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I am using MNP package for using multinomial porbit model.
model<-mnp(choice ~ 1 + AUS + Sub + AFR + BEG + CAM, base = 4, invcdf = FALSE, trace = FALSE, n.draws = 15000, p.var = 0.25, p.scale = 1, coef.start = 0, cov.start = 1, burnin = 0, thin = 0, verbose = TRUE) The part of my data is: AUS Sub AFR BEG CAM choice 1 1 3.178054 14.74568 1 4.00 2 2 0 3.401197 14.56130 1 5.00 4 3 0 3.178054 14.70488 1 7.00 2 4 0 5.940171 11.42076 0 0.00 4 5 0 2.484907 14.14389 1 9.00 3 6 0 2.772589 14.34799 1 4.00 3 7 1 4.204693 14.64967 1 10.70 1 8 1 3.988984 14.57800 1 4.50 2 9 1 4.317488 14.46320 1 5.00 1 10 1 4.204693 13.55739 1 3.52 1 And the summary of the results is Call: mnp(formula = choice ~ 1 + AUS + Sub + AFR + BEG + CAM, base = 4, invcdf = FALSE, trace = FALSE, n.draws = 1400, p.var = 0.25, p.scale = 1, coef.start = 0, cov.start = 1, burnin = 0, thin = 0, verbose = TRUE) Coefficients: mean std.dev. 2.5% 97.5% (Intercept):1 -0.82756 0.13703 -1.09865 -0.572 (Intercept):2 -0.18204 0.12870 -0.47696 0.017 (Intercept):3 -1.09682 0.19403 -1.47622 -0.752 AUS:1 0.02495 0.11110 -0.19510 0.241 AUS:2 0.41382 0.16693 0.15241 0.808 AUS:3 0.18614 0.12702 -0.05246 0.459 Sub:1 0.45978 0.13664 0.20319 0.736 Sub:2 0.12482 0.13152 -0.15545 0.377 Sub:3 0.11627 0.14044 -0.16041 0.381 AFR:1 -0.08944 0.15293 -0.39566 0.202 AFR:2 0.72140 0.27844 0.26318 1.350 AFR:3 0.18651 0.18408 -0.14078 0.573 BEG:1 0.77832 0.14485 0.51547 1.083 BEG:2 -0.48614 0.25192 -1.07826 -0.088 BEG:3 0.21806 0.18540 -0.12307 0.613 CAM:1 -0.51257 0.21925 -0.94448 -0.069 CAM:2 -0.86292 0.29425 -1.48575 -0.393 CAM:3 -0.434 0.23812 -0.92518 0.036 Covariances: mean std.dev. 2.5% 97.5% 1:1 1.00000 0.00000 1.00000 1.000 1:2 0.31050 0.20152 -0.11133 0.688 1:3 0.05604 0.31220 -0.52210 0.563 2:2 1.28810 1.03709 0.17646 4.266 2:3 0.23577 0.39763 -0.26753 1.319 3:3 0.63592 0.25132 0.33837 1.283 I want to retain the summary of results with (Intercept):1, (Intercept):2, (Intercept):3, AUS:1, AUS:2, Sub:1, Sub:2, AFR:1, AFR:2, BEG:1, BEG:2, CAM:1, CAM:2 and the covariance matrix as well. I am anxiously seeking some help regarding this problem. Thanks Saqlain |
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Did you try
summary(model)$coefficients summary(model)$covariances names(model) would help you as well. ozgur |
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In reply to this post by Saqlain
If I use summary(model)$coef.table, it gives me all the coefficients but as I said I want to retain a part of coefficients not all in the summary.
I may further go ahead a step. This is the data of trading companies in which they use two quality signals for their products. The choice variable has been created as: (1 if company hold first quality signal), (2 if company hold second quality signal), (3 if company hold both quality signals) and (4 if company hold none of the quality signals). I need only the coefficients regarding choice==1 and choice==2 but intercepts for all three choices. |
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What about
summary(model)$coefficients[,1] ? ozgur |
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yes, it gives me the required summary statistics with the following command
summary(Mnp)$coef.table[c(1:3,5,6,8,9,11,12,14,15,17,18),] Normally if we compare the two summary results before and after skipping some variables, we note bit change in the estimates of the rest of the variables. But in this case, it shows the same. I want to ask are the results in this case valid and publishable? Thanks |
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>yes, it gives me the required summary statistics with the following command
>summary(Mnp)$coef.table[c(1:3,5,6,8,9,11,12,14,15,17,18),] >Normally if we compare the two summary results before and after skipping some variables, we note bit change in >the estimates of the rest of the variables. But in this case, it shows the same. I want to ask are the results in this case >valid and publishable? I expect difference as well. You can try assigning two models to different objects such as Mnp1 and Mnp2, then try to extract interested statistics from them and compare the results. >Thanks |
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