我想计算一系列Managers回报的均值，最小值和最大值，如下所示：
ManagerRet <-data.frame(diff(Managerprices)/lag(Managerprices,k=-1))
然后我用NaN替换return = 0，因为从数据库中提取数据而不是填充所有日期。
ManagerRet = replace(ManagerRet,ManagerRet==0,NaN)
我有以下3个功能
> min(ManagerRet,na.rm = TRUE)
[1] -0.0091716
> max(ManagerRet,na.rm = TRUE)
[1] 0.007565
> mean(ManagerRet,na.rm = TRUE)*252
[1] NaN
为什么均值函数会在最小和最大性能计算正确的情况下返回NaN值？
下面你可以找到动物园对象MangerRet
> ManagerRet
Manager
2011-10-04 NaN
2011-10-05 NaN
2011-10-06 NaN
2011-10-07 NaN
2011-10-11 NaN
2011-10-12 NaN
2011-10-13 NaN
2011-10-14 NaN
2011-10-17 NaN
2011-10-18 NaN
2011-10-19 NaN
2011-10-20 NaN
2011-10-21 NaN
2011-10-24 NaN
2011-10-25 NaN
2011-10-26 NaN
2011-10-27 NaN
2011-10-28 NaN
2011-10-31 6.3832e-04
2011-11-01 -4.4625e-06
2011-11-02 2.8142e-03
2011-11-03 5.1114e-04
2011-11-04 -1.0105e-03
2011-11-07 7.5650e-03
2011-11-08 2.1002e-03
2011-11-09 -9.1716e-03
2011-11-10 1.1173e-03
2011-11-14 -6.9207e-03
2011-11-15 2.6241e-04
2011-11-16 1.7520e-03
2011-11-17 -2.6443e-05
2011-11-18 -1.4169e-03
2011-11-21 3.7602e-04
2011-11-22 4.3982e-05
2011-11-23 -6.7328e-06
2011-11-25 1.1571e-05
2011-11-28 1.4016e-07
2011-11-29 -2.0426e-07
所需的其他信息
> sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: i386-w64-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=Italian_Italy.1252 LC_CTYPE=Italian_Italy.1252
[3] LC_MONETARY=Italian_Italy.1252 LC_NUMERIC=C
[5] LC_TIME=Italian_Italy.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] gWidgetsRGtk2_0.0-81 gWidgets_0.0-52
[3] RGtk2_2.20.25 lattice_0.20-15
[5] moments_0.13 data.table_1.8.8
[7] tseries_0.10-30 timeDate_2160.97
[9] PerformanceAnalytics_1.1.0 xts_0.9-3
[11] zoo_1.7-9 RODBC_1.3-6
loaded via a namespace (and not attached):
[1] grid_2.15.2 quadprog_1.5-4
答案 0 :(得分：4)
您应该使用colMeans：
colMeans(ManagerRet, na.rm=TRUE)
## Manager
## -6.826297e-05
如果这是一个data.frame，你会收到一个警告（但输出正确）。
在这里，您发现data.frame和zoo对象与[的子集具有逻辑矩阵索引的方式不一致。这似乎是[.zoo中的错误。我已经通过电子邮件发送了维护者。
问题发生在mean.default：
if (na.rm)
x <- x[!is.na(x)]
这是它出错的地方：
ManagerRet[!is.na(ManagerRet)]
## 1
## NaN
!is.na(ManagerRet)看起来像预期的那样，但不是：
class(!is.na(ManagerRet))
[1] "matrix"
此类在[.zoo中意外。这些行存在：
if (all(class(i) == "logical"))
i <- which(rep(i, length.out = n2))
else if (inherits(i, "zoo") && all(class(coredata(i)) ==
"logical")) {
i <- which(coredata(merge(zoo(, time(x)), i)))
}
else if (!((all(class(i) == "numeric") || all(class(i) ==
"integer"))))
i <- which(MATCH(index(x), i, nomatch = 0L) > 0L)
此处的最后一行实际上是在这种情况下运行，产生不正确的结果。
结构：
> dput(ManagerRet)
structure(c(NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN,
NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, 0.00063832, -4.4625e-06,
0.0028142, 0.00051114, -0.0010105, 0.007565, 0.0021002, -0.0091716,
0.0011173, -0.0069207, 0.00026241, 0.001752, -2.6443e-05, -0.0014169,
0.00037602, 4.3982e-05, -6.7328e-06, 1.1571e-05, 1.4016e-07,
-2.0426e-07), .Dim = c(38L, 1L), .Dimnames = list(c("2011-10-04",
"2011-10-05", "2011-10-06", "2011-10-07", "2011-10-11", "2011-10-12",
"2011-10-13", "2011-10-14", "2011-10-17", "2011-10-18", "2011-10-19",
"2011-10-20", "2011-10-21", "2011-10-24", "2011-10-25", "2011-10-26",
"2011-10-27", "2011-10-28", "2011-10-31", "2011-11-01", "2011-11-02",
"2011-11-03", "2011-11-04", "2011-11-07", "2011-11-08", "2011-11-09",
"2011-11-10", "2011-11-14", "2011-11-15", "2011-11-16", "2011-11-17",
"2011-11-18", "2011-11-21", "2011-11-22", "2011-11-23", "2011-11-25",
"2011-11-28", "2011-11-29"), "Manager"), index = 1:38, class = "zoo")
旧代码 - colMeans是执行此操作的正确方法：
使用$指定“列”可以解决这个问题：
mean(ManagerRet, na.rm=TRUE)
## [1] NaN
mean(ManagerRet$Manager, na.rm=TRUE)
## [1] -6.826297e-05