我想知道是否有data.table选项批量处理来自数据集的线性模型，并先进行检查。
我需要在每个唯一标识符上运行一堆线性模型，但首先我需要进行检查。对于每个唯一身份证和年份，我需要检查以前每月数据至少有24个月，但不超过60个月。因此，当我进行回归时，它应包括每个人每年24至60次上个月（年）数据的观察结果。如果该年度的数据少于24个月，则该年度的数据将被删除，但如果超过60个，则仅使用60个月。
感谢这个（感谢@akrun）帖子，我能够为每个人设置线性模型，运行它们，然后输出beta作为两个beta的总和。问题是，这只会在当前年度（12个障碍物）而不是之前的24-60个曲线上进行回归。
上一篇文章：dplyr and previous observations
我希望有一个dplyr选项，但它似乎不会起作用，并且帖子及其中的ddply方法需要数小时才能运行。但是，我需要在110万个障碍物的各种数据集上多次运行。
输入示例：
tdata <- structure(list(cusip = c(101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L, 101L), date = c(19901130L, 19901031L, 19900928L,
19900831L, 19900731L, 19900629L, 19900531L, 19900430L, 19900330L,
19900228L, 19900131L, 19891229L, 19891130L, 19891031L, 19890929L,
19890831L, 19890731L, 19890630L, 19890531L, 19890428L, 19890331L,
19890228L, 19890131L, 19881230L, 19881130L, 19881031L, 19880930L,
19880831L, 19880729L, 19880630L, 19880531L, 19880429L, 19880331L,
19880229L, 19880129L, 19871231L, 19871130L, 19871030L, 19870930L,
19870831L, 19870731L, 19870630L, 19870529L, 19870430L, 19870331L,
19870227L, 19870130L, 19861231L, 19861128L, 19861031L, 19860930L,
19860829L, 19860731L), fyear = c("1990", "1990", "1990", "1990",
"1990", "1990", "1990", "1990", "1990", "1990", "1990", "1989",
"1989", "1989", "1989", "1989", "1989", "1989", "1989", "1989",
"1989", "1989", "1989", "1988", "1988", "1988", "1988", "1988",
"1988", "1988", "1988", "1988", "1988", "1988", "1988", "1987",
"1987", "1987", "1987", "1987", "1987", "1987", "1987", "1987",
"1987", "1987", "1987", "1986", "1986", "1986", "1986", "1986",
"1986"), month = c("11", "10", "09", "08", "07", "06", "05",
"04", "03", "02", "01", "12", "11", "10", "09", "08", "07", "06",
"05", "04", "03", "02", "01", "12", "11", "10", "09", "08", "07",
"06", "05", "04", "03", "02", "01", "12", "11", "10", "09", "08",
"07", "06", "05", "04", "03", "02", "01", "12", "11", "10", "09",
"08", "07"), ret = c("0.117647", "0.030303", "-0.161017", "-0.186207",
"-0.131737", "0.128378", "0.027778", "-0.162791", "0.131579",
"0.178295", "-0.091549", "0.163934", "-0.089552", "0.007519",
"0.117647", "0.155340", "0.211765", "0.024096", "0.338710", "0.377778",
"0.071429", "-0.176471", "0.378378", "-0.026316", "-0.050000",
"-0.047619", "-0.086957", "-0.061224", "0.088889", "-0.062500",
"-0.040000", "-0.056604", "0.081633", "0.042553", "-0.096154",
"0.238095", "-0.263158", "-0.393617", "-0.160714", "0.400000",
"-0.090909", "-0.200000", "-0.098361", "-0.152778", "0.000000",
"0.107692", "0.460674", "-0.101010", "-0.019802", "0.246914",
"-0.052632", "0.179310", "-0.064516"), ewretd = c(0.035468, -0.057155,
-0.080468, -0.108911, -0.025732, 0.005359, 0.045675, -0.028117,
0.021315, 0.015434, -0.046408, -0.012375, -0.0058, -0.049934,
0.005532, 0.018626, 0.031017, -0.007744, 0.025054, 0.029089,
0.01806, 0.002988, 0.062124, 0.018872, -0.036484, -0.011485,
0.016951, -0.025001, 0.000289, 0.047677, -0.017671, 0.014016,
0.03569, 0.060265, 0.077392, 0.026065, -0.05085, -0.272248, -0.015876,
0.014544, 0.035123, 0.021487, 0.000573, -0.017709, 0.036283,
0.074612, 0.117565, -0.034609, -0.006263, 0.023777, -0.059071,
0.023269, -0.073128), lagewretd = c(-0.004526, 0.035468, -0.057155,
-0.080468, -0.108911, -0.025732, 0.005359, 0.045675, -0.028117,
0.021315, 0.015434, -0.046408, -0.012375, -0.0058, -0.049934,
0.005532, 0.018626, 0.031017, -0.007744, 0.025054, 0.029089,
0.01806, 0.002988, 0.062124, 0.018872, -0.036484, -0.011485,
0.016951, -0.025001, 0.000289, 0.047677, -0.017671, 0.014016,
0.03569, 0.060265, 0.077392, 0.026065, -0.05085, -0.272248, -0.015876,
0.014544, 0.035123, 0.021487, 0.000573, -0.017709, 0.036283,
0.074612, 0.117565, -0.034609, -0.006263, 0.023777, -0.059071,
0.023269)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,
-53L), .Names = c("cusip", "date", "fyear", "month", "ret", "ewretd",
"lagewretd"))
ddply方法：
library(dplyr)
## convert fyear to a proper number and then exploit for sorting
tdata <- tdata %>%
mutate(fyear = fyear %>% as.integer) %>%
arrange(fyear, month)
## figure out cumulative months available for each year (for each cusip)
yearstuff <- tdata %>%
group_by(cusip, fyear) %>%
summarize(n = n()) %>%
mutate(n_cum = cumsum(n))
## iterate over rows of yearstuff (for each cusip)
models <- plyr::ddply(yearstuff, ~ cusip + fyear, function(y) {
if(y$n_cum < 24) {
c('(Intercept)' = NA_real_, ewretd = NA_real_, lagewretd = NA_real_)
} else {
my_dat <- tdata %>%
filter(cusip == y$cusip, fyear <= y$fyear) %>%
mutate(rn = row_number(desc(date)))
lm(ret ~ ewretd + lagewretd, my_dat, subset = rn < 61) %>% coef
}
})
答案 0 :(得分：2)
我会为你获得系数的所有计算编写一个单独的函数。然后，您可以使用plyr，dplyr或data.table。您可能应该使用更大的数据集重新运行下面的基准测试。
# function to get coefficients
# (further optimization should probably focus on improving this function)
get_coefs <- function(.cusip, .fyear, .n_cum){
if(.n_cum < 24) {
data_frame(`(Intercept)` = NA_real_, ewretd = NA_real_, lagewretd = NA_real_)
} else {
my_dat <- tdata %>%
filter(cusip == .cusip, fyear <= .fyear) %>%
mutate(rn = row_number(desc(date)))
lm(ret ~ ewretd + lagewretd, my_dat, subset = rn < 61) %>%
coef %>%
as.list %>%
as_data_frame
}
}
require(microbenchmark)
microbenchmark(
models_plyr <- plyr::ddply(yearstuff, ~ cusip + fyear, function(y)
get_coefs(y$cusip, y$fyear, y$n_cum))
,
models_dplyr <- yearstuff %>%
group_by(cusip, fyear) %>%
do(get_coefs(.$cusip, .$fyear, .$n_cum))
,
models_dt <- as.data.table(as.data.frame(yearstuff))[, get_coefs(cusip, fyear, n_cum), by = list(cusip, fyear)]
)
## min lq mean median uq max neval cld
## 12.69178 13.29136 13.62600 13.45849 13.67471 16.73910 100 c
## 12.45302 12.94036 13.33589 13.14721 13.59907 14.73485 100 b
## 10.66120 11.09856 11.43126 11.21593 11.45625 13.69591 100 a
all.equal(models_plyr %>% data.frame,
models_dplyr %>% data.frame)
## [1] TRUE
all.equal(models_plyr %>% data.frame,
models_dt %>% data.frame)
## [1] TRUE