在 R 中填充空的 xts 对象

Question

我有空的 xts 对象，我想用简单的计算来填充列（预测日期 - xts 索引（日期）/ 365）。我已经能够填写第一个，问题是我有 46 列，将来还会有更多列，所以我这样做的方式不是最佳的。这是我能做的。如何填充其余 4 列（实际示例中为 46 列），而不必像本例中那样合并每一列。

创建空 xts

    xts <- xts(order.by=index(xts))
    merge(xts, col1 = (dt[1] - index(xts))/365)
              col1
2010-12-31 6.512329
2011-01-03 6.504110
2011-01-04 6.501370
2011-01-05 6.498630
2011-01-06 6.495890
2011-01-07 6.493151

最终结果应该是这样的。

               col1     col2     col3     col4     col5
2010-12-31 6.512329 6.789041 7.016438 7.153425 7.287671
2011-01-03 6.504110 6.780822 7.008219 7.145205 7.279452
2011-01-04 6.501370 6.778082 7.005479 7.142466 7.276712
2011-01-05 6.498630 6.775342 7.002740 7.139726 7.273973
2011-01-06 6.495890 6.772603 7.000000 7.136986 7.271233
2011-01-07 6.493151 6.769863 6.997260 7.134247 7.268493

这里是带有 5 个预定日期的 dt 变量的数据。

dput(xts)
structure(numeric(0), index = structure(c(1293753600, 1294012800, 
1294099200, 1294185600, 1294272000, 1294358400), tzone = "UTC", tclass = "Date"), class = c("xts", 
"zoo"), .indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC")

dput(dt)
structure(c(17351L, 17452L, 17535L, 17585L, 17634L), class = "Date")

Answer 1

关键是使用Reduce合并大列表对象

#Read Data

#main index for first series
mainIndex = as.Date(c("2010-12-31","2011-01-03","2011-01-04","2011-01-05","2011-01-06","2011-01-07"),format="%Y-%m-%d")

referenceDates = as.Date(c("2017-07-04","2017-10-13","2018-01-04","2018-02-23","2018-04-13"),format="%Y-%m-%d")


#Create subsequent xts objects and save as list object

TS_List = lapply(1:length(referenceDates),function(x) {

tsObj =xts((referenceDates[x] - mainIndex)/365,order.by=mainIndex);
colnames(tsObj)=paste0("col",x);
return(tsObj) 
}) 


#General syntax for Reduce : function(x, y) merge(x, y,by="column_column")
#here merge uses merge.xts and common column  is index of xts objects

mergeXTSfun = function(x, y) merge(x, y)

merged_TS = Reduce(mergeXTSfun, TS_List )
merged_TS

#               col1     col2     col3     col4     col5
#2010-12-31 6.512329 6.789041 7.016438 7.153425 7.287671
#2011-01-03 6.504110 6.780822 7.008219 7.145205 7.279452
#2011-01-04 6.501370 6.778082 7.005479 7.142466 7.276712
#2011-01-05 6.498630 6.775342 7.002740 7.139726 7.273973
#2011-01-06 6.495890 6.772603 7.000000 7.136986 7.271233
#2011-01-07 6.493151 6.769863 6.997260 7.134247 7.268493



DesiredOutput= read.table(text="col1     col2     col3     col4     col5
2010-12-31 6.512329 6.789041 7.016438 7.153425 7.287671
2011-01-03 6.504110 6.780822 7.008219 7.145205 7.279452
2011-01-04 6.501370 6.778082 7.005479 7.142466 7.276712
2011-01-05 6.498630 6.775342 7.002740 7.139726 7.273973
2011-01-06 6.495890 6.772603 7.000000 7.136986 7.271233
2011-01-07 6.493151 6.769863 6.997260 7.134247 7.268493",header=TRUE,stringsAsFactors=FALSE)


DesiredOutput = xts(DesiredOutput,order.by=as.Date(rownames(DesiredOutput),format="%Y-%m-%d"))



all.equal(merged_TS,DesiredOutput)
#[1] "Mean relative difference: 3.67637e-08"

Answer 2

与其创建一堆 xts 对象然后通过 Reduce 递归合并它们，不如直接创建一个 xts 对象。

mat <- sapply(dt, function(d) (d-index(x))/365)
res <- xts(mat, index(x))
colnames(res) <- paste0("col", seq(ncol(res)))

我个人觉得这更直接。

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