在时间序列上平均组

Question

我正在研究 Fama-McBeth 回归，并尝试平均投资组合组。为此，我最大化了投资组合组的方差，并确定 ABC、FHI、GED 是合适的分组。现在我需要跨时间序列获取平均回报并将它们组合在一起。我正在使用单独的for 循环来完成此操作，但我正在寻找一种更“R”的方式来通过dplyr 或lapply 或任何其他适当的方法来执行此操作，但我不确定如何完成此操作。我正在尝试摆脱 for 循环思维模式，希望能提供任何帮助。

输入：

    data <- structure(list(`NA` = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
), Index = c(29.65, -11.91, 14.73, 27.68, 5.18, 25.97, 10.64, 
1.02, 18.82, 23.92, -41.61, -6.64), Factor = c(-9.84, 6.46, 16.12, 
-16.51, 17.82, -13.31, -3.52, 8.43, 8.23, 7.06, -15.74, 2.03), 
    A = c(33.88, -49.87, 65.14, 14.46, 15.67, -32.17, -31.55, 
    -23.79, -4.59, -8.03, 78.22, 4.75), B = c(-25.2, 24.7, -25.04, 
    -38.64, 61.93, 44.94, -74.65, 47.02, 28.69, 48.61, -85.02, 
    42.95), C = c(36.48, -25.11, 18.91, -23.31, 63.95, -19.56, 
    50.18, -42.28, -0.54, 23.65, -0.79, -48.6), D = c(42.89, 
    -54.39, -39.86, -0.72, -32.82, 69.42, 74.52, 28.61, 2.32, 
    26.26, -68.7, 26.27), E = c(-39.89, 44.92, -3.91, -3.21, 
    44.26, 90.43, 15.38, -17.64, 42.36, -3.65, -85.71, 13.24), 
    F = c(39.67, -54.33, -5.69, 92.39, -42.96, 76.72, 21.95, 
    28.83, 18.93, 23.31, -45.64, -34.34), G = c(74.57, -79.76, 
    26.73, -3.82, 101.67, 1.72, -43.95, 98.01, -2.45, 15.36, 
    2.27, -54.47), H = c(40.22, -71.58, 14.49, 13.74, 24.24, 
    77.22, -13.4, 28.12, 37.65, 80.59, -72.47, -1.5), I = c(90.19, 
    -26.64, 18.14, 0.09, 8.98, 72.38, 28.95, 39.41, 94.67, 52.51, 
    -80.26, -24.46)), .Names = c("NA", "Index", "Factor", "A", 
"B", "C", "D", "E", "F", "G", "H", "I"), row.names = c(NA, -12L
), class = "data.frame")

For 循环：

# Get Group 1 (ABC) averages
vec <- NULL
ABC <- NULL
for (i in 1:12) {
  avg <- (data[i,4] + data[i,5] + data[i,6])/3
  vec <- append(vec, avg)
  assign(paste(stocks[1], stocks[2], stocks[3], sep = ""), vec)
}

# Get Group 2 (FHI) averages
vec <- NULL
FHI <- NULL
for (i in 1:12) {
  avg <- (data[i,9] + data[i,11] + data[i,12])/3
  vec <- append(vec, avg)
  assign(paste(stocks[6], stocks[8], stocks[9], sep = ""), vec)
}

# Get Group 1 (GED) averages
vec <- NULL
GED <- NULL
for (i in 1:12) {
  avg <- mean(data[i,7] + data[i,8] + data[i,10])/3
  vec <- append(vec, avg)
  assign(paste(stocks[7], stocks[5], stocks[4], sep = ""), vec)
}

group.df <- data.frame(ABC = ABC, FHI = FHI, GED = GED, index = data$Index, factor = data$Factor)    # Build data.frame with averages, index, and factor

当前（期望）输出：

          ABC        FHI        GED  index factor
1   15.053333  56.693333  25.856667  29.65  -9.84
2  -16.760000 -50.850000 -29.743333 -11.91   6.46
3   19.670000   8.980000  -5.680000  14.73  16.12
4  -15.830000  35.406667  -2.583333  27.68 -16.51
5   47.183333  -3.246667  37.703333   5.18  17.82
6   -2.263333  75.440000  53.856667  25.97 -13.31
7  -18.673333  12.500000  15.316667  10.64  -3.52
8   -6.350000  32.120000  36.326667   1.02   8.43
9    7.853333  50.416667  14.076667  18.82   8.23
10  21.410000  52.136667  12.656667  23.92   7.06
11  -2.530000 -66.123333 -50.713333 -41.61 -15.74
12  -0.300000 -20.100000  -4.986667  -6.64   2.03

Answer 1

简单的答案是

library(dplyr)
data %>% mutate(ABC = (A + B + C) / 3,
                FHI = (F + H + I) / 3,
                GED = (G + E + D) / 3) %>%
    select(ABC, FHI, GED, Index, Factor)

这或多或少以相同的方式完成您所拥有的。 （我省略了 GED 代码中单个值的 mean。）它不会自动将名称粘贴在一起。如果您想这样做，那么我希望您提供一个分组键，例如

grouping = data.frame(stock = c("A", "B", "C", "D", "E", "F", "G", "H", "I"),
                      group = c(1, 1, 1, 3, 3, 2, 3, 2, 2))

然后事情可以更自动地完成。由于您的示例在看似任意的分组中手动硬编码列号，因此使列名成为唯一的编程部分并不会增加太多。

要自动执行操作，您需要将 melt 数据转换为长格式，然后加入分组 df，然后您可以在 group_by(group, NA) 子句之后计算平均值，最后使用dcast。 （顺便说一句，"NA" 是列名的糟糕选择，我建议使用“row.number”或“time”或任何非特殊字符。）

清理基地R

我喜欢上面的 dplyr，但你做了很多不必要的事情。即使在基础 R 中，您也可以更轻松地到达目的地，例如

data$ABC = with(data, (A + B + C) / 3)
data$FHI = with(data, (F + H + I) / 3)
data$GED = with(data, (G + E + D) / 3)
data = data[, c("ABC", "FHI", "GED", "Index", "Factor")]