我是R的新手，我还不完全理解其计算的逻辑......
在以前的帖子的帮助下，我无法解决我的问题。
我有一个关于11个变量的大约600个观测数据集。我已成功运行多项模型，但由于我的mean()命令获得了NA，我无法实现边际效应：
数据集：
> head(data,n=50)
ID time CHINN DEBT ERA INFL MONEY OPENNESS RESERVES RGDP RSVS
1 POL 1993 -1.8639720 NA 0 32.8815343 33.47353 41.05223 4276726138 100.00000 4.2767261
2 POL 1994 -0.8081098 NA 0 30.7586977 31.98628 41.26984 6023061884 105.13912 6.0230619
3 POL 1995 -0.1129656 NA 0 26.0622777 31.69277 44.24652 14957024390 112.15091 14.9570244
4 POL 1996 -1.1688280 NA 0 19.1794515 33.19738 46.04792 18018686050 119.00579 18.0186860
5 POL 1997 -1.1688280 NA 0 14.8181395 35.11881 50.74269 20669498248 127.43530 20.6694982
6 POL 1998 -1.1688280 NA 4 11.1353669 37.70208 56.79412 28276350644 133.65597 28.2763506
7 POL 1999 -1.1688280 NA 4 7.0830285 40.69941 54.29511 27314254555 139.74225 27.3142546
8 POL 2000 -1.1688280 32.96343 1 10.7321436 40.60334 60.67079 27469379821 145.66606 27.4693798
9 POL 2001 -1.1688280 41.37857 1 4.3657643 44.43624 57.77678 26563086922 147.25896 26.5630869
10 POL 2002 0.0644257 NA 1 0.4689145 41.98616 60.73279 29783861006 149.32464 29.7838610
11 POL 2003 0.0644257 NA 1 -0.3092132 42.42629 69.31849 33958963841 154.95895 33.9589638
12 POL 2004 0.0644257 NA 1 2.5056441 40.21144 77.33549 36772764782 162.98411 36.7727648
13 POL 2005 0.0644257 NA 1 1.4888388 43.50525 74.91127 42560657450 168.54956 42.5606575
14 POL 2006 0.0644257 NA 1 0.8029315 46.92580 82.51674 48473947849 178.83656 48.4739478
15 POL 2007 0.0644257 NA 1 0.7577241 47.82156 84.38853 65724834811 190.53326 65.7248348
16 POL 2008 0.0644257 NA 1 3.5756117 52.36352 83.76331 62183606786 199.83465 62.1836068
17 POL 2009 0.0644257 NA 1 2.6459580 53.71374 78.80573 79521598778 203.40385 79.5215988
18 POL 2010 0.0644257 NA 1 1.6733804 55.42878 85.68774 93472496388 211.50103 93.4724964
19 POL 2011 0.0644257 NA 1 3.0274845 57.87980 91.29239 97712443397 220.71189 97.7124434
20 BGR 1993 NA NA 1 68.8945207 78.96788 84.04149 1052450357 100.00000 1.0524504
21 BGR 1994 -1.1688280 NA 1 93.5639075 79.70282 90.72013 1396927258 101.92207 1.3969273
22 BGR 1995 -1.1688280 NA 1 60.0454149 67.02978 101.82717 1635188166 104.48391 1.6351882
23 BGR 1996 -0.9050840 NA 1 120.9697303 83.30492 116.19712 864262494 95.38304 0.8642625
24 BGR 1997 -0.9050840 NA 5 1058.1103740 35.17444 112.09386 2485359931 93.68869 2.4853599
25 BGR 1998 -0.9050840 NA 6 18.0824730 28.90920 117.17808 3056954172 97.96299 3.0569542
26 BGR 1999 -0.9050840 NA 6 2.3810375 30.99876 116.55936 3264673405 99.87455 3.2646734
27 BGR 2000 -0.9050840 NA 6 10.9885883 35.97481 106.26211 3507199103 105.59192 3.5071991
28 BGR 2001 -1.1688280 NA 6 6.2354669 40.92721 106.86605 3646131855 109.94781 3.6461319
29 BGR 2002 -1.1688280 NA 6 4.3788710 41.75304 102.98033 4846429828 114.94803 4.8464298
30 BGR 2003 -0.9050840 NA 6 1.0599753 46.25984 107.38463 6825720096 120.93287 6.8257201
31 BGR 2004 -0.6413398 NA 6 5.2752300 51.21173 115.32294 9337108247 128.76534 9.3371082
32 BGR 2005 -0.3775955 NA 6 4.4206316 55.53535 96.16404 8697081229 136.82237 8.6970812
33 BGR 2006 2.1752650 NA 6 6.9496009 61.91483 140.00543 11756001804 145.51154 11.7560018
34 BGR 2007 2.4390090 NA 6 6.7721512 69.93833 138.64999 17544560083 154.43412 17.5445601
35 BGR 2008 2.4390090 NA 6 11.5750056 66.12314 136.94939 17930378450 163.65291 17.9303784
36 BGR 2009 2.4390090 13.82140 6 1.5731668 69.86327 103.84837 18522120691 154.75483 18.5221207
37 BGR 2010 2.4390090 14.93887 6 1.4049372 71.96152 116.71516 17223201500 155.40066 17.2232015
38 BGR 2011 2.4390090 15.44620 6 2.9890214 75.58550 132.27417 17215734344 158.00846 17.2157343
39 CYP 1993 -0.1129656 NA 5 0.8698960 129.36758 95.41637 1276350873 100.00000 1.2763509
40 CYP 1994 -0.1129656 NA 5 2.2051585 131.48065 95.77615 1640273818 105.77324 1.6402738
41 CYP 1995 -0.1129656 NA 5 0.6063991 128.64850 100.10371 1294935266 111.96553 1.2949353
42 CYP 1996 -1.1688280 NA 5 2.3411552 136.94871 104.44163 1704484151 114.02390 1.7044842
43 CYP 1997 -1.1688280 NA 5 3.3418671 145.38644 105.39959 1525648570 116.75559 1.5256486
44 CYP 1998 -1.1688280 261.72872 4 1.6379206 147.00670 100.52824 1512684182 122.55893 1.5126842
45 CYP 1999 -1.1688280 155.75745 4 1.4380285 159.30150 101.75789 1967461070 128.47057 1.9674611
46 CYP 2000 -1.1688280 156.25023 4 4.8139861 161.29472 109.87129 1868515552 135.00970 1.8685156
47 CYP 2001 -1.1688280 160.87807 4 0.8515728 169.52762 109.90206 2396061109 140.39178 2.3960611
48 CYP 2002 -1.1688280 170.52011 4 1.3698956 175.62242 103.06500 3181358048 143.20035 3.1813580
49 CYP 2003 -0.1129656 179.11164 4 3.0419336 174.50803 95.17382 3450932640 145.59528 3.4509326
50 CYP 2004 1.3840320 180.22661 4 1.2153140 171.42676 98.02155 4113824175 151.47324 4.1138242
转换数据的命令：
pdata<-plm.data(data,index=c("ID","time"))
mldata<-mlogit.data(pdata,choice="ERA",shape="wide")
指示平均值矩阵的命令：
z<-with(mldata,data.frame(CHINN=mean(CHINN),DEBT=mean(DEBT),INFL=mean(INFL),MONEY=mean(MONEY),OPENNESS=mean(OPENNESS),RGDP=mean(RGDP),RSVS=mean(RSVS)))
我得到的输出：
CHINN DEBT INFL MONEY OPENNESS RGDP RSVS
1 NA NA NA NA NA 133.4633 NA
你能否就这个命令的NA的原因提出建议？
我会理解它是否会为DEBT提供正确的输出，主要是NAs，但为什么它不计算RSVS，CHINN和其他人的平均值？
为什么计算RGDP的平均值？这两个变量都有：
> class(mldata$CHINN)
[1] "numeric"
> class(mldata$RGDP)
[1] "numeric"
如何克服这个问题？
更新
感谢David Arenburg的评论，我设法计算了手段：
> z
CHINN DEBT INFL MONEY OPENNESS RGDP RSVS
1 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
然而，在计算边际效应时会发生另一个错误：
effects(mlogit.data2,covariate="CHINN",data=z)
Error in predict.mlogit(object, data) :
the number of rows of the data.frame should be a multiple of the number of alternatives
复制平均值5次以获得6行的矩阵zz，用于6种选择：
> zz
CHINN DEBT INFL MONEY OPENNESS RGDP RSVS
1 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
2 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
3 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
4 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
5 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
6 1.342326 59.85562 33.88494 58.2304 95.88219 133.4633 29.21734
effects（）命令给出一个输出：
> effects(mlogit.data2,covariate=c("CHINN","DEBT","INFL","MONEY","OPENNESS","RGDP","RSVS"),data=zz)
2 0 1 4 5 6
-1.288185e-02 -1.933325e-02 -8.801676e-06 1.095252e-02 2.120814e-02 6.324152e-05
但是我无法对此输出得出任何结论 - 我不知道哪个变量可以分配这些边际效应。
当我尝试逐个运行时，我得到另一个错误：
> effects(mlogit.data2,covariate=c("CHINN"),data=zz)
Error in if (rhs %in% c(1, 3)) { : argument is of length zero
模型命令：
mlogit.data2<-mlogit(ERA~1|CHINN+INFL+MONEY+OPENNESS+RGDP+RSVS+DEBT,data=mldata,reflevel="4")
我正在使用的套餐：
library(Formula)
library(miscTools)
library(lattice)
library(zoo)
library(sandwich)
library(maxLik)
library(lmtest)
library(statmod)
library(mlogit)
library(plm)
提前谢谢你， Zyta
答案 0 :(得分：2)
好的，似乎我在以下方面有一些解决方案：
http://www.talkstats.com/showthread.php/44314-calculate-marginal-effects-using-mlogit-package
和
How can I view the source code for a function?
显然足以看到：
> getAnywhere(effects.mlogit)
A single object matching ‘effects.mlogit’ was found
It was found in the following places
registered S3 method for effects from namespace mlogit
namespace:mlogit
with value
function (object, covariate = NULL, type = c("aa", "ar", "rr",
"ra"), data = NULL, ...)
{
type <- match.arg(type)
if (is.null(data)) {
P <- predict(object, returnData = TRUE)
data <- attr(P, "data")
attr(P, "data") <- NULL
}
else P <- predict(object, data)
newdata <- data
J <- length(P)
alt.levels <- names(P)
pVar <- substr(type, 1, 1)
xVar <- substr(type, 2, 2)
cov.list <- lapply(attr(formula(object), "rhs"), as.character)
rhs <- sapply(cov.list, function(x) length(na.omit(match(x,
covariate))) > 0)
rhs <- (1:length(cov.list))[rhs]
eps <- 1e-05
if (rhs %in% c(1, 3)) {
if (rhs == 3) {
theCoef <- paste(alt.levels, covariate, sep = ":")
theCoef <- coef(object)[theCoef]
}
else theCoef <- coef(object)[covariate]
me <- c()
for (l in 1:J) {
newdata[l, covariate] <- data[l, covariate] + eps
newP <- predict(object, newdata)
me <- rbind(me, (newP - P)/eps)
newdata <- data
}
if (pVar == "r")
me <- t(t(me)/P)
if (xVar == "r")
me <- me * matrix(rep(data[[covariate]], J), J)
dimnames(me) <- list(alt.levels, alt.levels)
}
if (rhs == 2) {
newdata[, covariate] <- data[, covariate] + eps
newP <- predict(object, newdata)
me <- (newP - P)/eps
if (pVar == "r")
me <- me/P
if (xVar == "r")
me <- me * data[[covariate]]
names(me) <- alt.levels
}
me
}
<environment: namespace:mlogit>
然后复制该函数并调整其16行：
myeffects<-function (object, covariate = NULL, type = c("aa", "ar", "rr",
"ra"), data = NULL, ...)
{
type <- match.arg(type)
if (is.null(data)) {
P <- predict(object, returnData = TRUE)
data <- attr(P, "data")
attr(P, "data") <- NULL
}
else P <- predict(object, data)
newdata <- data
J <- length(P)
alt.levels <- names(P)
pVar <- substr(type, 1, 1)
xVar <- substr(type, 2, 2)
cov.list <- strsplit(as.character(attr(formula(object), "rhs")), " + ", fixed = TRUE)
rhs <- sapply(cov.list, function(x) length(na.omit(match(x,
covariate))) > 0)
rhs <- (1:length(cov.list))[rhs]
eps <- 1e-05
if (rhs %in% c(1, 3)) {
if (rhs == 3) {
theCoef <- paste(alt.levels, covariate, sep = ":")
theCoef <- coef(object)[theCoef]
}
else theCoef <- coef(object)[covariate]
me <- c()
for (l in 1:J) {
newdata[l, covariate] <- data[l, covariate] + eps
newP <- predict(object, newdata)
me <- rbind(me, (newP - P)/eps)
newdata <- data
}
if (pVar == "r")
me <- t(t(me)/P)
if (xVar == "r")
me <- me * matrix(rep(data[[covariate]], J), J)
dimnames(me) <- list(alt.levels, alt.levels)
}
if (rhs == 2) {
newdata[, covariate] <- data[, covariate] + eps
newP <- predict(object, newdata)
me <- (newP - P)/eps
if (pVar == "r")
me <- me/P
if (xVar == "r")
me <- me * data[[covariate]]
names(me) <- alt.levels
}
me
}
现在结果如下：
> myeffects(mlogit.data2,covariate="RSVS",data=zz)
2 0 1 4 5 6
3.612318e-03 5.368693e-04 -4.903995e-08 -5.382731e-03 1.238768e-03 -5.175053e-06
答案 1 :(得分：1)
您可以使用colMeans
op <- options(scipen= 100, digits=2)
colMeans(mldata[,3:11], na.rm=TRUE)
# CHINN DEBT ERA INFL MONEY
# -0.27 115.25 0.20 33.66 74.66
#OPENNESS RESERVES RGDP RSVS
# 92.06 18809465124.14 136.56 18.81
options(op)
summarise_each 的或dplyr
library(dplyr)
mldata %>%
summarise_each(funs(round(mean(., na.rm=TRUE),2)), CHINN:RSVS)
# CHINN DEBT ERA INFL MONEY OPENNESS RESERVES RGDP RSVS
#1 -0.27 115.25 0.2 33.66 74.66 92.06 18809465124 136.56 18.81