我有一个类似于此虚拟数据的数据帧（时间序列）：
df <- data.frame(stringsAsFactors=FALSE,
symbol = c("N2", "NJ", "K-Kl", "K-P3", "K-N", "KP+", "K13", "KS",
"KTotal", "P500", "P800", "P23", "P55", "PA", "PKA"),
date = c("2017-10-12", "2017-10-12", "2017-10-12", "2017-10-12",
"2017-10-12", "2017-10-12", "2017-10-12", "2017-10-12",
"2017-10-12", "2017-10-12", "2017-10-12", "2017-10-12", "2017-10-12",
"2017-10-12", "2017-10-12"),
open_pr = c(10.2, 2.7, 0.5, 4.5, 2.9, 8.1, 2.3, 1, 43.2, 28.5, 5.8, 6.7,
5.7, 0.1, 10),
gross = c(460L, 121L, 21L, 203L, 130L, 363L, 102L, 45L, 1946L, 1282L,
262L, 303L, 256L, 6L, 449L),
avg_aud = c(19L, 3L, 0L, 5L, 5L, 21L, 4L, 1L, 153L, 92L, 10L, 14L, 6L, 0L,
27L),
ts = c(59L, 32L, 31L, 34L, 57L, 83L, 59L, 28L, 113L, 103L, 53L, 69L,
33L, 4L, 87L),
tv = c(6L, 1L, 0L, 2L, 2L, 7L, 1L, 0L, 49L, 29L, 3L, 5L, 2L, 0L, 9L)
)
head（df）
symbol date open_pr gross avg_aud ts tv
1 N2 2017-10-12 10.2 460 19 59 6
2 NJ 2017-10-12 2.7 121 3 32 1
3 K-Kl 2017-10-12 0.5 21 0 31 0
4 K-P3 2017-10-12 4.5 203 5 34 2
5 K-N 2017-10-12 2.9 130 5 57 2
我的代码段
df %>%
as.tbl() %>%
mutate(date = ymd(date)) %>%
as.xts(date_col = date)
错误消息
Error in as.POSIXlt.character(x, tz, ...) :
character string is not in a standard unambiguous format
我想将此数据帧转换为xts对象。与股市数据类似的东西
library(quamtmod)
x <- getSymbols("GOOG", auto.assign = FALSE)
结果：
GOOG.Open GOOG.High GOOG.Low GOOG.Close GOOG.Volume GOOG.Adjusted
2007-01-03 231.4944 236.7899 229.0652 232.2842 15513200 232.2842
2007-01-04 232.9847 240.4114 232.6618 240.0686 15877700 240.0686
2007-01-05 239.6910 242.1749 237.5102 242.0209 13833500 242.0209
2007-01-08 242.2693 243.3522 239.5420 240.2276 9570600 240.2276
2007-01-09 241.1565 242.5475 239.0452 241.1814 10832700 241.1814
答案 0 :(得分：3)
下面的代码将为您提供dplyr和管道所需的功能。我不确定为什么所有事情都需要用管道来完成，因为不是每个功能都是针对magrittr管道构建的。对于as.xts，如果要使用管道，则需要使用.$引用日期列。
但是结果将不会有用。 xts转换矩阵中的数据，由于Symbol和date在矩阵中，因此整个矩阵将是字符矩阵。
library(xts)
library(dplyr)
df %>%
mutate(date = as.Date(date)) %>%
as.xts(order.by = .$date)
symbol date open_pr gross avg_aud ts tv
2017-10-12 "N2" "2017-10-12" "10.2" " 460" " 19" " 59" " 6"
2017-10-12 "NJ" "2017-10-12" " 2.7" " 121" " 3" " 32" " 1"
2017-10-12 "K-Kl" "2017-10-12" " 0.5" " 21" " 0" " 31" " 0"
2017-10-12 "K-P3" "2017-10-12" " 4.5" " 203" " 5" " 34" " 2"
2017-10-12 "K-N" "2017-10-12" " 2.9" " 130" " 5" " 57" " 2"
2017-10-12 "KP+" "2017-10-12" " 8.1" " 363" " 21" " 83" " 7"
2017-10-12 "K13" "2017-10-12" " 2.3" " 102" " 4" " 59" " 1"
2017-10-12 "KS" "2017-10-12" " 1.0" " 45" " 1" " 28" " 0"
2017-10-12 "KTotal" "2017-10-12" "43.2" "1946" "153" "113" "49"
2017-10-12 "P500" "2017-10-12" "28.5" "1282" " 92" "103" "29"
2017-10-12 "P800" "2017-10-12" " 5.8" " 262" " 10" " 53" " 3"
2017-10-12 "P23" "2017-10-12" " 6.7" " 303" " 14" " 69" " 5"
2017-10-12 "P55" "2017-10-12" " 5.7" " 256" " 6" " 33" " 2"
2017-10-12 "PA" "2017-10-12" " 0.1" " 6" " 0" " 4" " 0"
2017-10-12 "PKA" "2017-10-12" "10.0" " 449" " 27" " 87" " 9"
但是，如果您想在Google底部找到类似您的示例的内容，请使用以下类似内容。
第1步是创建一个函数，以创建xts时间序列，并在列名前面加上符号。步骤2按符号分割原始数据，并创建一个列表，以将所有数据包含在命名列表中。步骤3是使用Map将功能应用于数据。之后，您可以访问my_data列表中的所有数据。
my_func <- function(x, symbol){
index <- as.Date(x[["date"]])
x <- x[, setdiff(colnames(x), c("symbol", "date"))]
x <- xts::as.xts(x, order.by = index)
colnames(x) <- paste0(symbol, ".", colnames(x))
return(x)
}
my_data <- split(df, df$symbol)
my_data <- Map(my_func, my_data, names(my_data))
head(my_data, 2)
$`K-Kl`
K-Kl.open_pr K-Kl.gross K-Kl.avg_aud K-Kl.ts K-Kl.tv
2017-10-12 0.5 21 0 31 0
$`K-N`
K-N.open_pr K-N.gross K-N.avg_aud K-N.ts K-N.tv
2017-10-12 2.9 130 5 57 2
答案 1 :(得分：1)
您可以简单地编写以下代码
x <- xts(df[,c(1,3:7)],df$date)
为我工作
答案 2 :(得分：0)
这项工作可以吗？
library(lubridate)
df$date <- date(df$date)
library(timetk)
df <- tk_xts(df, date_col = date)
保持管道功能：</ p>
library(lubridate)
dat <- df %>%
as.tbl() %>%
mutate(date = date(date)) %>%
tk_xts(date_col = date)