作为R中的乞丐，我有一个简单的问题。
我对此规范进行了线性回归：
X1 = X1_t-h + X2_t-h
h for等于1,2,3,4,5：
例如，当h = 1时，我运行此代码：
Modelo11 <- dynlm(X1 ~ L(X1,1) + L(X2, 1)-1, data = GDP)
这是一个简单的回归。
我想实现一个函数，它给出了有和没有HAC异方差估计的五个线性回归（h = 1,2,3,4和5）：
我这样做了，并没有奏效：
for(h in 1:5){
Modelo1[h] <- dynlm(GDPTrimestralemT ~ L(SpreademT,h) + L(GDPTrimestralemT, h)-1, data = MatrizDadosUS)
coeftest(Modelo1[h], df = Inf, vcov = parzenHAC)
return(list(summary(Modelo1[h])))
}
其中一条错误消息是：
number of items to replace is not a multiple of replacement length
这是我的data.frame：
GDP <- data.frame(data )
GDP
X1 X2
1 0.542952690 0.226341364
2 0.102328393 0.743360185
3 0.166345969 0.186533485
4 1.406733422 1.392420181
5 -0.469811005 -0.114609464
6 -0.509268267 0.687555461
7 1.470439930 0.298655018
8 1.046456428 -1.056387597
9 -0.492462197 -0.530284962
10 -0.516065519 0.645957530
11 0.624638996 1.044731264
12 0.213616470 -1.652979785
13 0.669747432 1.398602289
14 0.552089131 -0.821013792
15 0.452715216 1.420094663
16 -0.892063248 -1.436600779
17 1.429284965 0.559738610
18 0.853740565 -0.898976767
19 0.741864168 1.352012831
20 0.171494650 1.704764705
21 0.422326351 -0.267064235
22 -1.261643503 -2.090694608
23 -1.321086283 -0.273954212
24 0.365226000 1.965167113
25 -0.080888690 -0.594498893
26 -0.183293801 -0.483053404
27 -1.033792032 0.586491772
28 0.718322432 1.776210145
29 -2.822693790 -0.731509917
30 -1.251740437 -1.918124078
31 1.184256949 -0.016548037
32 2.255202675 0.303438286
33 -0.930446147 0.803126180
34 -1.691383225 -0.157839283
35 -1.081643279 -0.006652717
36 1.034162006 -1.970063305
37 -0.716827488 0.306792930
38 0.098471514 0.338333164
39 0.343536547 0.389775011
40 1.442117465 -0.668885360
41 0.095131066 -0.298356861
42 0.222524607 0.291485267
43 -0.499969717 1.308312472
44 0.588162304 0.026539575
45 0.581215173 0.167710855
46 0.629343124 -0.052835206
47 0.811618963 0.716913172
48 1.463610069 -0.356369304
49 -2.000576321 1.226446201
50 1.278233553 0.313606888
51 -0.700373666 0.770273988
52 -1.206455648 0.344628878
53 0.024602262 1.001621886
54 0.858933385 -0.865771777
55 -1.592291995 -0.384908852
56 -0.833758365 -1.184682199
57 -0.281305858 2.070391729
58 -0.122848757 -0.308397782
59 -0.661013984 1.590741535
60 1.887869805 -1.240283364
61 -0.313677463 -1.393252994
62 1.142864110 -1.150916732
63 -0.633380499 -0.223923970
64 -0.158729527 -1.245647224
65 0.928619010 -1.050636078
66 0.424317087 0.593892028
67 1.108704956 -1.792833100
68 -1.338231248 1.138684394
69 -0.647492569 0.181495183
70 0.295906675 -0.101823172
71 -0.079827607 0.825158278
72 0.050353111 -0.448453121
73 0.129068772 0.205619797
74 -0.221450137 0.051349511
75 -1.300967949 1.639063824
76 -0.861963677 1.273104220
77 -1.691001610 0.746514122
78 0.365888734 -0.055308006
79 1.297349754 1.146102001
80 -0.652382297 -1.095031447
81 0.165682952 -0.012926971
82 0.127996446 0.510673745
83 0.338743162 -3.141650682
84 -0.266916587 -2.483389321
85 0.148135154 -1.239997153
86 1.256591385 0.051984536
87 -0.646281986 0.468210275
88 0.180472423 0.393014848
89 0.231892902 -0.545305005
90 -0.709986273 0.104969765
91 1.231712844 -1.703489840
92 0.435378714 0.876505107
93 -1.880394798 -0.885893722
94 1.083580732 0.117560662
95 -0.499072654 -1.039222894
96 1.850756855 -1.308752222
97 1.653952857 0.440405804
98 -1.057618294 -1.611779530
99 -0.021821282 -0.807071503
100 0.682923562 -2.358596342
101 -1.132293845 -1.488806929
102 0.319237353 0.706203968
103 -2.393105781 -1.562111727
104 0.188653972 -0.637073832
105 0.667003685 0.047694037
106 -0.534018861 1.366826933
107 -2.240330371 -0.071797320
108 -0.220633546 1.612879694
109 -0.022442941 1.172582601
110 -1.542418139 0.635161458
111 -0.684128812 -0.334973482
112 0.688849615 0.056557966
113 0.848602803 0.785297518
114 -0.874157558 -0.434518305
115 -0.404999060 -0.078893114
116 0.735896917 1.637873669
117 -0.174398836 0.542952690
118 0.222418628 0.102328393
119 0.419461884 0.166345969
120 -0.042602368 1.406733422
121 2.135670836 -0.469811005
122 1.197644287 -0.509268267
123 0.395951293 1.470439930
124 0.141327444 1.046456428
125 0.691575897 -0.492462197
126 -0.490708151 -0.516065519
127 -0.358903359 0.624638996
128 -0.227550909 0.213616470
129 -0.766692832 0.669747432
130 -0.001690915 0.552089131
131 -1.786701123 0.452715216
132 -1.251495762 -0.892063248
133 1.123462446 1.429284965
134 0.237862653 0.853740565
感谢。
答案 0 :(得分：3)
您的变量Modelo1是一个无法存储lm个对象的向量。当Modelo1是一个列表时，它应该有效。
library(dynlm)
df<-data.frame(rnorm(50),rnorm(50))
names(df)<-c("a","b")
c<-list()
for(h in 1:5){
c[[h]] <- dynlm(a ~ L(a,h) + L(b, h)-1, data = df)
}
要获取摘要，您必须访问单个列表元素。例如：
summary(c[[1]])
*编辑以回应Richard Scriven评论
获得所有摘要的最有效方法是：
lapply(c, summary)
这将摘要函数应用于列表的每个元素，并返回包含结果的列表。