R:如何通过前一天的信息更改证券交易所每日指数的时间序列中的差距(假期)？

Question

我是R语言,使用来自不同国家的时间序列每日股票指数.为了对不同的索引进行比较(如相关性,因果关系等),我需要所有系列都有相同数量的行,但由于不同国家的差异假,因此每个系列中的行数会发生变化.

我正在使用yahoo finance提取的文件,格式为.csv,就像......

> head(sp)
>           Date    Open    High     Low   Close     Volume Adj.Close
>1288 2010-01-04 1116.56 1133.87 1116.56 1132.99 3991400000   1132.99
>1287 2010-01-05 1132.66 1136.63 1129.66 1136.52 2491020000   1136.52
>1286 2010-01-06 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14

我需要...例如,假设当天2010-01-07是假日,在这种情况下,文件中的下一行(第1285行)是2010-01-08日:

> head(sp)
>           Date    Open    High     Low   Close     Volume Adj.Close
>1288 2010-01-04 1116.56 1133.87 1116.56 1132.99 3991400000   1132.99
>1287 2010-01-05 1132.66 1136.63 1129.66 1136.52 2491020000   1136.52
>1286 2010-01-06 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
>1285 2010-01-08 1140.52 1145.39 1136.22 1144.98 4389590000   1144.98

需要填补2010-01-07与当前日数据的差距,如:

> head(sp)
>           Date    Open    High     Low   Close     Volume Adj.Close
>1288 2010-01-04 1116.56 1133.87 1116.56 1132.99 3991400000   1132.99
>1287 2010-01-05 1132.66 1136.63 1129.66 1136.52 2491020000   1136.52
>1286 2010-01-06 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
>1285 2010-01-07 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
>1284 2010-01-08 1140.52 1145.39 1136.22 1144.98 4389590000   1144.98

我怎么能这样做？

我的代码是(查看我尝试使用的所有库来解决我的问题kkk)

>library(PerformanceAnalytics)
>library(tseries)
>library(urca)
>library(zoo)
>library(lmtest)
>library(timeDate)
>library(timeSeries)

>setwd("C:/Users/Fatima/Documents/R")

>sp = read.csv("SP500.csv", header = TRUE, stringsAsFactors = FALSE)
>sp$Date = as.Date(sp$Date)
>sp = sp[order(sp$Date), ]

抱歉我的英语不好

Answer 1

xts 包在这里很有用：

DF <- read.table(text = "           Date    Open    High     Low   Close     Volume Adj.Close
1288 2010-01-04 1116.56 1133.87 1116.56 1132.99 3991400000   1132.99
1287 2010-01-05 1132.66 1136.63 1129.66 1136.52 2491020000   1136.52
1286 2010-01-06 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
1285 2010-01-08 1140.52 1145.39 1136.22 1144.98 4389590000   1144.98", header = TRUE)

DF$Date <- as.Date(DF$Date)

library(xts)
X <- as.xts(DF[,-1], order.by = DF$Date)
na.locf(merge(X, seq(min(DF$Date), max(DF$Date), by = 1)))
#              Open    High     Low   Close     Volume Adj.Close
#2010-01-04 1116.56 1133.87 1116.56 1132.99 3991400000   1132.99
#2010-01-05 1132.66 1136.63 1129.66 1136.52 2491020000   1136.52
#2010-01-06 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
#2010-01-07 1135.71 1139.19 1133.95 1137.14 4972660000   1137.14
#2010-01-08 1140.52 1145.39 1136.22 1144.98 4389590000   1144.98

编辑：

回应您的评论：您可以排除这样的周末：

dates <- seq(min(DF$Date), max(DF$Date), by = 1)
#you might have to adjust the following to the translations in your locale
dates <- dates[!(weekdays(dates) %in% c("Saturday", "Sunday"))]
na.locf(merge(X, dates))