如何矢量化依赖于 R 中先前计算的函数？
How to vectorize a function that depends on a previous calculation in R?
我有一些随机游走。我的目标是创建一个函数，向此 data.table 添加一列，根据其累积百分比收益和百分比缩减标记其所在的“区域”。
library(data.table)
set.seed(1)
# generate random returns with drift
df <- data.table(
"date" = 1:50,
"ret" = rnorm(50, mean = .002, sd = .01)
)
# calculate the value of the random-walk over-time
df[, val := cumprod(1 + ret)]
df[, draw_down := val / cummax(val) - 1]
第一个区域出现在第一行并一直上升，直到出现 5% cumulative gain 或 2% drawdown。
第二个区域在第一个区域结束后开始一行，一直持续到同样的情况再次发生，5% cumulative gain 或 2% drawdown
重复此过程，直到这两种情况都没有发生，在这种情况下，区域会继续到最后一行。
这是一个可重现的例子：
# start with the first row and zone of 1
idx <- 1
count <- 1
res <- data.table()
while (idx <= nrow(df)) {
# grab the start of the zone and all future rows
tmp <- df[idx:.N]
# calculate the necessary things
tmp[, val := cumprod(1 + ret)]
tmp[, draw_down := val / cummax(val) - 1]
# find out if we crossed our drawdown threshold
loss_idx <- which(
tmp$draw_down == min(tmp$draw_down[tmp$draw_down <= -.02])
)
# find out if we crossed gain threshold
gain_idx <- which(tmp$val == min(tmp$val[tmp$val >= 1.05]))
# if we have no thresholds, label the rest of the zones
# and exit
if (length(loss_idx) == 0 & length(gain_idx) == 0) {
tmp[, zone := count]
res <- rbind(res, tmp)
break
}
# mark the zone
tmp[1:min(gain_idx, loss_idx), zone := count]
# increment our index
idx <- tmp[min(gain_idx, loss_idx)]$date + 1
print(idx)
# increment our zone
count <- count + 1
res <- rbind(res, tmp[!is.na(zone)])
}
我已经尝试获取这些区域点出现位置的索引。但是后来我 运行 遇到了需要根据最后一个区域的索引重新计算 val 和 drawdown 的问题。我想不出一种方法来对其进行矢量化。也许在这里使用 roll 函数会有效？
问题归结为知道按区域划分的回撤，但需要前一个区域才能计算回撤。与累积 return 类似。如果这个函数依赖于先前的值，是否可以对其进行矢量化？
在尝试实现以下所需输出时，我们将不胜感激任何方向的帮助。
所需的输出：
> res
date ret val draw_down zone
<int> <dbl> <dbl> <dbl> <dbl>
1 -0.0042645381 0.9957355 0.0000000000 1
2 0.0038364332 0.9995555 0.0000000000 1
3 -0.0063562861 0.9932021 -0.0063562861 1
4 0.0179528080 1.0110328 0.0000000000 1
5 0.0052950777 1.0163863 0.0000000000 1
6 -0.0062046838 1.0100800 -0.0062046838 1
7 0.0068742905 1.0170236 0.0000000000 1
8 0.0093832471 1.0265665 0.0000000000 1
9 0.0077578135 1.0345305 0.0000000000 1
10 -0.0010538839 1.0334402 -0.0010538839 1
11 0.0171178117 1.0511304 0.0000000000 1
12 0.0058984324 1.0058984 0.0000000000 2
13 -0.0042124058 1.0016612 -0.0042124058 2
14 -0.0201469989 0.9814807 -0.0242745373 2
15 0.0132493092 1.0132493 0.0000000000 3
16 0.0015506639 1.0148205 0.0000000000 3
17 0.0018380974 1.0166859 0.0000000000 3
18 0.0114383621 1.0283151 0.0000000000 3
19 0.0102122120 1.0388164 0.0000000000 3
20 0.0079390132 1.0470636 0.0000000000 3
21 0.0111897737 1.0587800 0.0000000000 3
22 0.0098213630 1.0691787 0.0000000000 3
23 0.0027456498 1.0721143 0.0000000000 3
24 -0.0178935170 1.0529304 -0.0178935170 3
25 0.0081982575 1.0615626 -0.0098419551 3
26 0.0014387126 1.0630899 -0.0084174023 3
27 0.0004420449 1.0635598 -0.0079790782 3
28 -0.0127075238 1.0500446 -0.0205852077 3
29 -0.0027815006 0.9972185 0.0000000000 4
30 0.0061794156 1.0033807 0.0000000000 4
31 0.0155867955 1.0190202 0.0000000000 4
32 0.0009721227 1.0200108 0.0000000000 4
33 0.0058767161 1.0260051 0.0000000000 4
34 0.0014619496 1.0275051 0.0000000000 4
35 -0.0117705956 1.0154108 -0.0117705956 4
36 -0.0021499456 1.0132277 -0.0138952351 4
37 -0.0019428995 1.0112591 -0.0158111376 4
38 0.0014068660 1.0126818 -0.0144265157 4
39 0.0130002537 1.0258469 -0.0016138103 4
40 0.0096317575 1.0357276 0.0000000000 4
41 0.0003547640 1.0360951 0.0000000000 4
42 -0.0005336168 1.0355422 -0.0005336168 4
43 0.0089696338 1.0448306 0.0000000000 4
44 0.0075666320 1.0527365 0.0000000000 4
45 -0.0048875569 0.9951124 0.0000000000 5
46 -0.0050749516 0.9900623 -0.0050749516 5
47 0.0056458196 0.9956520 0.0000000000 5
48 0.0096853292 1.0052952 0.0000000000 5
49 0.0008765379 1.0061764 0.0000000000 5
50 0.0108110773 1.0170543 0.0000000000 5
我不认为滚动计算是正确的方法：通常他们已经固定 windows，而这更动态一些。类似地，累积运算（例如 cumsum）出于类似原因将不起作用。 （这并不是说我不能扭曲 zoo::rollapply 方法来执行此操作，但我认为它比推荐的方法效率低得多。）
这是一个简单的 while 循环，似乎可以提供您所要求的 zone：
breaks <- integer(0)
rn <- 1L
while (rn <= nrow(df)) {
theserows <- seq(rn, nrow(df))
ratios <- df$val[theserows] / df$val[theserows][1]
upordown <- which(ratios >= 1.05 | ratios <= 0.98)
if (!length(upordown)) break
breaks <- c(breaks, upordown[1] + rn)
rn <- rn + upordown[1]
}
df[, zone := cumsum(seq_len(.N) %in% breaks)]
# date ret val draw_down zone
# <int> <num> <num> <num> <int>
# 1: 1 -0.0042645381 0.9957355 0.0000000000 0
# 2: 2 0.0038364332 0.9995555 0.0000000000 0
# 3: 3 -0.0063562861 0.9932021 -0.0063562861 0
# 4: 4 0.0179528080 1.0110328 0.0000000000 0
# 5: 5 0.0052950777 1.0163863 0.0000000000 0
# 6: 6 -0.0062046838 1.0100800 -0.0062046838 0
# 7: 7 0.0068742905 1.0170236 0.0000000000 0
# 8: 8 0.0093832471 1.0265665 0.0000000000 0
# 9: 9 0.0077578135 1.0345305 0.0000000000 0
# 10: 10 -0.0010538839 1.0334402 -0.0010538839 0
# 11: 11 0.0171178117 1.0511304 0.0000000000 0
# 12: 12 0.0058984324 1.0573304 0.0000000000 1
# 13: 13 -0.0042124058 1.0528765 -0.0042124058 1
# 14: 14 -0.0201469989 1.0316642 -0.0242745373 1
# 15: 15 0.0132493092 1.0453331 -0.0113468490 2
# 16: 16 0.0015506639 1.0469540 -0.0098137803 2
# 17: 17 0.0018380974 1.0488784 -0.0079937216 2
# 18: 18 0.0114383621 1.0608759 0.0000000000 2
# 19: 19 0.0102122120 1.0717098 0.0000000000 2
# 20: 20 0.0079390132 1.0802181 0.0000000000 2
# 21: 21 0.0111897737 1.0923055 0.0000000000 2
# 22: 22 0.0098213630 1.1030334 0.0000000000 2
# 23: 23 0.0027456498 1.1060620 0.0000000000 3
# 24: 24 -0.0178935170 1.0862706 -0.0178935170 3
# 25: 25 0.0081982575 1.0951762 -0.0098419551 3
# 26: 26 0.0014387126 1.0967518 -0.0084174023 3
# 27: 27 0.0004420449 1.0972366 -0.0079790782 3
# 28: 28 -0.0127075238 1.0832934 -0.0205852077 3
# 29: 29 -0.0027815006 1.0802803 -0.0233094505 4
# 30: 30 0.0061794156 1.0869558 -0.0172740737 4
# 31: 31 0.0155867955 1.1038979 -0.0019565256 4
# 32: 32 0.0009721227 1.1049710 -0.0009863049 4
# 33: 33 0.0058767161 1.1114646 0.0000000000 4
# 34: 34 0.0014619496 1.1130896 0.0000000000 4
# 35: 35 -0.0117705956 1.0999878 -0.0117705956 4
# 36: 36 -0.0021499456 1.0976229 -0.0138952351 4
# 37: 37 -0.0019428995 1.0954903 -0.0158111376 4
# 38: 38 0.0014068660 1.0970316 -0.0144265157 4
# 39: 39 0.0130002537 1.1112932 -0.0016138103 4
# 40: 40 0.0096317575 1.1219969 0.0000000000 4
# 41: 41 0.0003547640 1.1223950 0.0000000000 4
# 42: 42 -0.0005336168 1.1217961 -0.0005336168 4
# 43: 43 0.0089696338 1.1318582 0.0000000000 4
# 44: 44 0.0075666320 1.1404225 0.0000000000 4
# 45: 45 -0.0048875569 1.1348486 -0.0048875569 5
# 46: 46 -0.0050749516 1.1290893 -0.0099377044 5
# 47: 47 0.0056458196 1.1354640 -0.0043479913 5
# 48: 48 0.0096853292 1.1464613 0.0000000000 5
# 49: 49 0.0008765379 1.1474662 0.0000000000 5
# 50: 50 0.0108110773 1.1598716 0.0000000000 5
# date ret val draw_down zone
还有一个简单的函数来做同样的事情：
func <- function(x, up = 1.05, down = 0.98) {
breaks <- integer(0)
if (!length(x)) return(breaks)
ind <- 1L
while (ind <= length(x)) {
theseind <- seq(ind, length(x))
ratios <- x[theseind] / x[theseind][1]
upordown <- which(ratios >= up | ratios <= down)
if (!length(upordown)) break
breaks <- c(breaks, upordown[1] + ind)
ind <- ind + upordown[1]
}
return(cumsum(seq_along(x) %in% breaks))
}
df[, zone := func(val, 1.05, 0.98) ]
假设您正在探索矢量化以加速计算，这里是另一个使用 Rccp 加速计算的选项：
library(Rcpp)
cppFunction("IntegerVector zoning(NumericVector idx) {
int zone = 1, n = idx.size();
IntegerVector res = IntegerVector(n);
double x0 = idx[0];
for (int i = 1; i < n; i++) {
res[i] = zone;
if (idx[i]/x0 < 0.98 || idx[i]/x0 > 1.05) {
if (i+1 < n) {
x0 = idx[i+1];
}
zone++;
}
}
return res;
}")
df[, zone := zoning(c(1, val))[-1L]]
输出：
date ret val zone
1: 1 -0.0042645381 0.9957355 1
2: 2 0.0038364332 0.9995555 1
3: 3 -0.0063562861 0.9932021 1
4: 4 0.0179528080 1.0110328 1
5: 5 0.0052950777 1.0163863 1
6: 6 -0.0062046838 1.0100800 1
7: 7 0.0068742905 1.0170236 1
8: 8 0.0093832471 1.0265665 1
9: 9 0.0077578135 1.0345305 1
10: 10 -0.0010538839 1.0334402 1
11: 11 0.0171178117 1.0511304 1
12: 12 0.0058984324 1.0573304 2
13: 13 -0.0042124058 1.0528765 2
14: 14 -0.0201469989 1.0316642 2
15: 15 0.0132493092 1.0453331 3
16: 16 0.0015506639 1.0469540 3
17: 17 0.0018380974 1.0488784 3
18: 18 0.0114383621 1.0608759 3
19: 19 0.0102122120 1.0717098 3
20: 20 0.0079390132 1.0802181 3
21: 21 0.0111897737 1.0923055 3
22: 22 0.0098213630 1.1030334 3
23: 23 0.0027456498 1.1060620 4
24: 24 -0.0178935170 1.0862706 4
25: 25 0.0081982575 1.0951762 4
26: 26 0.0014387126 1.0967518 4
27: 27 0.0004420449 1.0972366 4
28: 28 -0.0127075238 1.0832934 4
29: 29 -0.0027815006 1.0802803 5
30: 30 0.0061794156 1.0869558 5
31: 31 0.0155867955 1.1038979 5
32: 32 0.0009721227 1.1049710 5
33: 33 0.0058767161 1.1114646 5
34: 34 0.0014619496 1.1130896 5
35: 35 -0.0117705956 1.0999878 5
36: 36 -0.0021499456 1.0976229 5
37: 37 -0.0019428995 1.0954903 5
38: 38 0.0014068660 1.0970316 5
39: 39 0.0130002537 1.1112932 5
40: 40 0.0096317575 1.1219969 5
41: 41 0.0003547640 1.1223950 5
42: 42 -0.0005336168 1.1217961 5
43: 43 0.0089696338 1.1318582 5
44: 44 0.0075666320 1.1404225 5
45: 45 -0.0048875569 1.1348486 6
46: 46 -0.0050749516 1.1290893 6
47: 47 0.0056458196 1.1354640 6
48: 48 0.0096853292 1.1464613 6
49: 49 0.0008765379 1.1474662 6
50: 50 0.0108110773 1.1598716 6
date ret val zone
我有一些随机游走。我的目标是创建一个函数，向此 data.table 添加一列，根据其累积百分比收益和百分比缩减标记其所在的“区域”。
library(data.table)
set.seed(1)
# generate random returns with drift
df <- data.table(
"date" = 1:50,
"ret" = rnorm(50, mean = .002, sd = .01)
)
# calculate the value of the random-walk over-time
df[, val := cumprod(1 + ret)]
df[, draw_down := val / cummax(val) - 1]
第一个区域出现在第一行并一直上升，直到出现 5% cumulative gain 或 2% drawdown。
第二个区域在第一个区域结束后开始一行，一直持续到同样的情况再次发生，5% cumulative gain 或 2% drawdown
重复此过程，直到这两种情况都没有发生，在这种情况下，区域会继续到最后一行。
这是一个可重现的例子：
# start with the first row and zone of 1
idx <- 1
count <- 1
res <- data.table()
while (idx <= nrow(df)) {
# grab the start of the zone and all future rows
tmp <- df[idx:.N]
# calculate the necessary things
tmp[, val := cumprod(1 + ret)]
tmp[, draw_down := val / cummax(val) - 1]
# find out if we crossed our drawdown threshold
loss_idx <- which(
tmp$draw_down == min(tmp$draw_down[tmp$draw_down <= -.02])
)
# find out if we crossed gain threshold
gain_idx <- which(tmp$val == min(tmp$val[tmp$val >= 1.05]))
# if we have no thresholds, label the rest of the zones
# and exit
if (length(loss_idx) == 0 & length(gain_idx) == 0) {
tmp[, zone := count]
res <- rbind(res, tmp)
break
}
# mark the zone
tmp[1:min(gain_idx, loss_idx), zone := count]
# increment our index
idx <- tmp[min(gain_idx, loss_idx)]$date + 1
print(idx)
# increment our zone
count <- count + 1
res <- rbind(res, tmp[!is.na(zone)])
}
我已经尝试获取这些区域点出现位置的索引。但是后来我 运行 遇到了需要根据最后一个区域的索引重新计算 val 和 drawdown 的问题。我想不出一种方法来对其进行矢量化。也许在这里使用 roll 函数会有效？
问题归结为知道按区域划分的回撤，但需要前一个区域才能计算回撤。与累积 return 类似。如果这个函数依赖于先前的值，是否可以对其进行矢量化？
在尝试实现以下所需输出时，我们将不胜感激任何方向的帮助。
所需的输出：
> res
date ret val draw_down zone
<int> <dbl> <dbl> <dbl> <dbl>
1 -0.0042645381 0.9957355 0.0000000000 1
2 0.0038364332 0.9995555 0.0000000000 1
3 -0.0063562861 0.9932021 -0.0063562861 1
4 0.0179528080 1.0110328 0.0000000000 1
5 0.0052950777 1.0163863 0.0000000000 1
6 -0.0062046838 1.0100800 -0.0062046838 1
7 0.0068742905 1.0170236 0.0000000000 1
8 0.0093832471 1.0265665 0.0000000000 1
9 0.0077578135 1.0345305 0.0000000000 1
10 -0.0010538839 1.0334402 -0.0010538839 1
11 0.0171178117 1.0511304 0.0000000000 1
12 0.0058984324 1.0058984 0.0000000000 2
13 -0.0042124058 1.0016612 -0.0042124058 2
14 -0.0201469989 0.9814807 -0.0242745373 2
15 0.0132493092 1.0132493 0.0000000000 3
16 0.0015506639 1.0148205 0.0000000000 3
17 0.0018380974 1.0166859 0.0000000000 3
18 0.0114383621 1.0283151 0.0000000000 3
19 0.0102122120 1.0388164 0.0000000000 3
20 0.0079390132 1.0470636 0.0000000000 3
21 0.0111897737 1.0587800 0.0000000000 3
22 0.0098213630 1.0691787 0.0000000000 3
23 0.0027456498 1.0721143 0.0000000000 3
24 -0.0178935170 1.0529304 -0.0178935170 3
25 0.0081982575 1.0615626 -0.0098419551 3
26 0.0014387126 1.0630899 -0.0084174023 3
27 0.0004420449 1.0635598 -0.0079790782 3
28 -0.0127075238 1.0500446 -0.0205852077 3
29 -0.0027815006 0.9972185 0.0000000000 4
30 0.0061794156 1.0033807 0.0000000000 4
31 0.0155867955 1.0190202 0.0000000000 4
32 0.0009721227 1.0200108 0.0000000000 4
33 0.0058767161 1.0260051 0.0000000000 4
34 0.0014619496 1.0275051 0.0000000000 4
35 -0.0117705956 1.0154108 -0.0117705956 4
36 -0.0021499456 1.0132277 -0.0138952351 4
37 -0.0019428995 1.0112591 -0.0158111376 4
38 0.0014068660 1.0126818 -0.0144265157 4
39 0.0130002537 1.0258469 -0.0016138103 4
40 0.0096317575 1.0357276 0.0000000000 4
41 0.0003547640 1.0360951 0.0000000000 4
42 -0.0005336168 1.0355422 -0.0005336168 4
43 0.0089696338 1.0448306 0.0000000000 4
44 0.0075666320 1.0527365 0.0000000000 4
45 -0.0048875569 0.9951124 0.0000000000 5
46 -0.0050749516 0.9900623 -0.0050749516 5
47 0.0056458196 0.9956520 0.0000000000 5
48 0.0096853292 1.0052952 0.0000000000 5
49 0.0008765379 1.0061764 0.0000000000 5
50 0.0108110773 1.0170543 0.0000000000 5
我不认为滚动计算是正确的方法：通常他们已经固定 windows，而这更动态一些。类似地，累积运算（例如 cumsum）出于类似原因将不起作用。 （这并不是说我不能扭曲 zoo::rollapply 方法来执行此操作，但我认为它比推荐的方法效率低得多。）
这是一个简单的 while 循环，似乎可以提供您所要求的 zone：
breaks <- integer(0)
rn <- 1L
while (rn <= nrow(df)) {
theserows <- seq(rn, nrow(df))
ratios <- df$val[theserows] / df$val[theserows][1]
upordown <- which(ratios >= 1.05 | ratios <= 0.98)
if (!length(upordown)) break
breaks <- c(breaks, upordown[1] + rn)
rn <- rn + upordown[1]
}
df[, zone := cumsum(seq_len(.N) %in% breaks)]
# date ret val draw_down zone
# <int> <num> <num> <num> <int>
# 1: 1 -0.0042645381 0.9957355 0.0000000000 0
# 2: 2 0.0038364332 0.9995555 0.0000000000 0
# 3: 3 -0.0063562861 0.9932021 -0.0063562861 0
# 4: 4 0.0179528080 1.0110328 0.0000000000 0
# 5: 5 0.0052950777 1.0163863 0.0000000000 0
# 6: 6 -0.0062046838 1.0100800 -0.0062046838 0
# 7: 7 0.0068742905 1.0170236 0.0000000000 0
# 8: 8 0.0093832471 1.0265665 0.0000000000 0
# 9: 9 0.0077578135 1.0345305 0.0000000000 0
# 10: 10 -0.0010538839 1.0334402 -0.0010538839 0
# 11: 11 0.0171178117 1.0511304 0.0000000000 0
# 12: 12 0.0058984324 1.0573304 0.0000000000 1
# 13: 13 -0.0042124058 1.0528765 -0.0042124058 1
# 14: 14 -0.0201469989 1.0316642 -0.0242745373 1
# 15: 15 0.0132493092 1.0453331 -0.0113468490 2
# 16: 16 0.0015506639 1.0469540 -0.0098137803 2
# 17: 17 0.0018380974 1.0488784 -0.0079937216 2
# 18: 18 0.0114383621 1.0608759 0.0000000000 2
# 19: 19 0.0102122120 1.0717098 0.0000000000 2
# 20: 20 0.0079390132 1.0802181 0.0000000000 2
# 21: 21 0.0111897737 1.0923055 0.0000000000 2
# 22: 22 0.0098213630 1.1030334 0.0000000000 2
# 23: 23 0.0027456498 1.1060620 0.0000000000 3
# 24: 24 -0.0178935170 1.0862706 -0.0178935170 3
# 25: 25 0.0081982575 1.0951762 -0.0098419551 3
# 26: 26 0.0014387126 1.0967518 -0.0084174023 3
# 27: 27 0.0004420449 1.0972366 -0.0079790782 3
# 28: 28 -0.0127075238 1.0832934 -0.0205852077 3
# 29: 29 -0.0027815006 1.0802803 -0.0233094505 4
# 30: 30 0.0061794156 1.0869558 -0.0172740737 4
# 31: 31 0.0155867955 1.1038979 -0.0019565256 4
# 32: 32 0.0009721227 1.1049710 -0.0009863049 4
# 33: 33 0.0058767161 1.1114646 0.0000000000 4
# 34: 34 0.0014619496 1.1130896 0.0000000000 4
# 35: 35 -0.0117705956 1.0999878 -0.0117705956 4
# 36: 36 -0.0021499456 1.0976229 -0.0138952351 4
# 37: 37 -0.0019428995 1.0954903 -0.0158111376 4
# 38: 38 0.0014068660 1.0970316 -0.0144265157 4
# 39: 39 0.0130002537 1.1112932 -0.0016138103 4
# 40: 40 0.0096317575 1.1219969 0.0000000000 4
# 41: 41 0.0003547640 1.1223950 0.0000000000 4
# 42: 42 -0.0005336168 1.1217961 -0.0005336168 4
# 43: 43 0.0089696338 1.1318582 0.0000000000 4
# 44: 44 0.0075666320 1.1404225 0.0000000000 4
# 45: 45 -0.0048875569 1.1348486 -0.0048875569 5
# 46: 46 -0.0050749516 1.1290893 -0.0099377044 5
# 47: 47 0.0056458196 1.1354640 -0.0043479913 5
# 48: 48 0.0096853292 1.1464613 0.0000000000 5
# 49: 49 0.0008765379 1.1474662 0.0000000000 5
# 50: 50 0.0108110773 1.1598716 0.0000000000 5
# date ret val draw_down zone
还有一个简单的函数来做同样的事情：
func <- function(x, up = 1.05, down = 0.98) {
breaks <- integer(0)
if (!length(x)) return(breaks)
ind <- 1L
while (ind <= length(x)) {
theseind <- seq(ind, length(x))
ratios <- x[theseind] / x[theseind][1]
upordown <- which(ratios >= up | ratios <= down)
if (!length(upordown)) break
breaks <- c(breaks, upordown[1] + ind)
ind <- ind + upordown[1]
}
return(cumsum(seq_along(x) %in% breaks))
}
df[, zone := func(val, 1.05, 0.98) ]
假设您正在探索矢量化以加速计算，这里是另一个使用 Rccp 加速计算的选项：
library(Rcpp)
cppFunction("IntegerVector zoning(NumericVector idx) {
int zone = 1, n = idx.size();
IntegerVector res = IntegerVector(n);
double x0 = idx[0];
for (int i = 1; i < n; i++) {
res[i] = zone;
if (idx[i]/x0 < 0.98 || idx[i]/x0 > 1.05) {
if (i+1 < n) {
x0 = idx[i+1];
}
zone++;
}
}
return res;
}")
df[, zone := zoning(c(1, val))[-1L]]
输出：
date ret val zone
1: 1 -0.0042645381 0.9957355 1
2: 2 0.0038364332 0.9995555 1
3: 3 -0.0063562861 0.9932021 1
4: 4 0.0179528080 1.0110328 1
5: 5 0.0052950777 1.0163863 1
6: 6 -0.0062046838 1.0100800 1
7: 7 0.0068742905 1.0170236 1
8: 8 0.0093832471 1.0265665 1
9: 9 0.0077578135 1.0345305 1
10: 10 -0.0010538839 1.0334402 1
11: 11 0.0171178117 1.0511304 1
12: 12 0.0058984324 1.0573304 2
13: 13 -0.0042124058 1.0528765 2
14: 14 -0.0201469989 1.0316642 2
15: 15 0.0132493092 1.0453331 3
16: 16 0.0015506639 1.0469540 3
17: 17 0.0018380974 1.0488784 3
18: 18 0.0114383621 1.0608759 3
19: 19 0.0102122120 1.0717098 3
20: 20 0.0079390132 1.0802181 3
21: 21 0.0111897737 1.0923055 3
22: 22 0.0098213630 1.1030334 3
23: 23 0.0027456498 1.1060620 4
24: 24 -0.0178935170 1.0862706 4
25: 25 0.0081982575 1.0951762 4
26: 26 0.0014387126 1.0967518 4
27: 27 0.0004420449 1.0972366 4
28: 28 -0.0127075238 1.0832934 4
29: 29 -0.0027815006 1.0802803 5
30: 30 0.0061794156 1.0869558 5
31: 31 0.0155867955 1.1038979 5
32: 32 0.0009721227 1.1049710 5
33: 33 0.0058767161 1.1114646 5
34: 34 0.0014619496 1.1130896 5
35: 35 -0.0117705956 1.0999878 5
36: 36 -0.0021499456 1.0976229 5
37: 37 -0.0019428995 1.0954903 5
38: 38 0.0014068660 1.0970316 5
39: 39 0.0130002537 1.1112932 5
40: 40 0.0096317575 1.1219969 5
41: 41 0.0003547640 1.1223950 5
42: 42 -0.0005336168 1.1217961 5
43: 43 0.0089696338 1.1318582 5
44: 44 0.0075666320 1.1404225 5
45: 45 -0.0048875569 1.1348486 6
46: 46 -0.0050749516 1.1290893 6
47: 47 0.0056458196 1.1354640 6
48: 48 0.0096853292 1.1464613 6
49: 49 0.0008765379 1.1474662 6
50: 50 0.0108110773 1.1598716 6
date ret val zone