我有多个xts对象存储在列表中,每个列表都有1000行以上。它们代表股票滚动窗口回归数据。每个元素都有其独特的Ticker名称。在这里他们被称为Stock1,2 ...等用于测试目的。与xts格式一样,行按日期命名。每个元素的尺寸相同。每一个看起来是这样的:如何通过rownames将列表中的数据重新组织到新的数据框中? R
> 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
我需要做什么: 通过名称和通过聚集所有的数据在我的名单,以获得一组新数据的合并行。每个应该看起来像这样:
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是一个时间序列更多。但是是静态的,每只股票在时间“t”代表它的系数值。就大小而言,每个元素的行数应等于原始列表中的股票数量。
编辑如要求通过josilber
> 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"))
我在黑暗中我完全。我看了一些可能会用到的功能:lapply
/merge
函数似乎是合适的,但它只适用于2个元素。
我会继续更新这篇文章,因为我寻找答案。如果任何人有任何线索或已经做到了这一点,并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)
))
每当我需要出去行中的共同点是日期。我现在用Date来分割数据框。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
请问您能提供一个可重现的数据示例吗?例如,包含'dput(list(Stock1 = tail(testlist $ Stock1,n = 3),Stock2 = tail(testlist $ Stock2,n = 3)))'的输出可能就足够了。 – josliber
它现在包括在内。谢谢。 –