如何为R中的每个组填充一个循环列？
如何为R中的每个组填充一个循环列？
提问于 2020-10-20 15:21:13
我目前在图片中显示了一个数据帧：
在每个PERMNO的第一天有一个初始的sigma_t值，我想用以下公式填充每个PERMNO的sigma_t列的其余部分，其中lambda = 0.94：
我目前编写了一个for循环，它实现了这个公式：
for(i in 2:nrow(data)){data$sigma_t[i] <- 0.94*data$sigma_t[i-1]^2+0.06*data$RET[i-1]^2}但是，我应该如何将这个for循环应用到dataframe中的每个PERMNO组？
回答 1
Stack Overflow用户
回答已采纳
发布于 2020-10-20 16:04:57
假设您的数据如下(即第一个sigma_t是每个PERMNO的非NA值)
Date PERMNO RET sigma_t
1 2000-01-03 10806 0.913192312 0.9979223
2 2000-01-04 10806 -0.092526597 NA
3 2000-01-05 10806 0.280083864 NA
4 2000-01-06 10806 0.130769875 NA
5 2000-01-07 10806 0.098093785 NA
6 2000-01-08 10806 0.558508840 NA
7 2000-01-09 10806 2.181083768 NA
8 2000-01-10 10806 -1.076594408 NA
9 2000-01-11 10806 -0.255776524 NA
10 2000-01-12 10806 -0.660575251 NA
11 2000-01-13 10806 -1.561518300 NA
12 2000-01-14 10806 0.402352610 NA
13 2000-01-15 10806 0.051492486 NA
14 2000-01-03 10807 0.135825002 0.5288962
15 2000-01-04 10807 -1.589023433 NA
16 2000-01-05 10807 -0.122931603 NA然后你就可以做这样的事
library(purrr)
library(dplyr)
data %>% group_by(PERMNO) %>% mutate(sigma_t = accumulate(head(RET, -1L), ~0.94 * .x * .x + 0.06 * .y * .y, .init = sigma_t[[1L]]))输出
# A tibble: 65 x 4
# Groups: PERMNO [5]
Date PERMNO RET sigma_t
<date> <int> <dbl> <dbl>
1 2000-01-03 10806 0.913 0.998
2 2000-01-04 10806 -0.0925 0.986
3 2000-01-05 10806 0.280 0.915
4 2000-01-06 10806 0.131 0.791
5 2000-01-07 10806 0.0981 0.589
6 2000-01-08 10806 0.559 0.327
7 2000-01-09 10806 2.18 0.119
8 2000-01-10 10806 -1.08 0.299
9 2000-01-11 10806 -0.256 0.153
10 2000-01-12 10806 -0.661 0.0261
# ... with 55 more rows检查结果是否与for循环生成的结果相同：
> res1 <- accumulate(head(data$RET, -1L), ~0.94 * .x * .x + 0.06 * .y * .y, .init = data$sigma_t[[1L]])
> res2 <- double(nrow(data))
> res2[[1L]] <- data$sigma_t[[1L]]
> for(i in 2:nrow(data)){
+ res2[[i]] <- 0.94*res2[[i-1]]*res2[[i-1]]+0.06*data$RET[[i-1]]*data$RET[[i-1]]
+ }
> all(res1 == res2)
[1] TRUE我用的数据
structure(list(Date = structure(c(10959, 10960, 10961, 10962,
10963, 10964, 10965, 10966, 10967, 10968, 10969, 10970, 10971,
10959, 10960, 10961, 10962, 10963, 10964, 10965, 10966, 10967,
10968, 10969, 10970, 10971, 10959, 10960, 10961, 10962, 10963,
10964, 10965, 10966, 10967, 10968, 10969, 10970, 10971, 10959,
10960, 10961, 10962, 10963, 10964, 10965, 10966, 10967, 10968,
10969, 10970, 10971, 10959, 10960, 10961, 10962, 10963, 10964,
10965, 10966, 10967, 10968, 10969, 10970, 10971), class = "Date"),
PERMNO = c(10806L, 10806L, 10806L, 10806L, 10806L, 10806L,
10806L, 10806L, 10806L, 10806L, 10806L, 10806L, 10806L, 10807L,
10807L, 10807L, 10807L, 10807L, 10807L, 10807L, 10807L, 10807L,
10807L, 10807L, 10807L, 10807L, 10808L, 10808L, 10808L, 10808L,
10808L, 10808L, 10808L, 10808L, 10808L, 10808L, 10808L, 10808L,
10808L, 10809L, 10809L, 10809L, 10809L, 10809L, 10809L, 10809L,
10809L, 10809L, 10809L, 10809L, 10809L, 10809L, 10810L, 10810L,
10810L, 10810L, 10810L, 10810L, 10810L, 10810L, 10810L, 10810L,
10810L, 10810L, 10810L), RET = c(0.913192312238358, -0.092526596846852,
0.280083863779238, 0.130769874502966, 0.0980937848761638,
0.558508839970204, 2.18108376768683, -1.07659440811492, -0.255776524083785,
-0.660575251329072, -1.56151829951539, 0.402352610100085,
0.0514924859261968, 0.135825001773724, -1.58902343332535,
-0.122931603238164, 0.452006216926532, 0.693114146318084,
1.92191254928755, -1.46908545801166, 0.913386433733295, 0.58897079826165,
-0.734672875439902, -0.316864004253191, -1.5823359900577,
-0.640666477381628, 1.08418139895388, -1.14921141675937,
0.162609558626544, -1.9716659701064, 0.104005665525902, -0.795358745231712,
-0.2893381621628, 0.458189703352807, 0.00872023089334384,
0.00814797415178989, -0.67721320720348, 0.204413384165412,
-1.076872789543, 0.0945359885708477, -0.930160781560965,
-0.105321201055104, -0.315010497487111, -1.61634033850791,
2.524839463943, 0.290994112062563, -0.849772864569264, -1.65201573210097,
1.89042297661439, 1.33084303096302, 0.161021482912505, -1.31376297235415,
0.15668445036454, -0.549574920114175, 1.36530405318811, -0.0940911313529308,
-0.870085847088321, -0.388873127012174, -0.19829361512545,
-0.615723371631421, -0.290458500882665, -0.87058694446336,
0.498150842211666, -1.22058065663026, -1.13199591809083),
sigma_t = c(0.997922302223742, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, 0.528896240284666, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, 0.527911589480937, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.760049085598439, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 0.531686891801655,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA,
-65L), groups = structure(list(PERMNO = 10806:10810, .rows = structure(list(
1:13, 14:26, 27:39, 40:52, 53:65), ptype = integer(0), class = c("vctrs_list_of",
"vctrs_vctr", "list"))), row.names = c(NA, 5L), class = c("tbl_df",
"tbl", "data.frame"), .drop = TRUE), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))更新
如果您希望使用PERMNO更改第三个参数，那么
data %>%
group_by(PERMNO) %>%
mutate(
sigma_t = accumulate(
head(RET, -1L), ~..3 * .x * .x + (1 - ..3) * .y * .y,
c("10806" = 0.94, "10807" = 0.90)[[as.character(PERMNO[[1L]])]],
.init = sigma_t[[1L]]
)
)要使上述管道更通用，只需使用变量替换此部件c("10806" = 0.94, "10807" = 0.90)即可。假设您的data有100个唯一的PERMNO组，它们对应于介于1到100之间的lambda，然后只需设置
lambdas <- setNames(1:100, unique(df$PERMNO))然后跑
data %>%
group_by(PERMNO) %>%
mutate(
sigma_t = accumulate(
head(RET, -1L), ~..3 * .x * .x + (1 - ..3) * .y * .y,
lambdas[[as.character(PERMNO[[1L]])]],
.init = sigma_t[[1L]]
)
)页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/64448255
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