r - 使用 ggplot 绘制方差分解结果

Question

我有以下xts（south.africa）

structure(c(0.711700911871027, 2.18508552562202, 0.101828936775838, 
0.559018189973415, 0.791510887118818, 0.0702572653418443, 0.545643651047278, 
1.96973394058411, 1.13702436984386, -0.110342572242494, 0.895562224513213, 
1.14344682000116, 1.24275400501261, -0.27136808242183, 0.263419830113065, 
2.34411425782337, 1.33985361856794, 1.01841436556569, 0.866157306469795, 
1.93932019030232, 1.27825249354618, 1.30223261455793, 0.375324910356056, 
1.90149463122147, 1.17893305264101, 1.78893495881187, 2.01345172908223, 
1.15135439226659, 0.429493055158936, 1.30897897154671, 1.59811481462455, 
0.384501108705138, 1.33152847641842, -0.0756364961798539, -0.625567867488819, 
-1.63277895723031, -0.302044860024786, 0.683501644311235, 0.579148593539358, 
1.35072894035346, 0.375566525011983, 1.0714726283032, 0.500230467477891, 
1.45616977552994, 0.289398525763396, 0.863566266240667, 0.508353713803089, 
0.475394391746953, 0.820846564990596, 0.340994768769143, 0.168979388512014, 
0.839854291774778, 1.14128072836799, 0.0124180977749688, 0.967066943385397, 
-0.0617993469917977, 0.765586687222708, 0.271147530795801, 0.648173777980432, 
0.68234658026185, -0.0897710114593053, -0.708258282061003, 0.448438992679812, 
-0.16686150038101, 1.34207945665406, -1.17408818506828, 0.765914568114034, 
0.133331283880445, 2.05881770081913, -1.49604288251695, 0.64064064153655, 
-0.568876991290068, 1.73070039441132, -0.636868179734762, 0.511232708705478, 
-1.58975642052397, 2.70424824256459, -1.52879441672304, -0.0794733364552158, 
-1.02884935859766, -16.1951626038544, 10.9740288232354, 2.0276108060183, 
1.6996167612223, 1.51306289759408, 1.90213612695742, 1.11098177244755, 
0.144156689942943, 1.04824032192408, 3.29391128227785, 3.07063103030947, 
2.79156130819782, 3.54522073492278, 1.36531761632988, 0.273561477276285, 
-0.889626720355619, -1.47407679728495, 0.309985806569024, 0.325381422046789, 
-0.0591542128929312, 1.06059875690399, 0.580413813579916, 0.289504123768669, 
0.414777339697103, 0.756406499801709, 0.557780657388651, 0.806178986320827, 
1.55429806729295, 1.57762672232398, 1.08603082661238, 1.61211690910754, 
1.82455024510801, 2.45718798396846, 2.68426228226435, 2.50978429363213, 
3.02373059291385, 1.36301343061644, 1.56005214831358, 1.82910637591691, 
1.44037258065417, 0.881529697550931, 1.05274032728078, 0.708190049422974, 
0.674295163204164, 0.934457611055617, 1.36025819164436, 1.50804232025443, 
1.48581974120869, 1.73882008402959, 1.27679490711952, 1.12343976330829, 
0.925436930729973, 2.24990447458007, 1.35913624746395, 0.885006549446121, 
1.62977131236151, 1.3877570813734, 1.80222261823024, 1.54945562263986, 
1.44302994559631, 0.761493713563954, 0.350108153311623, 1.86807656796626, 
1.42175556001272, 1.04910912422067, 2.01205755468221, 1.86891746075357, 
1.23492256958846, 1.55102685266497, 1.61000293519873, 0.695481642542095, 
0.625493793378151, 1.48196760068915, 1.05708735900016, 1.0857422205409, 
1.1577949394427, 1.41164544510541, 0.428380530156947, 1.34355343917676, 
0.890871006592686, 1.01595011760134, 0.991035049882694, -0.561127940846706, 
1.55278035816462), .Dim = c(82L, 2L), .Dimnames = list(NULL, 
    c("GDP_SA", "CPI_SA")), index = structure(c(954547200, 962409600, 
970358400, 978307200, 986083200, 993945600, 1001894400, 1009843200, 
1017619200, 1025481600, 1033430400, 1041379200, 1049155200, 1057017600, 
1064966400, 1072915200, 1080777600, 1088640000, 1096588800, 1104537600, 
1112313600, 1120176000, 1128124800, 1136073600, 1143849600, 1151712000, 
1159660800, 1167609600, 1175385600, 1183248000, 1191196800, 1199145600, 
1207008000, 1214870400, 1222819200, 1230768000, 1238544000, 1246406400, 
1254355200, 1262304000, 1270080000, 1277942400, 1285891200, 1293840000, 
1301616000, 1309478400, 1317427200, 1325376000, 1333238400, 1341100800, 
1349049600, 1356998400, 1364774400, 1372636800, 1380585600, 1388534400, 
1396310400, 1404172800, 1412121600, 1420070400, 1427846400, 1435708800, 
1443657600, 1451606400, 1459468800, 1467331200, 1475280000, 1483228800, 
1491004800, 1498867200, 1506816000, 1514764800, 1522540800, 1530403200, 
1538352000, 1546300800, 1554076800, 1561939200, 1569888000, 1577836800, 
1585699200, 1593561600), tzone = "UTC", tclass = "Date"), class = c("xts", 
"zoo"))

由此，我使用vars包创建了一个 SVAR 模型并运行方差分解。寻找一种方便的方式来创建结果的堆叠decomposition条形图ggplot2

library(vars)
library(tidyverse)

var.south.africa <- south.africa %>% VAR(.,p=1,type="both",season=NULL)
SVAR.south.africa <- BQ(var.south.africa)

decomposition <- fevd(SVAR.south.africa, n.ahead = 10)

标签： rggplot2

Answer 1

我们可以按照这个 Stack Overflow对一个半不相关的问题的回答（以及之后更长时间的旋转）的方法来将数据重塑为更方便的格式：

dat <- map_df(decomposition, ~as.data.frame(.x), .id="id") %>%
    mutate(horizon = rep(1:10, 2)) %>%
    pivot_longer(names_to = "var", cols = c(GDP_SA, CPI_SA))
head(dat)
# # A tibble: 6 x 4
#   id     horizon var     value
#   <chr>    <int> <chr>   <dbl>
# 1 GDP_SA       1 GDP_SA 0.987 
# 2 GDP_SA       1 CPI_SA 0.0128
# 3 GDP_SA       2 GDP_SA 0.982 
# 4 GDP_SA       2 CPI_SA 0.0178
# 5 GDP_SA       3 GDP_SA 0.983 
# 6 GDP_SA       3 CPI_SA 0.0173

那里id告诉您该行的哪个元素decomposition来自，并给出该行的观察来自var该元素的矩阵的列。value然后只需要做一个简单的ggplot堆积条形图：

ggplot(data = dat, mapping = aes(x = horizon, y = value, fill = var)) +
    facet_wrap(~id) +
    geom_bar(stat = "identity") +
    theme_bw()