Question

我试图在R中合并两个不同的时间序列，具有以下特征：

数据必须每天在08:30至15:00之间。
数据跨越数周，而不仅仅是某一天。
数据中随机存在间隙。
这两个数据集必须在相同的时间间隔内没有间隙

我想合并两个数据集，所有时间都在08:30到15:00的序列中，并且每个数据集中都有一个间隙，我希望结转前一个值（或后面的值）。 / p>

# I have verified that the csv files are imported correctly
# The first column contains dates. and the strptime
# function can convert strings into Date/Time objects.
#
sec1_dates <- strptime(sec1[,1], "%m/%d/%Y %H:%M:%S")
sec2_dates <- strptime(sec2[,1], "%m/%d/%Y %H:%M:%S")

# The second column contains the close.
# I use the zoo function to create zoo objects from that data.
# But for some reason this ends up creating duplicates PROBLEM 1
#
a <- zoo(sec1[,2], sec1_dates)
b <- zoo(sec2[,2], sec2_dates)

# I know that I need use seq to fill in gaps but I am clueless as to how
# Once I have the proper seq I can just use na.locf to fill the appropriate values
# HOWEVER seq(start(sec1_dates), end(sec1_dates), "min") would end up returning
# every minute for each day, and I only want 08:30 to 15:30. PROBLEM 2

# The merge function can combine two zoo objects, in union
# Obviously this fails because the two index sizes don't match PROBLEM 3
#
t.zoo <- merge(a, b, all=TRUE)

詹姆斯你对问题1是正确的。谢谢。我验证了csv文件是两次拉数据并删除数据修复了问题。我也使用了你的问题2的解决方案，但我不确定这是做我正在尝试做的最有效的方法。最终，我可能希望使用它来运行回归，并且在那时可能需要某种循环来拉动任意数量的数据集。我可能会做出任何优化。

更新的解决方案

library(zoo)
library(tseries)

# Read the CSV files into data frames
sec1 <- read.csv("C:\\exportdata\\sec1.csv", stringsAsFactors=F, header=F)
sec2 <- read.csv("C:\\exportdata\\sec2.csv", stringsAsFactors=F, header=F)

# The first column contains dates.  
# I use strptime to tell it what format these appear in.
sec1_dates <- strptime(sec1[,1], "%m/%d/%Y %H:%M:%S")
sec2_dates <- strptime(sec2[,1], "%m/%d/%Y %H:%M:%S")

# The second column contains the close prices for the securities.
# I use the zoo function to create zoo objects from that data.
# Input =  a vector of data and a vector of dates.
a <- zoo(sec1[,2], sec1_dates)
b <- zoo(sec2[,2], sec2_dates)

# create a discrete time-series with the exact time frame desired
# per tip from James
template <- zoo(NULL, seq(sec1_dates[1], tail(sec1_dates, 1), "min"))
template <- template[which(strftime(time(template),"%H:%M")>"08:30" & strftime(time(template),"%H:%M")<"15:00")]

# The merge function is then used to merge
# 1) each security to the template (uses the discrete date/time range)
# 2) remove the column of data from template (used only for dates)
# 3) each security to one another (this was the ultimate goal anyway.
a.zoo <- merge(a, template, all=TRUE)
a.zoo$template <- NULL
b.zoo <- merge(b, template, all=TRUE)
b.zoo$template <- NULL
t.zoo <- merge(a.zoo, b.zoo, all=TRUE)

# Fill all NA elements with the closest non NA value.
t <- na.locf(t.zoo)

Answer 1

问题1

?zoo详细介绍了如何处理重复项，但这可能是因为您在strptime创建的日期中有重复项。

问题2

您可以使用[，which和time与zoo个对象进行子集时间，请参阅?zoo，例如：

t.zoo[which(strftime(time(t.zoo),"%H:%M")>"08:30" & strftime(time(t.zoo),"%H:%M")<"15:30")]

问题3

使用c合并：t.zoo <- c(a,b)

R时间序列，序列复杂

1 个答案: