 -0.009132420 -0.004838710 0.0
or as prices where simple return would be last prices in series minus first / first或者作为简单回报将是系列中的最后价格减去第一个/第一个的价格
GOLD PA PL SLV
2020-10-09 1920 2454 888 25
I've tried some performance analytic packages but I know the returns are wrong.我尝试了一些性能分析包,但我知道返回值是错误的。
Assuming the input data shown reproducibly in the Note at the end and using returns:假设输入数据在末尾的注释中可重复显示并使用返回值:
apply(rets + 1, 2, prod) - 1
## GOLD PA PL SLV
## 0.00000000 -0.02714782 -0.01444600 0.00000000
or要么
library(PerformanceAnalytics)
Return.cumulative(rets)
## GOLD PA PL SLV
## Cumulative Return 0 -0.02714782 -0.014446 0
or approximating using sums:或使用总和进行近似:
colSums(rets)
## GOLD PA PL SLV
## 0.00000000 -0.02739726 -0.01451613 0.00000000
or using prices:或使用价格:
n <- nrow(prices)
diff(prices[c(1, n)], arith = FALSE) - 1
## GOLD PA PL SLV
## 2020-10-11 0 0 0.002252252 0.08
or using n
from above:或从上面使用
n
:
exp(diff(log(prices[c(1, n)]))) - 1
## GOLD PA PL SLV
## 2020-10-11 0 0 0.002252252 0.08
Lines <- "
Date GOLD PA PL SLV
2001-05-22 0.000000000 -0.009132420 -0.004838710 0.0
2001-05-23 0.000000000 -0.009132420 -0.004838710 0.0
2001-05-24 0.000000000 -0.009132420 -0.004838710 0.0"
Lines2 <- "
Date GOLD PA PL SLV
2020-10-09 1920 2454 888 25
2020-10-10 1900 2454 899 26
2020-10-11 1920 2454 890 27"
library(zoo)
rets <- read.zoo(text = Lines, header = TRUE)
prices <- read.zoo(text = Lines2, header = TRUE)
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