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我有多个 xts 对象存储在一个列表中，每个对象有 1000 多行。它们代表股票滚动窗口回归数据。每个元素都有其唯一的 Ticker 名称。在这里，出于测试目的，它们被称为 Stock1、2 ...等。行按日期命名，xts 格式也是如此。每个元素的维度都是相等的。每个看起来像这样：

  > tail(testlist$Stock1, n = 3)
            (Intercept)     rmrf         smb        hml        rmw      cma
2014-12-29 0.0003223177 1.010215 -0.02164844 -0.3322500 0.07819563 1.106934
2014-12-30 0.0002631315 1.002356 -0.02351438 -0.3465390 0.05954400 1.118506
2014-12-31 0.0002837304 1.000084 -0.01619536 -0.3494401 0.06121434 1.124845
> tail(testlist$Stock2, n = 3)
            (Intercept)      rmrf        smb         hml        rmw       cma
2014-12-29 0.0003308951 0.7503819 -0.1967255 -0.10242616 -0.2264914 0.8329570
2014-12-30 0.0003051495 0.7409709 -0.1899856 -0.07461764 -0.2240448 0.7921883
2014-12-31 0.0002614874 0.7478099 -0.1833077 -0.06197362 -0.2056615 0.7550211
> tail(testlist$Stock3, n = 3)
             (Intercept)      rmrf        smb       hml        rmw        cma
2014-12-29 -0.0003803988 0.8363603 -0.4153470 0.7459769 -0.7981382 -0.2839360
2014-12-30 -0.0004121386 0.8352243 -0.4224404 0.7405976 -0.8114066 -0.2790438
2014-12-31 -0.0004660716 0.8355641 -0.4343012 0.7571033 -0.8057412 -0.3026019
> tail(testlist$Stock4, n = 3)
             (Intercept)      rmrf         smb         hml        rmw       cma
2014-12-29 -0.0008295692 0.9296299 -0.07776571 0.007084297 -0.1377356 0.8038542
2014-12-30 -0.0007734696 0.9383387 -0.08941983 0.011685507 -0.1092656 0.7863335
2014-12-31 -0.0007591168 0.9391670 -0.08782070 0.015619229 -0.1083707 0.7924232

我需要做的： 按名称合并行并聚合列表中的所有数据以获得一组新数据。每个应该如下所示：

 Name    Date      (Intercept)     rmrf        smb        hml        rmw         cma

Stock1 2014-12-29  0.0003223177 1.010215  -0.02164844 -0.3322500  0.07819563  1.106934
Stock2 2014-12-29  0.0003308951 0.7503819 -0.1967255  -0.10242616 -0.2264914  0.8329570
Stock3 2014-12-29 -0.0003803988 0.8363603 -0.4153470   0.7459769  -0.7981382 -0.2839360
Stock4 2014-12-29 -0.0008295692 0.9296299 -0.07776571  0.007084297 -0.1377356 0.8038542

每个这样的元素不应该再是一个时间序列。但是一个静态的，每只股票代表它在时间“t”的系数值。就大小而言，每个元素的行数应等于原始列表中的 Stocks 数。

按照josilber的要求进行编辑

   > dput(list(Stock1=tail(testlist$Stock1, n = 3), Stock2=tail(testlist$Stock2, n = 3)))

structure(list(Stock1 = structure(c(0.000322317700198485, 0.000263131488679374, 
0.000283730373928844, 1.01021497011709, 1.00235580055438, 1.00008407331697, 
-0.0216484434660844, -0.023514378867335, -0.0161953614672028, 
-0.332250031553704, -0.346538978804535, -0.349440052163927, 0.078195628743663, 
0.0595439997647003, 0.0612143446991752, 1.1069343396633, 1.11850626745067, 
1.12484530131584), class = c("xts", "zoo"), .indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC", index = structure(c(1419811200, 
1419897600, 1419984000), tzone = "UTC", tclass = "Date"), .Dim = c(3L, 
6L), .Dimnames = list(NULL, c("(Intercept)", "rmrf", "smb", "hml", 
"rmw", "cma"))), Stock2 = structure(c(0.000330895099805035, 0.000305149500450527, 
0.000261487411574969, 0.750381906747217, 0.740970893865186, 0.747809929767095, 
-0.1967254672836, -0.189985607343021, -0.183307667378927, -0.10242615734439, 
-0.0746176364711423, -0.0619736225998069, -0.226491384004977, 
-0.224044849587752, -0.205661480898329, 0.832956994676299, 0.792188348360969, 
0.755021100668421), class = c("xts", "zoo"), .indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC", index = structure(c(1419811200, 
1419897600, 1419984000), tzone = "UTC", tclass = "Date"), .Dim = c(3L, 
6L), .Dimnames = list(NULL, c("(Intercept)", "rmrf", "smb", "hml", 
"rmw", "cma")))), .Names = c("Stock1", "Stock2"))

我完全处于黑暗之中。我查看了一些可能会使用的功能：lapply/ 该merge功能似乎也很合适，但它只适用于 2 个元素。

我会在寻找答案时继续更新这篇文章。如果有人有任何线索或以前这样做过并且可以指出正确的方向，谢谢！

编辑

#Flatten data add one more name and put into one data frame 

all_coef_data<- do.call(rbind,Map(cbind,
                  Name=names(testlist),
                  Date=lapply(testlist,function(x) as.Date(as.POSIXct(c(attr(x,'index')),origin='1970-01-01'))),
                  lapply(testlist, as.data.frame)
))

我需要退出的每一行中的一个共同点是日期。我现在使用这个按日期拆分数据框。输出是一个列表。

out <- split(all_coef_data , f = all_coef_data$Date )

输出：

> head(out$'2011-05-23', n=3)
                    Name       Date   (Intercept)     rmrf         smb        hml         rmw        cma
Stock1.2011-05-23 Stock1 2011-05-23 -4.376389e-04 1.103582 -0.21747611 -0.1879211 -0.05849794 -0.1949192
Stock2.2011-05-23 Stock2 2011-05-23  1.115140e-04 1.198622  0.05422819  0.9998529  0.92141407 -0.8565260
Stock3.2011-05-23 Stock3 2011-05-23  5.457214e-05 1.303025  0.04705294  0.6897673 -0.19708983 -0.8247877
> tail(out$'2011-05-23', n=3)
                      Name       Date  (Intercept)     rmrf        smb         hml         rmw        cma
Stock48.2011-05-23 Stock48 2011-05-23 0.0007354997 0.505054  0.1774544 -0.38934089  0.71775909  0.5189329
Stock49.2011-05-23 Stock49 2011-05-23 0.0004224351 1.304719  0.4511903 -0.64937062 -0.08872941  0.1545058
Stock50.2011-05-23 Stock50 2011-05-23 0.0003851261 1.020434 -0.1107910 -0.03964192  0.09526658 -0.4961902
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2 回答 2

1

这是另一个解决方案。

library(xts)
library(magrittr)
library(dplyr)

stock1 = 
  data.frame(a = c(1, 2), b = c(1, 2) ) %>%
  xts(order.by = 
        c("2014-12-29", "2014-12-30") %>%
        as.Date)

stock2 = 
  data.frame(a = c(1, 2), b = c(1, 2) ) %>%
  xts(order.by = 
        c("2014-12-29", "2014-12-30") %>%
        as.Date)

test = list(stock1 = stock1, stock2 = stock2)

result = 
  1:length(test) %>%
  lapply(function(i)
    test[[i]] %>%
      as.data.frame %>%
      mutate(name = names(test)[i],
             date = rownames(.) %>% as.Date ) ) %>%
  bind_rows
于 2015-08-24T01:11:40.677 回答
1

听起来你想

  1. as.data.frame()使用,将 xts 对象展平为 data.frames
  2. cbind()新列Name（来自列表组件名称）和Date（来自 xts 行名称，实际上来自index属性），最后
  3. rbind()将所有内容合并到一个 data.frame 中。
do.call(rbind,Map(cbind,
    Name=names(testlist),
    Date=lapply(testlist,function(x) as.Date(as.POSIXct(c(attr(x,'index')),origin='1970-01-01'))),
    lapply(testlist,as.data.frame)
));
##                     Name       Date  (Intercept)      rmrf         smb         hml         rmw       cma
## Stock1.2014-12-29 Stock1 2014-12-29 0.0003223177 1.0102150 -0.02164844 -0.33225003  0.07819563 1.1069343
## Stock1.2014-12-30 Stock1 2014-12-30 0.0002631315 1.0023558 -0.02351438 -0.34653898  0.05954400 1.1185063
## Stock1.2014-12-31 Stock1 2014-12-31 0.0002837304 1.0000841 -0.01619536 -0.34944005  0.06121434 1.1248453
## Stock2.2014-12-29 Stock2 2014-12-29 0.0003308951 0.7503819 -0.19672547 -0.10242616 -0.22649138 0.8329570
## Stock2.2014-12-30 Stock2 2014-12-30 0.0003051495 0.7409709 -0.18998561 -0.07461764 -0.22404485 0.7921883
## Stock2.2014-12-31 Stock2 2014-12-31 0.0002614874 0.7478099 -0.18330767 -0.06197362 -0.20566148 0.7550211

如果您不喜欢新的行名，可以将该行换成`rownames<-`(...,NULL).

于 2015-08-24T00:42:55.807 回答