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H · ‎07-16-2014

Great options. Yesterday I ended up using the option that was removed from this thread. Though, Ksharp's seems to be very concise. Follow-up question, Is there a good way to do this for pulling observations based on the first three digits (going left to right) of the value. So I would like to do that exact samething but for numbers starting 958, which are actually list in the sample as 958.20, 958.3, 958.30, 958.92, etc.).

H · ‎07-15-2014

I am looking to pull all observations that have a particular numeric value in at least one of the listed variables. Though I do not want to have to write all of the variables out. I have tried dx1-50 and also a macro to no success. See following for examples: proc print data=test; where dx1-dx50 = 808.44; var sex; run; or %let vars = dx1-50; proc print data=test; where &vars = 808.44; var sex; run; quit; Perhaps it needs an "or"? I am still messing around with options but may need some help. I also believe this could be done in proc sql, but I am not experienced with that procedure.

H · ‎07-10-2014

See: Detry & Roger. The Intention-to-Treat Principle. JAMA. 312(1);2014:85.

H · ‎07-07-2014

Thanks, yes that is the approach I took and I finally got the graph I wanted out of it. Thanks to everyone here. I also had tried the proc genmod approach awhile back. I kept getting very close to a solution, but you all helped me get there and confirm my methods. Once I have some free time I will update this thread giving an overview of what I did. Thanks again - you all have been very helpful! P.S., The discrepancy I referenced in my prior post on the values jumpping up quite a bit was just a mis-type in my coding that I was able to clean up.

H · ‎07-03-2014

I just got back from a conference, so I have not had a chance to work on this until this afternoon. I centered the continuous variabe using proc sql and the proc nlmixed then ran. Per Denham's suggestion, I first requested the model estimates in proc nlmixed (which were exactly the same as the proc logistic). I have been working on writing the estimates. The continous variable mean was 11, so I place B4*-10 through 30 to get estimates for continuous variable 1-40, however between B4*0 and the B4*1 estimate there seem to be a bigger gap in intergers than between all of the other estimates. Not sure on a way to address that. I will keep working and updating this thread. Thank you all for your help so far. My next attempts will be to do a ods trace and try to pull these values out to eventually plot. Also, does anyone know how to insert a range or perhaps macro for the integers multiplied to the B4 term. Writing B4*-10, B4*-9,...,B4*30 for two groupings of estimates takes up quite a bit of line space. Lastly, StatDave - thanks, Funda had referenced that SAS document in a prior post on this thread. Was there a particular line you were referencing that may help? Also Bobderr, thanks for all of that scaling code, I ended up centering, which seemed to work - lucky for me, since I was getting a little lost in your code..

H · ‎06-20-2014

It runs fine until I add the continuous variable, then halts and the output stops prior to the "Interation History" with the following log message: NOTE: To assign starting values to parameters, use the PARMS statement. The default starting value of 1.0 is in effect for all parameters. ERROR: No valid parameter points were found. NOTE: PROCEDURE NLMIXED used (Total process time): real time 0.14 seconds cpu time 0.06 seconds This is the output: The NLMIXED Procedure Data Set WORK.TEST Dependent Variable y Distribution for Dependent Variable Binomial Optimization Technique Dual Quasi-Newton Integration Method None Observations Used 319 Observations Not Used 0 Total Observations 319 Parameters 5 1 1 1 1 1 1.15792E77 I had originally attempt to use this approach via: http://www.columbia.edu/~mmw2177/Testing%20for%20additive%20interactions%2029Feb2012.pdf Thanks, any help is appreciated. Once again, my original question was how to find the difference between two probabilities and the respective confidence interval for different levels of a continuous covariate only given the two probabilities their SE, which can be calculated for the continuous covariate levels. However, I am will to explore all options for the a solution.

H · ‎06-20-2014

Well it turns out I tried to go this route about 3 weeks ago to no avail. I kept botching things up when I added the continuous variable. Perhaps it is that I am not adding it correctly. So for example, I realize the following code is off but it would be something like this (see below), but the continuous variable is being treated as a binary variable. Ideally I want to run this and find the differences between the probabilities plus confidence intervals given specified values for the continuous variable (e.g., 1, 5, 10, 15,...). Any help would be appreciated. proc nlmixed data=test; p=1/(1+exp(-(Intercept + b1*x1 + b2*x2 = B3*x1*x2 + B4*x3)));/*B4 Continuous/ model y ~ binomial(1,p); estimate "Prob Diff x1 = 1 and X2 = 1 or 0" 1/(1+exp(-(Intercept+b1 + b2 +b3 + b4*x4))) - 1/(1+exp(-(Intercept + B1*X1 + B4))); run;

H · ‎06-20-2014

Yes, I am excited to try this approach (which is oddly enough is similar to another piece of code I have further down in my program). I am caught up in doing something else right now, so I will update this thread if I end up having difficulties or with the final code used once I run it. Thanks again every one - this is the third question I have posted on this forum, with great responses everytime! P.S., Unsure why I did not find that SAS document during my searches

H · ‎06-18-2014

Well I am on a stats forum, so lets please hear about the picky method. I had the code output the cov, which seems fairly splotchy with values just for the reference values of X1, X2, X1*X2, and for X3.

H · ‎06-18-2014

Additional option, If I just have the probabilities and SEs and want to find the difference between two of these probabilities with its confidence interval - is it just possible to do the following: ~ find the differences ~ square the individual SEs, then add them together and find the square root for the difference SE ~ calculate confindence intervals via difference +/- (SE * 1.96)

H · ‎06-18-2014

Steve Denham, I feel like I can get close to the results I am looking for but not quite there. The following gets me the individual probabilites needed for the subtraction: lsmeans X1*X2 / at X3 = 11.23 ilink; when I add "diff" as an option it calculates the differences, though these are differences in the beta coefficients, which when exponentiated and divided by 1 + them exponentiated, do not equal the same thing as differences in transformed values (probabilities) I am after (along with confidence intervals). Any suggestions would be appreciated! P.S., I could always pull out the probabilities and SE needed for the subtraction and perform the procedure in a data step, however - I do not know how to get the confidence intervals for this calculated difference based only on the probabilities and their respective SE.

H · ‎06-13-2014

I am running a logistic regression model with two binary categorical variables, their multiplicative interaction term, and a continuous variable. I am currently trying to get the differences in probabilities (along with respective confidence interval(CI)) for different combinations of independent variabls (e.g., for a strata) in an attempt to plot them with their CI. My goal is to generate graphs similar to those at the bottom of the following webpage: http://www.ats.ucla.edu/stat/stata/faq/logit2by2.htm My data is very similar to those in the STATA example, or what I mean is I have an X1, X2, X1*X2, and X3 (continuous variable). Using contrast statements in my proc logistic, I can get the probabilities for select groupings of my variables based on selected level of the continuous variable (see attached). However, I want to get the differences (and respective CI) of these based on the level of one of the categorical variables. The difference is basic subtraction, but I don't know how to get the CIs for them. Seems like something that can be done in a contrast statement, but I can't seem to get my code right, since based on simple subtraction, I know what the difference should be. I have tried things like "contrast X1 1 X1*X2 1", which are terms not in both contrast statements, to no avail. Thanks in advance for your time and help! proc logistic data=dataset; class X1 (ref='0') X2 (ref='0') / param=ref; model y(event='1') = X1 X2 X1*X2 X3 /*continuous*/ / link = logit expb clodds=wald; contrast 'X1 = 0 X2 = 1' X2 1 X3 11.23 intercept 1 /estimate = prob; /*11.23 mean of X3*/ contrast 'X1 = 1 X2 = 1' X1 1 X2 1 X3 11.23 intercept 1 X1*X2 1 /estimate = prob; contrast 'X1 = 0 X2 = 0' X3 11.23 intercept 1 /estimate = prob; contrast 'X1 = 1 X2 = 0' X1 1 X3 11.23 intercept 1 /estimate = prob; /*Want difference of first and second contrast plus confidence interval*/ /*Want difference of third and fourth contrast plus confidence interval*/ run;

H · ‎05-07-2014

Well prior to my last post I had marked PG's post as helpful and "Like", but did not see a "Correct Answer" option. Please let me know where that option is and I will mark it accordingly. Or is marking it correct not an option since I marked Linlin's original post as correct?

H · ‎05-06-2014

PGSTAT, Follow-up, I tried your code and it worked. I only had to switch the order of the datasets to get the nearest center to each of the observations listed in allcenters instead of selectCenters and change the data format to radians ("RM"). My follow-up question is I calculate the distances using geodist function (you provided) Great Circle Distance Formula (http://support.sas.com/kb/3/973.html) DISTANCE = 4000*ARCOS(arc); arc = SIN(Y2)*SIN(Y1) + COS(Y2)*COS(Y1)*COS(X2-X1) (http://support.sas.com/kb/3/973.html) Do you or anyone else have any preferences for these three methods, I looked up one distance that I knew in my head to be about two miles by foot in real life (not based on zip code centers) and all three were 1.6158, 1.6078, and 162.81, miles respectively. I think for all intents and purposes, I should be fine with any of the methods, but did not know if anyone knew if one was confirmed to be better (I am in Midwestern United States)?

H · ‎05-06-2014

Thank you to everyone, I was able to get my desired variable. I ended up using Linlin's last bit of code along with: comment by Dan at the following: http://blogs.sas.com/content/iml/2012/05/21/find-the-minumum-value-in-each-row/ Yes, PGSTAT, my data management skills usually leave me over complicating tasks. As I mentioned in a previous post, most of these tasks I perform so sporadically that they are not second nature. I will probably end up sitting on this code and reusing it about once a year, so I end-up being a sub-mediocre data management person. However, if time allows - I will try out your code option as well since as of now I also incorporate the general round distance code with all of its sines and cosines. Also, I am determining distances based on zip codes, which is less ideal than using addresses. Does anyone know of a way to use addresses, and pull their longitudes and latitudes from SAS to get more exact distances?

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