如何在没有来自高频包的 aggregateTrades 的情况下进行先前的报价聚合
How to do previous tick aggregation without aggregateTrades from highfrequency package
我需要每隔 5 分钟对我的报价数据集进行一次报价汇总。请注意我想要做的类似于高频包中的 aggregateTrades() 函数。但是由于其他一些数据处理问题，我需要在不使用高频包的情况下解决这个问题。这是我的数据集：
dput(tt)
structure(c(1371.25, NA, 1373.95, NA, NA, 1373, NA, 1373.95,
1373.9, NA, NA, 1374, 1374.15, NA, 1374, 1373.85, 1372.55, 1374.05,
1374.15, 1374.75, NA, NA, 1375.9, 1374.05, NA, NA, NA, NA, NA,
NA, NA, 1375, NA, NA, NA, NA, NA, 1376.35, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 1376.25, NA, 1378, 1376.5, NA, NA, NA, 1378,
1378, NA, NA, 1378.8, 231.9, 231.85, NA, 231.9, 231.85, 231.9,
231.8, 231.9, 232.6, 231.95, 232.35, 232, 232.1, 232.05, 232.05,
232.05, 231.5, 231.3, NA, NA, 231.1, 231.1, 231.1, 231, 231,
230.95, 230.6, 230.6, 230.7, 230.6, 231, NA, 231, 231, 231.45,
231.65, 231.4, 231.7, 231.3, 231.25, 231.25, 231.4, 231.4, 231.85,
231.75, 231.5, 231.55, 231.35, NA, 231.5, 231.5, NA, 231.5, 231.25,
231.15, 231, 231, 231, 231.05, NA), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta", index = structure(c(1459481850,
1459482301, 1459482302, 1459482303, 1459482304, 1459482305, 1459482306,
1459482307, 1459482309, 1459482310, 1459482311, 1459482312, 1459482314,
1459482315, 1459482316, 1459482317, 1459482318, 1459482319, 1459482320,
1459482321, 1459482322, 1459482323, 1459482324, 1459482326, 1459482328,
1459482329, 1459482330, 1459482331, 1459482332, 1459482336, 1459482337,
1459482338, 1459482339, 1459482342, 1459482344, 1459482346, 1459482347,
1459482348, 1459482349, 1459482350, 1459482351, 1459482354, 1459482355,
1459482356, 1459482357, 1459482358, 1459482359, 1459482362, 1459482363,
1459482364, 1459482369, 1459482370, 1459482371, 1459482372, 1459482373,
1459482378, 1459482379, 1459482380, 1459482382, 1459482388), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), .Dim = c(60L, 2L), .Dimnames = list(NULL, c("A",
"B")), class = c("xts", "zoo"))
这是我之前的分时聚合代码：
ag.5min.tt<-tt%>%filter(as.Date(index(tt)))%>%lapply(aggregate(by=cut(format(index(tt), format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
我想用上面的代码做的是每天 为 A 和 B 的价格设置 5 分钟的间隔。但是我遇到了错误。请建议如何修复此错误：
Error in UseMethod("filter_") :
no applicable method for 'filter_' applied to an object of class "c('xts', 'zoo')"
谢谢。
编辑：
将 xts 对象转换为数据帧：
tt<-as.data.frame(tt)
tt<-data.frame(Time=rownames(tt), coredata(tt))
ag.5min.tt<-tt%>% filter(as.Date(index(tt)))%>%lapply(aggregate(by=cut(format(index(tt), format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
新错误：
Error in eval(substitute(expr), envir, enclos) :
filter condition does not evaluate to a logical vector.
编辑：
尝试：
tt$Time<- as.POSIXct(tt$Time, format="%Y-%m-%d %H:%M:%S")
ag.5min.tt<-tt%>% group_by(Time==as.Date(tt$Time))%>%lapply(aggregate(by=cut(format(tt$Time, format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
错误：
Error in cut.default(format(tt$Time, format = "%H:%M:%S"), breaks = "5 mins", :
'x' must be numeric
In addition: Warning message:
In eval(substitute(expr), envir, enclos) :
Incompatible methods ("Ops.POSIXt", "Ops.Date") for "=="
结果应该是这样的。每五分钟时间戳将具有该特定时间戳的值，或者如果存在 NA，则该时间戳将具有股票 A 和 B 的最后一个非 NA 值
time A B
1 2016-04-01 09:00:00 NA NA
2 2016-04-01 09:05:00 NA NA
3 2016-04-01 09:10:00 NA NA
4 2016-04-01 09:15:00 1371.25 231.90
5 2016-04-01 09:20:00 1376.35 231.55
将 xts 对象更改为数据框以与 dplyr 一起使用
library(dplyr)
library(tibble)
library(xts)
library(tidyr)
dtf <- tt %>%
as.data.frame() %>%
# add time information
rownames_to_column("time") %>%
mutate(time = as.POSIXct(time))
生成要选择的时间向量
在最小和最大时间之间每 5 分钟（300 秒）
timepick <- seq(trunc(min(dtf$time),"hour"), # start at the hour
max(dtf$time)+300 , 300)
使用中断向量选择最后一个可用的
每5分钟观察一次。
ag.5min.tt <- dtf %>%
# Add missing interval
full_join(data_frame(time = timepick), by = "time") %>%
arrange(time) %>% # important to arrange by time here
# Replace each NA with the most recent non-NA
fill(-time) %>%
# take selected values only
filter(time %in% timepick)
转换回 xts 对象
ag.5min.tt <- ag.5min.tt %>%
as.data.frame() %>%
column_to_rownames("time") %>%
as.xts()
ag.5min.tt
A B
2016-04-01 09:00:00 NA NA
2016-04-01 09:05:00 NA NA
2016-04-01 09:10:00 1371.25 231.90
2016-04-01 09:15:00 1371.25 231.90
2016-04-01 09:20:00 1378.80 231.05
您可以使用 .indexmin 按分钟为您的时间序列编制索引，然后操纵该索引以对观察结果进行子集化：
ind <- which(diff(.indexmin(tt) %% 5) == -4)
res <- tt[ind]
此处，.indexmin(tt) %% 5 将 return 自上次第五分钟以来的分钟数。出于我们的目的，我们想从中提取每个 4 系列的最后一个索引，这是每五分钟前一分钟的最后一次观察。为此，我们可以使用 diff 并仅提取从 4 到 0 交叉的索引（导致 diff of -4) 使用 which.
为了说明，我们修改您发布的数据以添加实际满足您的提取条件的观察结果：
tt <- structure(c(1371.25, NA, 1373.95, NA, NA, 1373, NA, 1373.95,
1373.9, NA, NA, 1374, 1374.15, NA, 1374, 1373.85, 1372.55, 1374.05,
1374.15, 1374.75, NA, NA, 1375.9, 1374.05, NA, NA, NA, NA, NA,
NA, NA, 1375, NA, NA, NA, NA, NA, 1376.35, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 1376.25, NA, 1378, 1376.5, NA, NA, NA, 1378,
1378, NA, NA, 1378.8, 231.9, 231.85, NA, 231.9, 231.85, 231.9,
231.8, 231.9, 232.6, 231.95, 232.35, 232, 232.1, 232.05, 232.05,
232.05, 231.5, 231.3, NA, NA, 231.1, 231.1, 231.1, 231, 231,
230.95, 230.6, 230.6, 230.7, 230.6, 231, NA, 231, 231, 231.45,
231.65, 231.4, 231.7, 231.3, 231.25, 231.25, 231.4, 231.4, 231.85,
231.75, 231.5, 231.55, 231.35, NA, 231.5, 231.5, NA, 231.5, 231.25,
231.15, 231, 231, 231, 231.05, NA), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta", index = structure(c(1459482299,
1459482301, 1459482302, 1459482303, 1459482304, 1459482305, 1459482306,
1459482307, 1459482309, 1459482310, 1459482311, 1459482312, 1459482314,
1459482315, 1459482316, 1459482317, 1459482318, 1459482319, 1459482320,
1459482321, 1459482322, 1459482323, 1459482324, 1459482326, 1459482328,
1459482329, 1459482330, 1459482331, 1459482332, 1459482336, 1459482337,
1459482338, 1459482339, 1459482342, 1459482344, 1459482346, 1459482347,
1459482348, 1459482349, 1459482590, 1459482591, 1459482594, 1459482595,
1459482596, 1459482597, 1459482598, 1459482599, 1459482602, 1459482603,
1459482604, 1459482609, 1459482610, 1459482611, 1459482612, 1459482613,
1459482618, 1459482619, 1459482620, 1459482622, 1459482628), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), .Dim = c(60L, 2L), .Dimnames = list(NULL,c("A",
"B")), class = c("xts", "zoo"))
## A B
##2016-04-01 09:14:59 1371.25 231.90
##2016-04-01 09:15:01 NA 231.85
##2016-04-01 09:15:02 1373.95 NA
##2016-04-01 09:15:03 NA 231.90
##2016-04-01 09:15:04 NA 231.85
##2016-04-01 09:15:05 1373.00 231.90
##2016-04-01 09:15:06 NA 231.80
##2016-04-01 09:15:07 1373.95 231.90
##2016-04-01 09:15:09 1373.90 232.60
##2016-04-01 09:15:10 NA 231.95
##2016-04-01 09:15:11 NA 232.35
##2016-04-01 09:15:12 1374.00 232.00
##2016-04-01 09:15:14 1374.15 232.10
##2016-04-01 09:15:15 NA 232.05
##2016-04-01 09:15:16 1374.00 232.05
##2016-04-01 09:15:17 1373.85 232.05
##2016-04-01 09:15:18 1372.55 231.50
##2016-04-01 09:15:19 1374.05 231.30
##2016-04-01 09:15:20 1374.15 NA
##2016-04-01 09:15:21 1374.75 NA
##2016-04-01 09:15:22 NA 231.10
##2016-04-01 09:15:23 NA 231.10
##2016-04-01 09:15:24 1375.90 231.10
##2016-04-01 09:15:26 1374.05 231.00
##2016-04-01 09:15:28 NA 231.00
##2016-04-01 09:15:29 NA 230.95
##2016-04-01 09:15:30 NA 230.60
##2016-04-01 09:15:31 NA 230.60
##2016-04-01 09:15:32 NA 230.70
##2016-04-01 09:15:36 NA 230.60
##2016-04-01 09:15:37 NA 231.00
##2016-04-01 09:15:38 1375.00 NA
##2016-04-01 09:15:39 NA 231.00
##2016-04-01 09:15:42 NA 231.00
##2016-04-01 09:15:44 NA 231.45
##2016-04-01 09:15:46 NA 231.65
##2016-04-01 09:15:47 NA 231.40
##2016-04-01 09:15:48 1376.35 231.70
##2016-04-01 09:15:49 NA 231.30
##2016-04-01 09:19:50 NA 231.25
##2016-04-01 09:19:51 NA 231.25
##2016-04-01 09:19:54 NA 231.40
##2016-04-01 09:19:55 NA 231.40
##2016-04-01 09:19:56 NA 231.85
##2016-04-01 09:19:57 NA 231.75
##2016-04-01 09:19:58 NA 231.50
##2016-04-01 09:19:59 NA 231.55
##2016-04-01 09:20:02 NA 231.35
##2016-04-01 09:20:03 1376.25 NA
##2016-04-01 09:20:04 NA 231.50
##2016-04-01 09:20:09 1378.00 231.50
##2016-04-01 09:20:10 1376.50 NA
##2016-04-01 09:20:11 NA 231.50
##2016-04-01 09:20:12 NA 231.25
##2016-04-01 09:20:13 NA 231.15
##2016-04-01 09:20:18 1378.00 231.00
##2016-04-01 09:20:19 1378.00 231.00
##2016-04-01 09:20:20 NA 231.00
##2016-04-01 09:20:22 NA 231.05
##2016-04-01 09:20:28 1378.80 NA
根据这些数据，我们得到：
print(res)
## A B
##2016-04-01 09:14:59 1371.25 231.90
##2016-04-01 09:19:59 NA 231.55
要获得您发布的输出，您需要首先生成一个时间序列，其中包含您想要的每 5 分钟刻度的数据（设置为 NA）。对于此示例，此时间序列（仅适用于 2016-04-01 上从 09:00 到 09:20 的 5 分钟刻度）可以是：
every.5.min <- structure(c(NA, NA, NA, NA, NA), .Dim = c(5L, 1L), .Dimnames = list(
NULL, "Empty"), index = structure(c(1459481400, 1459481700,
1459482000, 1459482300, 1459482600), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), class = c("xts", "zoo"), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta")
## Empty
##2016-04-01 09:00:00 NA
##2016-04-01 09:05:00 NA
##2016-04-01 09:10:00 NA
##2016-04-01 09:15:00 NA
##2016-04-01 09:20:00 NA
然后，merge 这与 tt:
tt <- merge(tt, every.5.min, all=TRUE)[,1:ncol(tt)]
如果 tt 中不存在该行（即每 5 分钟），all=TRUE 将用 NA 填充原始 tt 中的行。请注意合并后，我们只保留原始 tt.
中的列
然后，在 tt 上，用之前的值填充所有 NA：
res <- do.call(merge, lapply(tt, na.locf))
最后，使用 .indexmin 和 .indexsec:
仅提取每 5 分钟刻度的那些行
res <- res[.indexmin(res) %% 5 == 0 & .indexsec(res) == 0]
## A B
##2016-04-01 09:00:00 NA NA
##2016-04-01 09:05:00 NA NA
##2016-04-01 09:10:00 NA NA
##2016-04-01 09:15:00 1371.25 231.90
##2016-04-01 09:20:00 1376.35 231.55
我需要每隔 5 分钟对我的报价数据集进行一次报价汇总。请注意我想要做的类似于高频包中的 aggregateTrades() 函数。但是由于其他一些数据处理问题，我需要在不使用高频包的情况下解决这个问题。这是我的数据集：
dput(tt)
structure(c(1371.25, NA, 1373.95, NA, NA, 1373, NA, 1373.95,
1373.9, NA, NA, 1374, 1374.15, NA, 1374, 1373.85, 1372.55, 1374.05,
1374.15, 1374.75, NA, NA, 1375.9, 1374.05, NA, NA, NA, NA, NA,
NA, NA, 1375, NA, NA, NA, NA, NA, 1376.35, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 1376.25, NA, 1378, 1376.5, NA, NA, NA, 1378,
1378, NA, NA, 1378.8, 231.9, 231.85, NA, 231.9, 231.85, 231.9,
231.8, 231.9, 232.6, 231.95, 232.35, 232, 232.1, 232.05, 232.05,
232.05, 231.5, 231.3, NA, NA, 231.1, 231.1, 231.1, 231, 231,
230.95, 230.6, 230.6, 230.7, 230.6, 231, NA, 231, 231, 231.45,
231.65, 231.4, 231.7, 231.3, 231.25, 231.25, 231.4, 231.4, 231.85,
231.75, 231.5, 231.55, 231.35, NA, 231.5, 231.5, NA, 231.5, 231.25,
231.15, 231, 231, 231, 231.05, NA), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta", index = structure(c(1459481850,
1459482301, 1459482302, 1459482303, 1459482304, 1459482305, 1459482306,
1459482307, 1459482309, 1459482310, 1459482311, 1459482312, 1459482314,
1459482315, 1459482316, 1459482317, 1459482318, 1459482319, 1459482320,
1459482321, 1459482322, 1459482323, 1459482324, 1459482326, 1459482328,
1459482329, 1459482330, 1459482331, 1459482332, 1459482336, 1459482337,
1459482338, 1459482339, 1459482342, 1459482344, 1459482346, 1459482347,
1459482348, 1459482349, 1459482350, 1459482351, 1459482354, 1459482355,
1459482356, 1459482357, 1459482358, 1459482359, 1459482362, 1459482363,
1459482364, 1459482369, 1459482370, 1459482371, 1459482372, 1459482373,
1459482378, 1459482379, 1459482380, 1459482382, 1459482388), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), .Dim = c(60L, 2L), .Dimnames = list(NULL, c("A",
"B")), class = c("xts", "zoo"))
这是我之前的分时聚合代码：
ag.5min.tt<-tt%>%filter(as.Date(index(tt)))%>%lapply(aggregate(by=cut(format(index(tt), format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
我想用上面的代码做的是每天 为 A 和 B 的价格设置 5 分钟的间隔。但是我遇到了错误。请建议如何修复此错误：
Error in UseMethod("filter_") :
no applicable method for 'filter_' applied to an object of class "c('xts', 'zoo')"
谢谢。
编辑： 将 xts 对象转换为数据帧：
tt<-as.data.frame(tt)
tt<-data.frame(Time=rownames(tt), coredata(tt))
ag.5min.tt<-tt%>% filter(as.Date(index(tt)))%>%lapply(aggregate(by=cut(format(index(tt), format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
新错误：
Error in eval(substitute(expr), envir, enclos) :
filter condition does not evaluate to a logical vector.
编辑： 尝试：
tt$Time<- as.POSIXct(tt$Time, format="%Y-%m-%d %H:%M:%S")
ag.5min.tt<-tt%>% group_by(Time==as.Date(tt$Time))%>%lapply(aggregate(by=cut(format(tt$Time, format = "%H:%M:%S"), breaks = "5 mins", Fun=tail)))
错误：
Error in cut.default(format(tt$Time, format = "%H:%M:%S"), breaks = "5 mins", :
'x' must be numeric
In addition: Warning message:
In eval(substitute(expr), envir, enclos) :
Incompatible methods ("Ops.POSIXt", "Ops.Date") for "=="
结果应该是这样的。每五分钟时间戳将具有该特定时间戳的值，或者如果存在 NA，则该时间戳将具有股票 A 和 B 的最后一个非 NA 值
time A B
1 2016-04-01 09:00:00 NA NA
2 2016-04-01 09:05:00 NA NA
3 2016-04-01 09:10:00 NA NA
4 2016-04-01 09:15:00 1371.25 231.90
5 2016-04-01 09:20:00 1376.35 231.55
将 xts 对象更改为数据框以与 dplyr 一起使用
library(dplyr)
library(tibble)
library(xts)
library(tidyr)
dtf <- tt %>%
as.data.frame() %>%
# add time information
rownames_to_column("time") %>%
mutate(time = as.POSIXct(time))
生成要选择的时间向量 在最小和最大时间之间每 5 分钟（300 秒）
timepick <- seq(trunc(min(dtf$time),"hour"), # start at the hour
max(dtf$time)+300 , 300)
使用中断向量选择最后一个可用的 每5分钟观察一次。
ag.5min.tt <- dtf %>%
# Add missing interval
full_join(data_frame(time = timepick), by = "time") %>%
arrange(time) %>% # important to arrange by time here
# Replace each NA with the most recent non-NA
fill(-time) %>%
# take selected values only
filter(time %in% timepick)
转换回 xts 对象
ag.5min.tt <- ag.5min.tt %>%
as.data.frame() %>%
column_to_rownames("time") %>%
as.xts()
ag.5min.tt
A B
2016-04-01 09:00:00 NA NA
2016-04-01 09:05:00 NA NA
2016-04-01 09:10:00 1371.25 231.90
2016-04-01 09:15:00 1371.25 231.90
2016-04-01 09:20:00 1378.80 231.05
您可以使用 .indexmin 按分钟为您的时间序列编制索引，然后操纵该索引以对观察结果进行子集化：
ind <- which(diff(.indexmin(tt) %% 5) == -4)
res <- tt[ind]
此处，.indexmin(tt) %% 5 将 return 自上次第五分钟以来的分钟数。出于我们的目的，我们想从中提取每个 4 系列的最后一个索引，这是每五分钟前一分钟的最后一次观察。为此，我们可以使用 diff 并仅提取从 4 到 0 交叉的索引（导致 diff of -4) 使用 which.
为了说明，我们修改您发布的数据以添加实际满足您的提取条件的观察结果：
tt <- structure(c(1371.25, NA, 1373.95, NA, NA, 1373, NA, 1373.95,
1373.9, NA, NA, 1374, 1374.15, NA, 1374, 1373.85, 1372.55, 1374.05,
1374.15, 1374.75, NA, NA, 1375.9, 1374.05, NA, NA, NA, NA, NA,
NA, NA, 1375, NA, NA, NA, NA, NA, 1376.35, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 1376.25, NA, 1378, 1376.5, NA, NA, NA, 1378,
1378, NA, NA, 1378.8, 231.9, 231.85, NA, 231.9, 231.85, 231.9,
231.8, 231.9, 232.6, 231.95, 232.35, 232, 232.1, 232.05, 232.05,
232.05, 231.5, 231.3, NA, NA, 231.1, 231.1, 231.1, 231, 231,
230.95, 230.6, 230.6, 230.7, 230.6, 231, NA, 231, 231, 231.45,
231.65, 231.4, 231.7, 231.3, 231.25, 231.25, 231.4, 231.4, 231.85,
231.75, 231.5, 231.55, 231.35, NA, 231.5, 231.5, NA, 231.5, 231.25,
231.15, 231, 231, 231, 231.05, NA), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta", index = structure(c(1459482299,
1459482301, 1459482302, 1459482303, 1459482304, 1459482305, 1459482306,
1459482307, 1459482309, 1459482310, 1459482311, 1459482312, 1459482314,
1459482315, 1459482316, 1459482317, 1459482318, 1459482319, 1459482320,
1459482321, 1459482322, 1459482323, 1459482324, 1459482326, 1459482328,
1459482329, 1459482330, 1459482331, 1459482332, 1459482336, 1459482337,
1459482338, 1459482339, 1459482342, 1459482344, 1459482346, 1459482347,
1459482348, 1459482349, 1459482590, 1459482591, 1459482594, 1459482595,
1459482596, 1459482597, 1459482598, 1459482599, 1459482602, 1459482603,
1459482604, 1459482609, 1459482610, 1459482611, 1459482612, 1459482613,
1459482618, 1459482619, 1459482620, 1459482622, 1459482628), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), .Dim = c(60L, 2L), .Dimnames = list(NULL,c("A",
"B")), class = c("xts", "zoo"))
## A B
##2016-04-01 09:14:59 1371.25 231.90
##2016-04-01 09:15:01 NA 231.85
##2016-04-01 09:15:02 1373.95 NA
##2016-04-01 09:15:03 NA 231.90
##2016-04-01 09:15:04 NA 231.85
##2016-04-01 09:15:05 1373.00 231.90
##2016-04-01 09:15:06 NA 231.80
##2016-04-01 09:15:07 1373.95 231.90
##2016-04-01 09:15:09 1373.90 232.60
##2016-04-01 09:15:10 NA 231.95
##2016-04-01 09:15:11 NA 232.35
##2016-04-01 09:15:12 1374.00 232.00
##2016-04-01 09:15:14 1374.15 232.10
##2016-04-01 09:15:15 NA 232.05
##2016-04-01 09:15:16 1374.00 232.05
##2016-04-01 09:15:17 1373.85 232.05
##2016-04-01 09:15:18 1372.55 231.50
##2016-04-01 09:15:19 1374.05 231.30
##2016-04-01 09:15:20 1374.15 NA
##2016-04-01 09:15:21 1374.75 NA
##2016-04-01 09:15:22 NA 231.10
##2016-04-01 09:15:23 NA 231.10
##2016-04-01 09:15:24 1375.90 231.10
##2016-04-01 09:15:26 1374.05 231.00
##2016-04-01 09:15:28 NA 231.00
##2016-04-01 09:15:29 NA 230.95
##2016-04-01 09:15:30 NA 230.60
##2016-04-01 09:15:31 NA 230.60
##2016-04-01 09:15:32 NA 230.70
##2016-04-01 09:15:36 NA 230.60
##2016-04-01 09:15:37 NA 231.00
##2016-04-01 09:15:38 1375.00 NA
##2016-04-01 09:15:39 NA 231.00
##2016-04-01 09:15:42 NA 231.00
##2016-04-01 09:15:44 NA 231.45
##2016-04-01 09:15:46 NA 231.65
##2016-04-01 09:15:47 NA 231.40
##2016-04-01 09:15:48 1376.35 231.70
##2016-04-01 09:15:49 NA 231.30
##2016-04-01 09:19:50 NA 231.25
##2016-04-01 09:19:51 NA 231.25
##2016-04-01 09:19:54 NA 231.40
##2016-04-01 09:19:55 NA 231.40
##2016-04-01 09:19:56 NA 231.85
##2016-04-01 09:19:57 NA 231.75
##2016-04-01 09:19:58 NA 231.50
##2016-04-01 09:19:59 NA 231.55
##2016-04-01 09:20:02 NA 231.35
##2016-04-01 09:20:03 1376.25 NA
##2016-04-01 09:20:04 NA 231.50
##2016-04-01 09:20:09 1378.00 231.50
##2016-04-01 09:20:10 1376.50 NA
##2016-04-01 09:20:11 NA 231.50
##2016-04-01 09:20:12 NA 231.25
##2016-04-01 09:20:13 NA 231.15
##2016-04-01 09:20:18 1378.00 231.00
##2016-04-01 09:20:19 1378.00 231.00
##2016-04-01 09:20:20 NA 231.00
##2016-04-01 09:20:22 NA 231.05
##2016-04-01 09:20:28 1378.80 NA
根据这些数据，我们得到：
print(res)
## A B
##2016-04-01 09:14:59 1371.25 231.90
##2016-04-01 09:19:59 NA 231.55
要获得您发布的输出，您需要首先生成一个时间序列，其中包含您想要的每 5 分钟刻度的数据（设置为 NA）。对于此示例，此时间序列（仅适用于 2016-04-01 上从 09:00 到 09:20 的 5 分钟刻度）可以是：
every.5.min <- structure(c(NA, NA, NA, NA, NA), .Dim = c(5L, 1L), .Dimnames = list(
NULL, "Empty"), index = structure(c(1459481400, 1459481700,
1459482000, 1459482300, 1459482600), tzone = "Asia/Calcutta", tclass = c("POSIXct",
"POSIXt")), class = c("xts", "zoo"), .indexCLASS = c("POSIXct",
"POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "Asia/Calcutta", tzone = "Asia/Calcutta")
## Empty
##2016-04-01 09:00:00 NA
##2016-04-01 09:05:00 NA
##2016-04-01 09:10:00 NA
##2016-04-01 09:15:00 NA
##2016-04-01 09:20:00 NA
然后，merge 这与 tt:
tt <- merge(tt, every.5.min, all=TRUE)[,1:ncol(tt)]
如果 tt 中不存在该行（即每 5 分钟），all=TRUE 将用 NA 填充原始 tt 中的行。请注意合并后，我们只保留原始 tt.
然后，在 tt 上，用之前的值填充所有 NA：
res <- do.call(merge, lapply(tt, na.locf))
最后，使用 .indexmin 和 .indexsec:
res <- res[.indexmin(res) %% 5 == 0 & .indexsec(res) == 0]
## A B
##2016-04-01 09:00:00 NA NA
##2016-04-01 09:05:00 NA NA
##2016-04-01 09:10:00 NA NA
##2016-04-01 09:15:00 1371.25 231.90
##2016-04-01 09:20:00 1376.35 231.55