Zero-inflated model predicting only a small range of values. I need help Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Please help me refine this zero-inflated negative binomial modelZero inflated Poisson modelWhen to use zero-inflated poisson regression and negative binomial distributionTransform response for hurdle modelInterpreting results from distribution fittingZero-inflated model does not produce zeroes in fitted values?Predict function on negative binomial produces strange fitted values when adding an offsetNeed for zero-inflated poisson even though model fits data?zero inflated poisson iregression AIC valuesDeviance residual for zero-inflated Poisson model
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Zero-inflated model predicting only a small range of values. I need help
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Please help me refine this zero-inflated negative binomial modelZero inflated Poisson modelWhen to use zero-inflated poisson regression and negative binomial distributionTransform response for hurdle modelInterpreting results from distribution fittingZero-inflated model does not produce zeroes in fitted values?Predict function on negative binomial produces strange fitted values when adding an offsetNeed for zero-inflated poisson even though model fits data?zero inflated poisson iregression AIC valuesDeviance residual for zero-inflated Poisson model
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I built a ZI model and it is producing predicted values that are from a very small range when compared to the observed values. Plus it does not produce any zeros. See the fitted vs. observed graph below.
http://tinypic.com/r/16bbrxs/9
My model structure is given below. My response is a count variable and the predictor is continuous water level measurements (scaled).
m.u.meanwl.gu <- zeroinfl(GU ~ MeanWLScaled | MeanWLScaled, data = uluabat_wl_fg, dist = "negbin", link = "logit")
I need help to figure out what's wrong and fix it. Thanks in advance.
r prediction zero-inflation
edited yesterday
Stephan Kolassa
48.1k8101180
asked yesterday
KO 88KO 88
697
$endgroup$
add a comment |
$begingroup$
I built a ZI model and it is producing predicted values that are from a very small range when compared to the observed values. Plus it does not produce any zeros. See the fitted vs. observed graph below.
http://tinypic.com/r/16bbrxs/9
My model structure is given below. My response is a count variable and the predictor is continuous water level measurements (scaled).
m.u.meanwl.gu <- zeroinfl(GU ~ MeanWLScaled | MeanWLScaled, data = uluabat_wl_fg, dist = "negbin", link = "logit")
I need help to figure out what's wrong and fix it. Thanks in advance.
r prediction zero-inflation
edited yesterday
Stephan Kolassa
48.1k8101180
asked yesterday
KO 88KO 88
697
$endgroup$
add a comment |
$begingroup$
I built a ZI model and it is producing predicted values that are from a very small range when compared to the observed values. Plus it does not produce any zeros. See the fitted vs. observed graph below.
http://tinypic.com/r/16bbrxs/9
My model structure is given below. My response is a count variable and the predictor is continuous water level measurements (scaled).
m.u.meanwl.gu <- zeroinfl(GU ~ MeanWLScaled | MeanWLScaled, data = uluabat_wl_fg, dist = "negbin", link = "logit")
I need help to figure out what's wrong and fix it. Thanks in advance.
r prediction zero-inflation
edited yesterday
Stephan Kolassa
48.1k8101180
asked yesterday
KO 88KO 88
697
$endgroup$
I built a ZI model and it is producing predicted values that are from a very small range when compared to the observed values. Plus it does not produce any zeros. See the fitted vs. observed graph below.
http://tinypic.com/r/16bbrxs/9
My model structure is given below. My response is a count variable and the predictor is continuous water level measurements (scaled).
m.u.meanwl.gu <- zeroinfl(GU ~ MeanWLScaled | MeanWLScaled, data = uluabat_wl_fg, dist = "negbin", link = "logit")
I need help to figure out what's wrong and fix it. Thanks in advance.
r prediction zero-inflation
r prediction zero-inflation
edited yesterday
Stephan Kolassa
48.1k8101180
asked yesterday
KO 88KO 88
697
edited yesterday
Stephan Kolassa
48.1k8101180
asked yesterday
KO 88KO 88
697
edited yesterday
Stephan Kolassa
48.1k8101180
edited yesterday
Stephan Kolassa
48.1k8101180
edited yesterday
Stephan Kolassa
48.1k8101180
48.1k8101180
asked yesterday
KO 88KO 88
697
asked yesterday
KO 88KO 88
697
asked yesterday
KO 88KO 88
697
697
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero.
To obtain a probabilistic prediction of the probabilities to observe specific actual count observations, use type="prob". An example:
> library(pscl)
> fm_zip <- zeroinfl(art ~ ., data = bioChemists)
> head(predict(fm_zip))
1 2 3 4 5 6
2.037956 1.323124 1.308704 1.439982 2.363233 0.854771
> head(predict(fm_zip,type="prob"))
0 1 2 3 4 5 6
1 0.2162690 0.1937560 0.2279639 0.17880748 0.10518810 0.049503684 0.019414543
2 0.3626973 0.2429423 0.2058847 0.11631983 0.04928840 0.016708046 0.004719819
3 0.3655549 0.2445794 0.2051107 0.11467420 0.04808441 0.016129943 0.004508999
4 0.3582308 0.2127514 0.2034182 0.12966290 0.06198734 0.023707199 0.007555727
5 0.1071273 0.2125181 0.2559524 0.20550920 0.12375553 0.059619449 0.023934814
6 0.5218005 0.2328010 0.1513963 0.06563783 0.02134296 0.005551943 0.001203522
7 8 9 10 11 12
1 0.006526345 0.0019196452 5.019025e-04 1.181028e-04 2.526437e-05 4.954139e-06
2 0.001142821 0.0002421248 4.559820e-05 7.728561e-06 1.190849e-06 1.682002e-07
3 0.001080390 0.0002265110 4.221291e-05 7.080172e-06 1.079567e-06 1.508923e-07
4 0.002064075 0.0004933813 1.048304e-04 2.004632e-05 3.484891e-06 5.553353e-07
5 0.008236171 0.0024798688 6.637118e-04 1.598722e-04 3.500850e-05 7.027254e-06
6 0.000223623 0.0000363569 5.254177e-06 6.833844e-07 8.080398e-08 8.758139e-09
13 14 15 16 17 18
1 8.967384e-07 1.507227e-07 2.364440e-08 3.477356e-09 4.813281e-10 6.292303e-11
2 2.192977e-08 2.654955e-09 2.999969e-10 3.177955e-11 3.168471e-12 2.983515e-13
3 1.946804e-08 2.332345e-09 2.607955e-10 2.733875e-11 2.697294e-12 2.513358e-13
4 8.168817e-08 1.115779e-08 1.422441e-09 1.700050e-10 1.912317e-11 2.031584e-12
5 1.302074e-06 2.240273e-07 3.597518e-08 5.415971e-09 7.673982e-10 1.026932e-10
6 8.762517e-10 8.140690e-11 7.058791e-12 5.738134e-13 4.390178e-14 3.172269e-15
19
1 7.792861e-12
2 2.661494e-14
3 2.218703e-14
4 2.044694e-13
5 1.301911e-11
6 2.171585e-16
>
Look at ?predict.zeroinfl for more information.
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
$endgroup$
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of"pscl"package. So I simply used the genericpredict()function withtype ="prob"command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for eitherzeroinflorglmmTMBobjects.
$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl()is available inpscl1.5.2, see p. 60 in the reference manual. ...
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
add a comment |
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero.
To obtain a probabilistic prediction of the probabilities to observe specific actual count observations, use type="prob". An example:
> library(pscl)
> fm_zip <- zeroinfl(art ~ ., data = bioChemists)
> head(predict(fm_zip))
1 2 3 4 5 6
2.037956 1.323124 1.308704 1.439982 2.363233 0.854771
> head(predict(fm_zip,type="prob"))
0 1 2 3 4 5 6
1 0.2162690 0.1937560 0.2279639 0.17880748 0.10518810 0.049503684 0.019414543
2 0.3626973 0.2429423 0.2058847 0.11631983 0.04928840 0.016708046 0.004719819
3 0.3655549 0.2445794 0.2051107 0.11467420 0.04808441 0.016129943 0.004508999
4 0.3582308 0.2127514 0.2034182 0.12966290 0.06198734 0.023707199 0.007555727
5 0.1071273 0.2125181 0.2559524 0.20550920 0.12375553 0.059619449 0.023934814
6 0.5218005 0.2328010 0.1513963 0.06563783 0.02134296 0.005551943 0.001203522
7 8 9 10 11 12
1 0.006526345 0.0019196452 5.019025e-04 1.181028e-04 2.526437e-05 4.954139e-06
2 0.001142821 0.0002421248 4.559820e-05 7.728561e-06 1.190849e-06 1.682002e-07
3 0.001080390 0.0002265110 4.221291e-05 7.080172e-06 1.079567e-06 1.508923e-07
4 0.002064075 0.0004933813 1.048304e-04 2.004632e-05 3.484891e-06 5.553353e-07
5 0.008236171 0.0024798688 6.637118e-04 1.598722e-04 3.500850e-05 7.027254e-06
6 0.000223623 0.0000363569 5.254177e-06 6.833844e-07 8.080398e-08 8.758139e-09
13 14 15 16 17 18
1 8.967384e-07 1.507227e-07 2.364440e-08 3.477356e-09 4.813281e-10 6.292303e-11
2 2.192977e-08 2.654955e-09 2.999969e-10 3.177955e-11 3.168471e-12 2.983515e-13
3 1.946804e-08 2.332345e-09 2.607955e-10 2.733875e-11 2.697294e-12 2.513358e-13
4 8.168817e-08 1.115779e-08 1.422441e-09 1.700050e-10 1.912317e-11 2.031584e-12
5 1.302074e-06 2.240273e-07 3.597518e-08 5.415971e-09 7.673982e-10 1.026932e-10
6 8.762517e-10 8.140690e-11 7.058791e-12 5.738134e-13 4.390178e-14 3.172269e-15
19
1 7.792861e-12
2 2.661494e-14
3 2.218703e-14
4 2.044694e-13
5 1.301911e-11
6 2.171585e-16
>
Look at ?predict.zeroinfl for more information.
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
$endgroup$
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of"pscl"package. So I simply used the genericpredict()function withtype ="prob"command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for eitherzeroinflorglmmTMBobjects.
$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl()is available inpscl1.5.2, see p. 60 in the reference manual. ...
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
add a comment |
$begingroup$
If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero.
To obtain a probabilistic prediction of the probabilities to observe specific actual count observations, use type="prob". An example:
> library(pscl)
> fm_zip <- zeroinfl(art ~ ., data = bioChemists)
> head(predict(fm_zip))
1 2 3 4 5 6
2.037956 1.323124 1.308704 1.439982 2.363233 0.854771
> head(predict(fm_zip,type="prob"))
0 1 2 3 4 5 6
1 0.2162690 0.1937560 0.2279639 0.17880748 0.10518810 0.049503684 0.019414543
2 0.3626973 0.2429423 0.2058847 0.11631983 0.04928840 0.016708046 0.004719819
3 0.3655549 0.2445794 0.2051107 0.11467420 0.04808441 0.016129943 0.004508999
4 0.3582308 0.2127514 0.2034182 0.12966290 0.06198734 0.023707199 0.007555727
5 0.1071273 0.2125181 0.2559524 0.20550920 0.12375553 0.059619449 0.023934814
6 0.5218005 0.2328010 0.1513963 0.06563783 0.02134296 0.005551943 0.001203522
7 8 9 10 11 12
1 0.006526345 0.0019196452 5.019025e-04 1.181028e-04 2.526437e-05 4.954139e-06
2 0.001142821 0.0002421248 4.559820e-05 7.728561e-06 1.190849e-06 1.682002e-07
3 0.001080390 0.0002265110 4.221291e-05 7.080172e-06 1.079567e-06 1.508923e-07
4 0.002064075 0.0004933813 1.048304e-04 2.004632e-05 3.484891e-06 5.553353e-07
5 0.008236171 0.0024798688 6.637118e-04 1.598722e-04 3.500850e-05 7.027254e-06
6 0.000223623 0.0000363569 5.254177e-06 6.833844e-07 8.080398e-08 8.758139e-09
13 14 15 16 17 18
1 8.967384e-07 1.507227e-07 2.364440e-08 3.477356e-09 4.813281e-10 6.292303e-11
2 2.192977e-08 2.654955e-09 2.999969e-10 3.177955e-11 3.168471e-12 2.983515e-13
3 1.946804e-08 2.332345e-09 2.607955e-10 2.733875e-11 2.697294e-12 2.513358e-13
4 8.168817e-08 1.115779e-08 1.422441e-09 1.700050e-10 1.912317e-11 2.031584e-12
5 1.302074e-06 2.240273e-07 3.597518e-08 5.415971e-09 7.673982e-10 1.026932e-10
6 8.762517e-10 8.140690e-11 7.058791e-12 5.738134e-13 4.390178e-14 3.172269e-15
19
1 7.792861e-12
2 2.661494e-14
3 2.218703e-14
4 2.044694e-13
5 1.301911e-11
6 2.171585e-16
>
Look at ?predict.zeroinfl for more information.
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
$endgroup$
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of"pscl"package. So I simply used the genericpredict()function withtype ="prob"command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for eitherzeroinflorglmmTMBobjects.
$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl()is available inpscl1.5.2, see p. 60 in the reference manual. ...
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
add a comment |
$begingroup$
If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero.
To obtain a probabilistic prediction of the probabilities to observe specific actual count observations, use type="prob". An example:
> library(pscl)
> fm_zip <- zeroinfl(art ~ ., data = bioChemists)
> head(predict(fm_zip))
1 2 3 4 5 6
2.037956 1.323124 1.308704 1.439982 2.363233 0.854771
> head(predict(fm_zip,type="prob"))
0 1 2 3 4 5 6
1 0.2162690 0.1937560 0.2279639 0.17880748 0.10518810 0.049503684 0.019414543
2 0.3626973 0.2429423 0.2058847 0.11631983 0.04928840 0.016708046 0.004719819
3 0.3655549 0.2445794 0.2051107 0.11467420 0.04808441 0.016129943 0.004508999
4 0.3582308 0.2127514 0.2034182 0.12966290 0.06198734 0.023707199 0.007555727
5 0.1071273 0.2125181 0.2559524 0.20550920 0.12375553 0.059619449 0.023934814
6 0.5218005 0.2328010 0.1513963 0.06563783 0.02134296 0.005551943 0.001203522
7 8 9 10 11 12
1 0.006526345 0.0019196452 5.019025e-04 1.181028e-04 2.526437e-05 4.954139e-06
2 0.001142821 0.0002421248 4.559820e-05 7.728561e-06 1.190849e-06 1.682002e-07
3 0.001080390 0.0002265110 4.221291e-05 7.080172e-06 1.079567e-06 1.508923e-07
4 0.002064075 0.0004933813 1.048304e-04 2.004632e-05 3.484891e-06 5.553353e-07
5 0.008236171 0.0024798688 6.637118e-04 1.598722e-04 3.500850e-05 7.027254e-06
6 0.000223623 0.0000363569 5.254177e-06 6.833844e-07 8.080398e-08 8.758139e-09
13 14 15 16 17 18
1 8.967384e-07 1.507227e-07 2.364440e-08 3.477356e-09 4.813281e-10 6.292303e-11
2 2.192977e-08 2.654955e-09 2.999969e-10 3.177955e-11 3.168471e-12 2.983515e-13
3 1.946804e-08 2.332345e-09 2.607955e-10 2.733875e-11 2.697294e-12 2.513358e-13
4 8.168817e-08 1.115779e-08 1.422441e-09 1.700050e-10 1.912317e-11 2.031584e-12
5 1.302074e-06 2.240273e-07 3.597518e-08 5.415971e-09 7.673982e-10 1.026932e-10
6 8.762517e-10 8.140690e-11 7.058791e-12 5.738134e-13 4.390178e-14 3.172269e-15
19
1 7.792861e-12
2 2.661494e-14
3 2.218703e-14
4 2.044694e-13
5 1.301911e-11
6 2.171585e-16
>
Look at ?predict.zeroinfl for more information.
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
$endgroup$
If you call predict.zeroinfl() without any parameters, it uses the default setting for the type parameter, which is type="response". You will then get a prediction for the mean, or the expected response. This expectation will typically vary much less than your actual observations, will not be integer, and will be larger than zero.
To obtain a probabilistic prediction of the probabilities to observe specific actual count observations, use type="prob". An example:
> library(pscl)
> fm_zip <- zeroinfl(art ~ ., data = bioChemists)
> head(predict(fm_zip))
1 2 3 4 5 6
2.037956 1.323124 1.308704 1.439982 2.363233 0.854771
> head(predict(fm_zip,type="prob"))
0 1 2 3 4 5 6
1 0.2162690 0.1937560 0.2279639 0.17880748 0.10518810 0.049503684 0.019414543
2 0.3626973 0.2429423 0.2058847 0.11631983 0.04928840 0.016708046 0.004719819
3 0.3655549 0.2445794 0.2051107 0.11467420 0.04808441 0.016129943 0.004508999
4 0.3582308 0.2127514 0.2034182 0.12966290 0.06198734 0.023707199 0.007555727
5 0.1071273 0.2125181 0.2559524 0.20550920 0.12375553 0.059619449 0.023934814
6 0.5218005 0.2328010 0.1513963 0.06563783 0.02134296 0.005551943 0.001203522
7 8 9 10 11 12
1 0.006526345 0.0019196452 5.019025e-04 1.181028e-04 2.526437e-05 4.954139e-06
2 0.001142821 0.0002421248 4.559820e-05 7.728561e-06 1.190849e-06 1.682002e-07
3 0.001080390 0.0002265110 4.221291e-05 7.080172e-06 1.079567e-06 1.508923e-07
4 0.002064075 0.0004933813 1.048304e-04 2.004632e-05 3.484891e-06 5.553353e-07
5 0.008236171 0.0024798688 6.637118e-04 1.598722e-04 3.500850e-05 7.027254e-06
6 0.000223623 0.0000363569 5.254177e-06 6.833844e-07 8.080398e-08 8.758139e-09
13 14 15 16 17 18
1 8.967384e-07 1.507227e-07 2.364440e-08 3.477356e-09 4.813281e-10 6.292303e-11
2 2.192977e-08 2.654955e-09 2.999969e-10 3.177955e-11 3.168471e-12 2.983515e-13
3 1.946804e-08 2.332345e-09 2.607955e-10 2.733875e-11 2.697294e-12 2.513358e-13
4 8.168817e-08 1.115779e-08 1.422441e-09 1.700050e-10 1.912317e-11 2.031584e-12
5 1.302074e-06 2.240273e-07 3.597518e-08 5.415971e-09 7.673982e-10 1.026932e-10
6 8.762517e-10 8.140690e-11 7.058791e-12 5.738134e-13 4.390178e-14 3.172269e-15
19
1 7.792861e-12
2 2.661494e-14
3 2.218703e-14
4 2.044694e-13
5 1.301911e-11
6 2.171585e-16
>
Look at ?predict.zeroinfl for more information.
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
answered yesterday
Stephan KolassaStephan Kolassa
48.1k8101180
48.1k8101180
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of"pscl"package. So I simply used the genericpredict()function withtype ="prob"command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for eitherzeroinflorglmmTMBobjects.
$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl()is available inpscl1.5.2, see p. 60 in the reference manual. ...
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
add a comment |
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of"pscl"package. So I simply used the genericpredict()function withtype ="prob"command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for eitherzeroinflorglmmTMBobjects.
$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl()is available inpscl1.5.2, see p. 60 in the reference manual. ...
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of
"pscl"package. So I simply used the generic predict() function with type ="prob" command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for either zeroinfl or glmmTMB objects.$endgroup$
– KO 88
yesterday
$begingroup$
Thank you for your reply, Sir. I understood what you said. However, this function is no longer available in the most recent form of
"pscl"package. So I simply used the generic predict() function with type ="prob" command. It gave me a huge object with tons of numbers in it. It basically has probabilities for each possible value I guess. All I want to do is to use a basic predicted vs observed graph to see how well my model is performing and I still can't do it. Is there a way to get this plot by the method you said for either zeroinfl or glmmTMB objects.$endgroup$
– KO 88
yesterday
$begingroup$
predict.zeroinfl() is available in pscl 1.5.2, see p. 60 in the reference manual. ...$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
predict.zeroinfl() is available in pscl 1.5.2, see p. 60 in the reference manual. ...$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
$begingroup$
... About getting a predicted vs. observed graph: there are different ways of predicting. What you had originally was a plot of predicted expectations. Which does not give an indication of variability. You could extract predicted probabilities, then calculate lower and upper quantiles and plot those. Then you could check whether, e.g., 80% of observations are between the 10% and the 90% quantile prediction.
$endgroup$
– Stephan Kolassa
yesterday
add a comment |
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