Question

我在文件夹中有多个文件，我用以下文件阅读这些文件：

files <- list.files( "PATH", pattern = '*csv' , full.names = TRUE)
(length(files))
for( i in length(files) ) { 
    df <- fread(files[i], header = TRUE, sep = ";",stringsAsFactors=FALSE)
}

我知道我一次又一次地覆盖df对象。我想要实现的是：

我希望将我的数据格式化为此示例数据：

> (str(StockPriceReturns))
'zoo' series from 2000-04-03 to 2013-03-28
  Data: num [1:3246, 1:30] NA NA NA NA NA NA NA NA NA NA ...
 - attr(*, "dimnames")=List of 2
  ..$ : NULL
  ..$ : chr [1:30] "Bajaj.Auto" "BHEL" "Bharti.Airtel" "Cipla" ...
  Index:  Date[1:3246], format: "2000-04-03" "2000-04-04" "2000-04-05" "2000-04-06" ...
NULL
> (head(StockPriceReturns))
           Bajaj.Auto       BHEL Bharti.Airtel     Cipla Coal.India   Dr.Reddy
2000-04-03         NA  4.9171044            NA  6.810041         NA -3.2541653
2000-04-04         NA -8.3348496            NA -3.368606         NA -8.3353739
2000-04-05         NA  0.3305788            NA  0.836825         NA  0.2616345
2000-04-06         NA -2.7605266            NA -2.466056         NA -1.8941289
2000-04-07         NA  3.2543548            NA  7.690426         NA  7.6961041
2000-04-10         NA  3.3107586            NA  6.154276         NA  6.4769648

我的数据实际上看起来就是这样：

> (dput(head(df1,10)))
structure(list(Name = c("C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT", 
"C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT", 
"C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT", 
"C-QUADRAT INVESTMENT", "C-QUADRAT INVESTMENT"), Date = c("01.01.2002", 
"02.01.2002", "03.01.2002", "04.01.2002", "07.01.2002", "08.01.2002", 
"09.01.2002", "10.01.2002", "11.01.2002", "14.01.2002"), Price = c("na", 
"na", "na", "na", "na", "na", "na", "na", "na", "na"), Currency = c("E", 
"E", "E", "E", "E", "E", "E", "E", "E", "E"), CDax = c("-0,260460226", 
"-1,827437365", "-0,814370143", "0,861279951", "-0,339133689", 
"-1,034650372", "0,713336597", "0,52727784", "2,893518519", "0,05790388"
), `Total Price Returns` = c("na", "na", "na", "na", "na", "na", 
"na", "na", "na", "na"), AbnormalReturns = c("na", "na", "na", 
"na", "na", "na", "na", "na", "na", "na")), .Names = c("Name", 
"Date", "Price", "Currency", "CDax", "Total Price Returns", "AbnormalReturns"
), class = c("data.table", "data.frame"), row.names = c(NA, -10L
), .internal.selfref = <pointer: 0x00000000002a0788>)
                    Name       Date Price Currency         CDax
 1: C-QUADRAT INVESTMENT 01.01.2002    na        E -0,260460226
 2: C-QUADRAT INVESTMENT 02.01.2002    na        E -1,827437365
 3: C-QUADRAT INVESTMENT 03.01.2002    na        E -0,814370143
 4: C-QUADRAT INVESTMENT 04.01.2002    na        E  0,861279951
 5: C-QUADRAT INVESTMENT 07.01.2002    na        E -0,339133689
 6: C-QUADRAT INVESTMENT 08.01.2002    na        E -1,034650372
 7: C-QUADRAT INVESTMENT 09.01.2002    na        E  0,713336597
 8: C-QUADRAT INVESTMENT 10.01.2002    na        E   0,52727784
 9: C-QUADRAT INVESTMENT 11.01.2002    na        E  2,893518519
10: C-QUADRAT INVESTMENT 14.01.2002    na        E   0,05790388
    Total Price Returns AbnormalReturns
 1:                  na              na
 2:                  na              na
 3:                  na              na
 4:                  na              na
 5:                  na              na
 6:                  na              na
 7:                  na              na
 8:                  na              na
 9:                  na              na
10:                  na              na

我可以为单变量情况执行此操作，我只需将data.frame转换为zoo对象，然后使用此函数将此对象转换为xts对象：

dfToZoo <- function(df) {
    require(zoo)
    date <- as.Date(df[, 1], format = '%d.%m.%Y')
    #TODO have a look if the column are rightly named
    with(df, zoo(TotalReturns, date))
}

但是如何为多变量情况做这个呢？

Answer 1

一种可能性是使用read.zoo读取您的数据，并使用split参数对其进行整形。

# some data
df <- data.frame(Date = 20140101 + 0:2,
                 Name = rep(c("Bajaj.Auto", "BHEL", "Bharti.Airtel"), each = 3),
                 CDAX = rnorm(9))
df
#       Date          Name       CDAX
# 1 20140101    Bajaj.Auto  0.4020118
# 2 20140102    Bajaj.Auto -0.7317482
# 3 20140103    Bajaj.Auto  0.8303732
# 4 20140101          BHEL -1.2080828
# 5 20140102          BHEL -1.0479844
# 6 20140103          BHEL  1.4411577
# 7 20140101 Bharti.Airtel -1.0158475
# 8 20140102 Bharti.Airtel  0.4119747
# 9 20140103 Bharti.Airtel -0.3810761

# convert to zoo object. Use 'split' to reshape, and 'format' date if necessary.
z <- read.zoo(file = df, format = "%Y%m%d", split = "Name")

str(z)
# ‘zoo’ series from 2002-01-01 to 2002-01-04
#   Data: num [1:4, 1:3] 2.308 0.1058 0.457 -0.0772 1.0274 ...
# - attr(*, "dimnames")=List of 2
#   ..$ : NULL
#   ..$ : chr [1:3] "Bajaj.Auto" "Bharti.Airtel" "BHEL"
#   Index:  Date[1:4], format: "2002-01-01" "2002-01-02" "2002-01-03" "2002-01-04"

z
#            Bajaj.Auto Bharti.Airtel      BHEL
# 2014-01-01  0.4020118    -1.0158475 -1.208083
# 2014-01-02 -0.7317482     0.4119747 -1.047984
# 2014-01-03  0.8303732    -0.3810761  1.441158

使用多列创建xts对象

1 个答案: