Graphing growth curves PreviousNext
Mplus Discussion > Growth Modeling of Longitudinal Data >
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 Anonymous posted on Sunday, July 04, 2004 - 8:04 pm
With your graphing function, I have generated the plots of 3 growth curves based on estimated means for 3 separate samples. Each of these curves is shown in a different .gph file because they came from different samples and were estimated with different models.

For presentation's sake, I would like one graph to show all three curves so they may be easily contrasted. I cannot figure out how to do this...essentially how to merge or cut/paste various curves.

Can you help?
 Linda K. Muthen posted on Monday, July 05, 2004 - 9:37 am
This cannot currently be done in Mplus. It should be available in the future.
 Hanno Petras posted on Friday, April 08, 2005 - 6:56 am
Dear Linda & Bengt,

I noticed that when I run a censored or censored inflated growth model that the plots for the estimated growth trajectories are not available. This is not the case in regular or two-part growth models. Is this intended or am I missing something? Thanks.

Best,

Hanno
 Thuy Nguyen posted on Sunday, April 10, 2005 - 12:40 pm
For a growth model with censored or censored inflated outcomes, you should get a plot of estimated means for the censored outcomes and a plot of estimated probabilities for the censored-inflated outcomes. Adding the PLOT option to example 6.3 on the Mplus CD gives us these 2 plots. Note that if there are covariates in the model, then the model estimated means cannot be computed. A note as to why will be printed in the RESIDUAL output section if RESIDUAL is requested.

If you are still not getting these plots without covariates in the model, please send input/output and data to support@statmodel.com.
 Hanno Petras posted on Thursday, April 14, 2005 - 4:36 am
Dear Thuy,

thank you for the information. Could you send me the example for a multiple class model with censored inflated outcomes with the graphics command added? My email is hpetras@jhsph.edu.
Also, in addition to defining a growth structure for the censored outcomes, you can run the censored inflated multiple class model allowing the censored inflated variables to be unstructured, i.e., not to define an intercept, slope etc. Both models run, but the first is very computational intensive. I am wondering if the model with unstructured censored variables is a legitimate model to run, or would you always define a growth structure for the censoring part as well? Are there any guidelines or suggestions for testing the correct specifications of the model?

Best,

Hanno
 BMuthen posted on Friday, April 15, 2005 - 1:50 am
Regardng your second question, the inflated parts of the outcomes need to be allowed to correlate in the model. A minimal model for accomplishing this is an intercept only growth model.
 Hanno Petras posted on Friday, April 15, 2005 - 5:54 am
Thank you. Do you have any model building recommedations? For example, when running a two part model, it makes sense to model the u part and the y-part separately before analyzing them in the same model. I am not quite sure how this would be translated into the censored inflated model.

Best,

Hanno
 BMuthen posted on Saturday, April 16, 2005 - 4:31 am
No, for the censored inflated model, the inflation part of the model is latent so you can't do them separately. So model building ideas have to be drawn on from regular growth modeling, that is, start with a simple model and go from there.
 Hanno Petras posted on Monday, April 18, 2005 - 6:07 am
OK, so does that mean that I would assume that for a growth model with intercept, slope and quadratic slope I would also specify the growth for the censored inflated part with an intercept, slope, and quadratic slope? In a second step, I could reduce the number of growth parameters for the censored inflated part if some turn out to be nonsignificant?

Best,

Hanno
 bmuthen posted on Tuesday, April 19, 2005 - 12:13 pm
You could start with the same growth factors in the inflation part but only allow the intercept to be random for the inflation part (so estimating only the means, not the variances, for s and q in the inflation part).
 Hanno Petras posted on Friday, January 05, 2007 - 10:39 am
Dear Linda & Bengt,

I am running a Two-Part model looking at delinquency data over 13 time points. The u-part relates to the probability of engagement in crime and the y-part relates to the count of crimes the person engaged in within that time period. The mean count of crimes range from 1.38 to 2.72. I noticed that treating it as count increases the computation immensely. In the Olson & Schaeffer paper the y-part is logged (if I remember correctly). Given that the outcome is relative rare would it be more appropriate to specify it as count instead of treating it as log continuous. If the latter is advisable, are the estimates accurate or as in the Poisson regression model (treating the outcome as log + an added constant) are they rough approximations. Sorry for the long question. Thanks.

Hanno
 Bengt O. Muthen posted on Monday, January 08, 2007 - 9:35 am
With two-part modeling, having y as a count variable would call for using a Poisson distribution that is truncated, that is, not including zero, which Mplus does not accomodate. So with counts, Mplus leads to ZIP modeling instead of two-part. It's hard to say which is best.
 Hanno Petras posted on Monday, January 08, 2007 - 11:02 am
Dear Bengt,

I just saw that I filed my question under the wrong heading. My apologies for that.

Thank you for your answer. So when treating the y part as count Mplus models it as Poisson and not truncated Poisson and that would mean that the estimates are biased. Therefore the log transformation may make more sense. Would you agree?

Best,

Hanno
 Bengt O. Muthen posted on Monday, January 08, 2007 - 11:49 am
Yes.
 Adam Darnell posted on Monday, February 05, 2007 - 10:28 am
Hello,
I have an LGM of 11 repeated observations, that includes an intercept, and linear and quadratic growth functions. I have several time invariant predictors of these growth parameters. I would like to display the estimated growth curves for two different levels of one of these predictors, in order to illustrate the effect of the predictor. Is there a way to display multiple estimated growth curves across conditions of a predictor? I have tried specifying covariate sets in the adjusted estimated means window, and just specifying two contrasting values of the predictor I am trying to illustrate. But the same curve is shown for both sets. Thanks in advance for your help,
 Bengt O. Muthen posted on Tuesday, February 06, 2007 - 10:31 am
This should work fine. Take ex6.10 as an example (you have that ex on your CD or find it on our web site), plotting against two values of x1, say -1 and +1. In the Adjusted means menu you enter say "low" in the upper right hand box and then click Name covariate set. Then you click on x1, click on Use and enter -1, plus say ok. Then put the cursor in the covariate set again and type "high" and click on Name covariate set (it is important to click again, otherwise you get 2 overlapping curves) and then enter +1. This gives 2 lines. Note that some models don't imply different curves for different covariate values (such as mixtures where x only influences c).
 Adam Darnell posted on Tuesday, February 06, 2007 - 6:16 pm
Thank you Dr. Muthen for your reply... I used example 6.10 as a guide. I was defining covariate sets in the same way that you described, but it seems that it works fine as long as you don't include multiple observed predictors with unanalyzed associations between them. If the predictors are uncorrelated, it is no problem to specify covariate sets (as is the case with ex6.10 -- x1 and x2 are uncorrelated). But if there are unanalyzed associations between the predictors and then you specify covariate sets the trajectories are always the same for the covariate sets. Could you explain this? In my model I am interested in illustrating the effect of one predictor on linear and quadratic growth, controlling for the effects of 4 observed covariates and the level of the intercept. All of the observed predictors have unanalyzed associations between them, and I am unable to specify covariate sets with any result as long as those correlations are there. Thank you for your assistance.
 Adam Darnell posted on Wednesday, February 07, 2007 - 8:33 am
Another question Dr. Muthen: as I described above, I am interested in the effect of one predictor on the slope controlling for the intercept, and on the quadratic controlling for the intercept and slope. So I have direct effects between the latents where appropriate. I gather that the adjusted means are not adjusted for these effects, correct? That is the effect of my predictor of interest will appear the same in the adjusted means graph regardless of whether the model specifies an unanalyzed association between intercept and slope, or a direct effect. Also, I am guessing that it is not possible to specify covariate sets based on these latent covariates? Thank you very much...
 Linda K. Muthen posted on Wednesday, February 07, 2007 - 9:25 am
Please send your input, data, output, and license number to support@statmodel.com. Give us the sequence you go through to get the graphs and explain where it goes wrong.
 Eva Van de gaer posted on Friday, February 16, 2007 - 5:56 am
Dear Linda and Bengt,

I have estimated a two level (level 1: students, level 2: schools) latent growth curve model and would like to draw the growth curves at level 2. Is this possible in Mplus? I would like to be able to draw growth curves of a number of schools but it seems to me that I only can draw growth curves at the student level.

This is part of my model specication:

ANALYSIS: TYPE = TWOLEVEL MISSING H1;
MODEL:%WITHIN%
iw1 sw1 qw1| aczc1@0 aczc2@1 aczc3@3 aczc4@5;
aczc1-aczc4 (1);
qw1@0;

%BETWEEN%
ib1 sb1 qb1| aczc1@0 aczc2@1 aczc3@3 aczc4@5;
aczc1-aczc4@0;

OUTPUT: modindices tech4 tech1;
SAVEDATA: results are results.dat;
PLOT: Type = plot1;
type = plot2;
Type = plot3;
Series = aczc1(0) aczc2(1) aczc3(3) aczc4(5);


Thank you!
 Bengt O. Muthen posted on Friday, February 16, 2007 - 7:43 pm
The plot that Mplus provides is for the estimated mean growth curve. There is only one mean per variable. If you want to plot curves for schools, you would have to ask for factor scores - which gives you the estimated school values for each time point. Mplus does not plot that for you, but you have the values in the graph file and can plot them with another program.
 Richard Rivera posted on Wednesday, October 24, 2007 - 12:32 pm
When, I run a two-part growth model with out covariates, can Mplus produce plots? If so, what is the code?
 Linda K. Muthen posted on Wednesday, October 24, 2007 - 5:21 pm
See the PLOT command.
 Julia Dmitrieva posted on Tuesday, October 14, 2008 - 11:29 pm
Dear Linda and Bengt,

I am trying to graph results of a growth curve model for censored inflated data. I am using a covariate, so plot command does not work. Can I manualy compute estimated values for a given value of a covariate (e.g., X=2)?

Here is my model:

i s q| Y1@0 Y2@.1 Y3@.2 Y4@.3
Y5@.4 Y6@.5 Y7@.6 Y8@.7 Y9@.8 Y10@.9 Y11@1.0
Y12@1.1 Y13@1.2;
ii si| Y1#1@0 Y2#1@.1 Y3#1@.2 Y4#1@.3
Y5#1@.4 Y6#1@.5 Y7#1@.6 Y8#1@.7 Y9#1@.8 Y10#1@.9
Y11#1@1.0 Y12#1@1.1 Y13#1@1.2;
ii-si@0; i-q@0;
Y1-Y13 ON X;

Thank you very much,
Julia
 Linda K. Muthen posted on Wednesday, October 15, 2008 - 8:07 am
This requires numerical integration so it cannot be calculated by hand.
 Julia Dmitrieva posted on Wednesday, October 15, 2008 - 11:15 am
Hi Linda,
Thank you. What would be your suggestion for reporting results? My ultimate model is a growth mixture model with one static and one time-varying control variable. I would need to somehow show the shape of the developmental trajectories. Should I graph results for trajectories without the covariates? Would that be an adequate representation?
Alternatively, should I run growth mixture models on the residuals (from the regression of the outcome variable on the control variables)? Are there any other options?

Thank you very much,
Julia
 Bengt O. Muthen posted on Wednesday, October 15, 2008 - 5:52 pm
I would show the graph without covariates.
 Wei Chun posted on Monday, February 02, 2009 - 6:21 pm
Dear Linda,

In Mplus output, following plots are usually available:
Histograms (sample values, estimated factor scores, estimated values)
Scatterplots (sample values, estimated factor scores, estimated values)
Sample means
Estimated means
Sample and estimated means
Observed individual values
Estimated individual values


But in my output, I can only get:
Histograms (sample values, estimated factor scores, estimated values)
Scatterplots (sample values, estimated factor scores, estimated values)
Sample proportions
Estimated probabilities
Item characteristic curves
Information curves
Estimated individual probability values

Although I have asked for the following:
OUTPUT: SAMPSTAT MODINDICES;
PLOT: SERIES ARE con1_94 con1_96
con1_98 con1_00 con1_02 con1_04(*);
TYPE IS PLOT3;


Could you please advise me on this issue? Many thanks.
 Linda K. Muthen posted on Tuesday, February 03, 2009 - 8:12 am
The plots that you get depend on the model that is estimated. For example, Observed individual values and estimated individual values are given for growth models. It sounds like you have an IRT model. If you have further questions regarding this, please send them along with your outputs and license number to support@statmodel.com.
 Wei Chun posted on Tuesday, February 03, 2009 - 2:24 pm
Thank you, Linda.

It is a 6 wave growth model with quadratic trend. I used WLSMV estimator because of my categorical outcome variables. If I use MLR estimator, I can get plots of
Sample and estimated means
Observed individual values
Estimated individual values
So do I use MLR for getting the plots and use WLSMV for reporting the outcomes?

Many thanks for your advice.
 Linda K. Muthen posted on Tuesday, February 03, 2009 - 3:01 pm
If you are not using Version 5.2, go to Support - Mplus Updates and download it. If you are, please send your input, data, output, and license number to support@statmodel.com.
 Kurt Beron posted on Tuesday, March 10, 2009 - 1:33 pm
Hi,

I'm trying to plot a censored normal outcome from a growth model. The plot that is produced seems to be the unconditional plot of the latent variables, which includes the zero values (lower bound in my case). However what I'd like is the plot of the more relevant, for me anyway, E(Y|Y>0). Is there a way to generate this in Mplus? Usually this requires computing the inverse Mills Ratio and this isn't obvious to me in Mplus.

Thanks....Kurt
 Bengt O. Muthen posted on Tuesday, March 10, 2009 - 4:10 pm
No, I don't see a way to plot this using the Mplus PLOT command. The normal density and distribution function would need to be available. Something to add.
 Dimiter M. Dimitrov posted on Tuesday, March 24, 2009 - 1:50 pm
Dear Linda,
In UG ex.19.6 (Individual development over time), which PLOT command(s) to add in order to plot individual trajectories over time?
 Linda K. Muthen posted on Tuesday, March 24, 2009 - 3:11 pm
You would need PLOT3.
 Dimiter  posted on Tuesday, March 24, 2009 - 4:09 pm
I used PLOT3, but it does not provide individual trajectories. Is there some additional syntax under the command: PLOT: TYPE IS PLOT3; ?
Also, how to make a decision about model fit based on the reported indices (Loglikelihood, AIC, and BIC)?
 Bengt O. Muthen posted on Tuesday, March 24, 2009 - 7:32 pm
There is no UG ex 19.6 so I can't tell which model you are considering. With growth models the PLOT command SERIES = is the one to use to look at estimated mean curves and individual trajectories.

Your second question concerns a big topic. Basically, these indices are meant to be used when comparing two models. For instance, you want the model with the lowest BIC.
 Dimiter  posted on Tuesday, March 24, 2009 - 7:35 pm
Sorry,
it's ex.9.16
 Bengt O. Muthen posted on Tuesday, March 24, 2009 - 7:48 pm
I think the individual plots may not be available for Type = Random.
 Nicole Nugent posted on Tuesday, May 11, 2010 - 10:50 am
Dear Bengt and Linda,

I have conducted a growth model with intercept and slope of count variables over 4 timepoints regressed on treatment (two groups; 0, 1) and covariates. I would like to plot (or use output to manually plot) the curves for each treatment group (0,1) as an illustration of the significant differences in counts over time. Any help with how to do this is much appreciated.

Thank you in advance,
Regards,
Nicole
 Linda K. Muthen posted on Wednesday, May 12, 2010 - 8:28 am
You can use TYPE=MIXTURE with the KNOWNCLASS option to obtain a curve for each of the two treatment groups.
 Deborah Bandalos posted on Wednesday, July 07, 2010 - 1:43 pm
I have run a censored-inflated growth model on data in which I know there is some missing data. However, in the censored-inflated output it states that there are 0 missing data patterns and I get no info on the amount of missing data. Why would this be?

My syntax is:

.......

USEVARIABLES = shop2 shop4 shop5-shop7 shop8 shop9;
CENSORED = shop2 shop4 shop5-shop7 shop8 shop9 (bi);
MISSING = ALL(-9);
ANALYSIS: ESTIMATOR = MLR;
MODEL: i s| shop2@0 shop4@1 shop5@2 shop6@3 shop7@4 shop8@5 shop9@6;
ii si | shop2#1@0 shop4#1@1 shop5#1@2 shop6#1@3 shop7#1@4 shop8#1@5 shop9#1@6;
si@0;
shop2-shop9 (1);! held = due to convergence problems;
OUTPUT: tech1 tech4;
plot: type is plot3; series = shop2-shop9(*);

Thanks,

Debbi
 Linda K. Muthen posted on Wednesday, July 07, 2010 - 2:59 pm
Please send the full output and your license number to support@statmodel.com.
 Diane Chen posted on Thursday, December 16, 2010 - 12:36 pm
Is it possible to graph growth curves for a 2nd process (e.g., aggression) within different classes of another process (e.g., peer rejection)? What would the plot/series input statement look like? Thanks!
 Linda K. Muthen posted on Thursday, December 16, 2010 - 2:50 pm
You use a | symbol to separate the two processes in the SERIES option.
 Isaac Petersen posted on Tuesday, February 15, 2011 - 4:55 pm
Is it possible to get confidence intervals or standard error bars for the adjusted estimated means?
 Linda K. Muthen posted on Tuesday, February 15, 2011 - 5:28 pm
No, this is not possible.
 EFried posted on Wednesday, February 22, 2012 - 7:54 am
In the online tutorial video Topic 4, second part, you mention that MPLUS is able to contrast the mean of the dependent variable on measurement point t of (1) people who will be present at t+1 in a longitudinal study vs. (2) people who will drop out at t+1.

I do not find this function, using the newest version of MPLUS, and wonder if you could hint me into the right direction.

Also, is there a paper covering methods to test whether longitudinal data are MAR/MCAR in MPLUS?

Thank you!!
 EFried posted on Wednesday, February 22, 2012 - 8:10 am
(My sample has 5 measurement points, missing data 5% at the first measurement point and 40% at the last. Not atypical for depression studies, but I don't quite know how to handle this)
 Linda K. Muthen posted on Wednesday, February 22, 2012 - 2:11 pm
See the DESCRIPTIVE option of the DATA MISSING command. See also the following paper on the website:

Muthén, B., Asparouhov, T., Hunter, A. & Leuchter, A. (2011). Growth modeling with non-ignorable dropout: Alternative analyses of the STAR*D antidepressant trial. Psychological Methods, 16, 17-33.

MAR cannot be tested. See the following paper on the website for MCAR:

Muthén, B., Kaplan, D. & Hollis, M. (1987). On structural equation modeling with data that are not missing completely at random. Psychometrika, 52:3, 431-462.
 EFried posted on Sunday, February 26, 2012 - 7:25 am
Thank you, Linda.

(1) The difference between means of the dependent variable between dropouts and non dropouts after measurement point 1 is highly significant p<.001, I guess that means that people with a higher value on my dependent variable dropout more readily.

(2) Does multiple imputation rely on the MAR assumption?

(3) Bringing covariates into the model with the "x1 x2 x3;" command so that MPLUS uses a different way to handle covariate missing values leads to non-identification very often. Are there any constraints that I would need to implement that MPLUS doesn't use by default and could cause these problems?
 Linda K. Muthen posted on Sunday, February 26, 2012 - 11:06 am
1. That sounds correct.

2. Yes.

3. When you bring a binary covariate into the model, it results in a non-identification message because the mean and variance of a binary variable are not orthogonal. When this is the reason, the message can be ignored.
 EFried posted on Monday, February 27, 2012 - 5:44 am
3.

I'm not sure whether the problems arise from this.

Using a GMM with

classes=c(4);
ANALYSIS:
type = MIXTURE;
Starts = 400 40;
STITERATIONS = 10;
MODEL:
%OVERALL%
i s q | phq_t0@0 phq_t1@1 phq_t2@2 phq_t3@3 phq_t4@4;
q@0;
i s ON sex hisdep;

leads to no problems, 27 free parameters, and a BIC of 14000.

Including "sex hisdep;" into the model (both binary) leads to 38 free parameters, a lot of perturbed starting value runs in both phases, and a BIC of 16000.

Entropy and class solutions are about the same, though.

What could be the reason for this?
 Linda K. Muthen posted on Monday, February 27, 2012 - 6:08 am
It is impossible to answer this question with the information given. Send the relevant outputs and your license number to support@statmodel.com.
 Andreas Stenling posted on Thursday, January 24, 2013 - 5:27 am
Hi,

We are trying to graphically display individual trajectories in latent growth curves. However, the plots only displays straight lines and no curves , although we request observed individual values. We have tried different syntax to obtain graphs as stated in the Mplus users guide but regardless of the syntax used we only receive straight lines.

So our question is how do we obtain a plot showing individual trajectories? We suspect that there is a simple solution to this problem but we are currently unable to solve this. Below is our model and plot syntax.

MODEL:
i s| Sjalvkansla1@0 Sjalvkansla2@1 Sjalvkansla3@2;

PLOT: SERIES = Sjalvkansla1 (0) Sjalvkansla2 (1) Sjalvkansla3 (2);
TYPE = PLOT3;
 Linda K. Muthen posted on Thursday, January 24, 2013 - 11:55 am
Please send your output, gph file, and license number to support@statmodel.com.
 Laia Becares posted on Thursday, May 30, 2013 - 4:36 am
Hello,

I have a piecewise latent growth curve which jointly models two outcomes. How can I plot both lines?

I have tried:
Series is czbmi3 (0) czbmi5 (1) czbmi7 (2)czght3 (0) czght5 (1) czhgt7 (2):
Type is plot3;

but don't get the expected graphs.

Thank you
 Bengt O. Muthen posted on Thursday, May 30, 2013 - 1:28 pm
You can try using | in the Series option to separate the two pieces.
 AshleyM posted on Sunday, November 17, 2013 - 1:04 pm
Hello,

I am trying to get PLOTS for the trajectories of classes for a series of LCGA models. However, I am not getting the expected plots in the output and the option is not available to view the plots.

Here are parts of my syntax that might be helpful:

VARIABLE:
USEVARIABLES ARE dintk1 dintk2 dintk3 dintk4 dintk5 dintk6 dintk7 dintk8 dintk9 dintk10 dintk11 OHFNP21 OHFNP22 OHFNP23 OHFNP24 OHFNP25 OHFNP26 OHFNP27 OHFNP28 OHFNP29 OHFNP210 OHFNP211;
MISSING ARE ALL (-9999);
TSCORES = dintk1-dintk11;
CLASSES = c(3);

ANALYSIS:
TYPE IS MIXTURE RANDOM;
ESTIMATOR IS ML;
STARTS = 100 10;
PROCESS = 4;

MODEL:
%OVERALL%
i s|OHFNP21-OHFNP211 AT dintk1 dintk11;
i-s@0;

PLOT:
SERIES = dintk1 - dintk11 (s) ;
TYPE = PLOT3;

As you can see, I am trying to use the TIMESCORES command to account for individually varying times of observation (UG 6.12) and the fact that the data are not time structured.

Is it possible to get plots while using TIMESCORES? Do you know if it is still possible to view the plots using this command? Or is there something else that I am doing incorrectly?

Thank you!
 Linda K. Muthen posted on Sunday, November 17, 2013 - 2:45 pm
Plots are not available for the AT and TIMESCORES options.
 AshleyM posted on Tuesday, November 19, 2013 - 10:27 am
Ok, thank you. If the plots, LMR-LRT, and BLRT options are not available for LCGA models using TIMESCORES, are there other indices of fit I can look to in order to make decisions about the best number of trajectory classes (aside from AIC, BIC, entropy, and posterior probabilities, and theory)? Do you have any suggestions about how to make such decisions?

Thank you!
 Sebastian Teran Hidalgo posted on Monday, December 09, 2013 - 10:59 am
Hello,

I am running a ZIP growth model. I have successfully run an unconditional version of the model. I requested the plots of the estimated and sample means and they look fine. When I do the same model but now conditional on some dummies related with race and request the sample means I only get a straight line. Why am I getting a straight line?

Thank you so much

MODEL:

i s q| DRINKS13@0
DRINKS14@0.1 ... DRINKS32@1.9;

ii si qi| DRINKS13#1@0
DRINKS14#1@0.1 ... DRINKS32#1@1.9;

[DRINKS13@0
DRINKS14@0 ... DRINKS32@0 i s q];

[DRINKS13#1@0
DRINKS14#1@0 ... DRINKS32#1@0 ii si qi];

i on race_h race_w race_b;
s on race_h race_w race_b;
q on race_h race_w race_b;
ii on race_h race_w race_b;
si on race_h race_w race_b;
qi on race_h race_w race_b;

qi@0;
q@0;

OUTPUT:TECH4;

Plot: type=plot2;
series = DRINKS13(13)
DRINKS14(14) ... DRINKS32(32);
 Linda K. Muthen posted on Monday, December 09, 2013 - 11:06 am
Please send the output and your license number to support@statmodel.com.
 xiaoyu bi posted on Monday, January 27, 2014 - 11:43 pm
Dear Dr. Muthen,
How to plot a variable using a long-format data? For example, for wide-format variables, x1 x2 x3 x4 x5, I can plot it by series = x1 x2 x3 x4 x5 (*);
But, how about in long-format data to plot variable X?
Thank you!
 Linda K. Muthen posted on Tuesday, January 28, 2014 - 11:53 am
The PLOT option is not available for long format.
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