计算 R 中每月超过 XTS 对象的滚动年回报

Question

我有一个跨多个列的月回报 XTS 对象，我正在尝试计算每列的滚动年回报（几何）。

Date            Manager 1   Manager 2   Manager 3   Manager 4   Manager 5
20160430        0.0152000   0.0100700   0.0102210   0.0046160   NA
20160531        0.0462000   0.0515240   0.0287490   0.0374920   NA
20160630        0.0007000   0.0126830   0.0156410   0.0130820   NA
20160731        0.0200000   0.0158810   0.0239540   0.0214950   NA
20160831        0.0339000   0.0531980   0.0021170   0.0476160   0.0457650
20160930        -0.0071000  0.0047540   -0.0088080  0.0031540   -0.0034070
20161031        -0.0224000  -0.0181930  0.0181410   -0.0048280  0.0170850
20161130        -0.0439000  -0.0131600  -0.0243030  -0.0064650  -0.0007180
20161231        -0.0051000  0.0200130   0.0204210   0.0160740   0.0172270
20170131        0.0083000   0.0146560   0.0247000   0.0203410   0.0227060
20170228        0.0211000   -0.0067120  0.0257530   0.0029940   0.0124730
20170331        0.0530000   0.0532190   0.0283950   0.0416190   0.0237900
20170430        0.0638300   0.0592280   0.0341340   0.0437430   0.0293500
20170531        0.0339000   0.0264270   0.0287670   0.0207810   0.0179080
20170630        NA          -0.0046950  -0.0091310  -0.0074520  -0.0137600
20170731        NA          0.0109280   0.0029630   0.0146560   0.0167990
20170831        NA          0.0290430   0.0372960   0.0284390   0.0229930
20170930        NA          0.0226390   0.0030190   0.0063850   -0.0087170

预期结果：

Date            Manager 1   Manager 2   Manager 3   Manager 4   Manager 5                       
20160430                        
20160531                        
20160630                        
20160731                        
20160831                        
20160930                        
20161031                        
20161130                        
20161231                        
20170131                        
20170228                        
20170331        0.121979182 0.212964432 0.176317288 0.213932804 
20170430        0.175724107 0.271996881 0.204161963 0.261212111 
20170531        0.161901314 0.241637796 0.204183032 0.240897626 
20170630                    0.220330851 0.174812396 0.215746067 
20170731                    0.214381041 0.150728807 0.207606539 0.200188843
20170831                    0.186529323 0.191124778 0.185500853 0.174054195
20170930                    0.207649992 0.205337395 0.189319163 0.167798654

我一直在使用 PerformanceAnalytics 包，但在每列应用该函数时遇到了一些问题：

apply.rolling(ManagerReturns, width = 12, trim = FALSE ,FUN = Return.annualized)

Answer 1

apply.rolling 是rollapply 的包装。出于某种原因，apply.rolling 无法正确处理您的数据，但使用 rollapply 可以解决问题。

使用rollapply 我可以接近你的结果，但是。但是 Return.annualized 删除了 NA 值但继续计算。您可以在 Manager1 和 Manager5 上看到这种情况。这不是因为rollapply，而是因为Return.annualized。例如，Return.annualized(my_data$Manager5[1:12]) 返回 0.2207884 的年化回报。

ra <- rollapply(my_data, width = 12, FUN = Return.annualized, fill = 0)     

            Manager1  Manager2  Manager3  Manager4  Manager5
2016-04-30 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-05-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-06-30 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-07-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-08-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-09-30 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-10-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-11-30 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2016-12-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2017-01-31 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2017-02-28 0.0000000 0.0000000 0.0000000 0.0000000 0.0000000
2017-03-31 0.1219792 0.2129644 0.1763173 0.2139328 0.2207884
2017-04-30 0.1757241 0.2719969 0.2041620 0.2612121 0.2409790
2017-05-31 0.1619013 0.2416378 0.2041830 0.2408976 0.2406184
2017-06-30 0.1769613 0.2203309 0.1748124 0.2157461 0.1982881
2017-07-31 0.1682027 0.2143810 0.1507288 0.2076065 0.2001888
2017-08-31 0.1368823 0.1865293 0.1911248 0.1855009 0.1740542
2017-09-30 0.1676742 0.2076500 0.2053374 0.1893192 0.1677987

现在您可以执行ra * !is.na(my_data) 之类的操作，如果是 NA，则将 ra 与 0 相乘，并删除 Manager1 的最后 4 条记录。但这对Manager5没有帮助。

数据：

my_data <- structure(c(0.0152, 0.0462, 7e-04, 0.02, 0.0339, -0.0071, -0.0224, 
-0.0439, -0.0051, 0.0083, 0.0211, 0.053, 0.06383, 0.0339, NA, 
NA, NA, NA, 0.01007, 0.051524, 0.012683, 0.015881, 0.053198, 
0.004754, -0.018193, -0.01316, 0.020013, 0.014656, -0.006712, 
0.053219, 0.059228, 0.026427, -0.004695, 0.010928, 0.029043, 
0.022639, 0.010221, 0.028749, 0.015641, 0.023954, 0.002117, -0.008808, 
0.018141, -0.024303, 0.020421, 0.0247, 0.025753, 0.028395, 0.034134, 
0.028767, -0.009131, 0.002963, 0.037296, 0.003019, 0.004616, 
0.037492, 0.013082, 0.021495, 0.047616, 0.003154, -0.004828, 
-0.006465, 0.016074, 0.020341, 0.002994, 0.041619, 0.043743, 
0.020781, -0.007452, 0.014656, 0.028439, 0.006385, NA, NA, NA, 
NA, 0.045765, -0.003407, 0.017085, -0.000718, 0.017227, 0.022706, 
0.012473, 0.02379, 0.02935, 0.017908, -0.01376, 0.016799, 0.022993, 
-0.008717), .Dim = c(18L, 5L), .Dimnames = list(NULL, c("Manager1", 
"Manager2", "Manager3", "Manager4", "Manager5")), index = structure(c(1461974400, 
1464652800, 1467244800, 1469923200, 1472601600, 1475193600, 1477872000, 
1480464000, 1483142400, 1485820800, 1488240000, 1490918400, 1493510400, 
1496188800, 1498780800, 1501459200, 1504137600, 1506729600), tzone = "UTC", tclass = "Date"), class = c("xts", 
"zoo"), .indexCLASS = "Date", tclass = "Date", .indexTZ = "UTC", tzone = "UTC")