我正在尝试根据许多替代特定特征以及一些公司特定特征来估计公司选址选择的嵌套 logit 模型,其中nests = countries
和。alternatives = provinces
我使用以下方法将数据格式化为“长”结构:
data <- mlogit.data(DB, choice="Occurrence", shape="long", chid.var="IDP", varying=6:ncol(DB), alt.var="Prov")
以下是数据示例:
IDP Occurrence From Prov ToC Dist Price Yield
5p1.APY 5p1 FALSE Sao Paulo APY PY 0.0000000 0.3698913 0.0000000
5p1.BOQ 5p1 FALSE Sao Paulo BOQ PY 0.6495493 0.3698913 0.0000000
5p1.CHA 5p1 FALSE Sao Paulo CHA AR 0.7870593 0.4622464 0.4461496
5p1.COR 5p1 FALSE Sao Paulo COR AR 0.3747480 0.4622464 0.5536546
5p1.FOR 5p1 FALSE Sao Paulo FOR AR 0.6822188 0.4622464 0.4402772
5p1.JUY 5p1 FALSE Sao Paulo JUY AR 1.0000000 0.4622464 0.3617038
请注意,为了清楚起见,我已将表格缩减为几个变量,但通常会使用更多。
我用于嵌套 logit 的代码如下:
nests <- list(Bolivia="SCZ",Paraguay=c("PHY","BOQ","APY"),Argentina=c("CHA","COR","FOR","JUY","SAL","SFE","SDE"))
nml <- mlogit(Occurrence ~ DistComp + PriceComp + YieldComp, data=data, nests=nests, unscaled=T)
summary(nml)
运行此模型时,我得到以下输出:
> summary(nml)
Call:
mlogit(formula = Occurrence ~ DistComp + PriceComp + YieldComp,
data = data, nests = nests, unscaled = T)
Frequencies of alternatives:
APY BOQ CHA COR FOR JUY PHY
SAL SCZ SDE SFE
0.1000000 0.0666667 0.1333333 0.0250000 0.0750000 0.0083333 0.0083333
0.1166667 0.2583333 0.1750000 0.0333333
bfgs method
1 iterations, 0h:0m:0s
g'(-H)^-1g = 1E+10
last step couldn't find higher value
Coefficients :
Estimate Std. Error t-value Pr(>|t|)
BOQ:(intercept) -0.29923 NA NA NA
CHA:(intercept) -1.25406 NA NA NA
COR:(intercept) -1.76020 NA NA NA
FOR:(intercept) -1.97083 NA NA NA
JUY:(intercept) -4.14476 NA NA NA
PHY:(intercept) -2.63961 NA NA NA
SAL:(intercept) -1.72047 NA NA NA
SCZ:(intercept) -0.15714 NA NA NA
SDE:(intercept) -0.57449 NA NA NA
SFE:(intercept) -2.47345 NA NA NA
DistComp 2.44322 NA NA NA
PriceComp 2.45202 NA NA NA
YieldComp 3.15611 NA NA NA
iv.Bolivia 1.00000 NA NA NA
iv.Paraguay 1.00000 NA NA NA
iv.Argentina 1.00000 NA NA NA
Log-Likelihood: -221.84
McFadden R^2: 0.10453
Likelihood ratio test : chisq = 51.79 (p.value = 2.0552e-09)
考虑到我使用mlogit.data()
. 对此的任何帮助将不胜感激。
最好的,
扬