When you say SAS Studio is required with SAS Viya, does that mean that there is no command-line way to submit a SAS program to Viya? (after, say, using SSH to log into a Viya server of some sort?)       ... View more

I don't see an issue with your contrast statement aside from the fact that it is not estimable.   The issue appears to be that your contrast is not estimable because there are no combinations of the rows of your model's X matrix that produce the contrast vector you want. In other words, the beta coefficients in your contrast vector that have non-zeroes cannot be estimated jointly.    If you want to compare the mean yield by levels of iron then using something like: lsmeans /pdiff;   will do that.   Just be aware that LSMEANS is an average effect (the terms in the underlying contrast vector applied to the model beta coefficients are, basically, proportions calculated using all 54 records.) The SLICE statement would let you estimate (a population average effect) of the yield by iron levels for some specified values of the terms with which IRON is interacted.           ... View more

It's not clear to me exactly how your data are structured and how you'd like to summarize your data.   If each individual choice is given a yes/no response, it is not clear how you could determine for a particular respondent, which item they ranked first, which they ranked 2nd, and which they ranked 3rd. Can you determine that from the data?   If you cannot, then you could create separate distributions for each y/n variable so see which item was selected the most (which doesn't mean that that item was ranked at the top the most.) If you can tell how each person ranked an item, then you could calculate three distributions: 1) Distribution of items that selected as a top choice, 2) Distribution of items that selected as a second choice, 3) Distribution of items selected as a third choice.    If those summaries don't appeal to you, you should think about what you mean when you say "frequency for the entire question for the top 3 choices."  One combined approach would be to define all possible combinations of the three top choice selections made by respondents and look at the distribution of those combinations. e.g. 3-1-2, 1-5-10 (in order of ranking: choice1 -choice 2 -choice 3).     In all of these cases, you can use proc surveyfreq. ... View more

@pmorel3 wrote: Hello everyone, I'm a beginner in sas and I'm wondering a question.   After my 2 regressions, I would like to know whether the difference between my 2 intercepts is statistically significant or not. Could you please help me to write the appropriate code ? All of my variables are numbers.   proc reg data=modelization outest=est; /*1*/ M1: model rp=rm / selection=rsquare b best=1; MFS1: model rp=rm rm_zdy rm_ztbl rm_ztms rm_zdfy / selection=rsquare b best=1; proc print data=est; run;     For nested models, you can compare model fit by calculating the difference in model fit (sums of squares) and appropriately standardizing that difference and use an F test. This will tell you if the larger model is, statistically speaking, explaining more of the variation of the outcome variable. See this paper:   https://www.stat.ncsu.edu/people/bloomfield/courses/st370/Slides/MandR-ch12-sec02-06.pdf   If you follow the paper, I think you should consider whether you can use a one-factor model to do the comparison of the intercept estimates.   I haven't explored how this would work but this may give you a starting point.     Just thinking of something. If all your model predictors are binary, then your estimate for the intercept in your models will simply be the average of your outcome variable (Y) among the records where all your binary predictors have a 0 value. At the end of they day, you need to compare the means between two probably overlapping groups, when comparing the intercepts beween two nested models. You, or others here, might think of a SAS procedure that can do this; as I was unable to think of a SAS procedure that let one construct separate means for overlapping groups and test their difference.   Best of luck.       ... View more

@lgslm wrote: I assume the weights are fixed since they are 1/(standard error of a parameter estimate) and these estimates are not changing?    How can I get variances of estimated model parameters using SAS? What will be the difference in using 1/se or 1/var as weighting values?   I didn't find an example yet on this approach, I guess that's why I'm struggling a bit with it. But for sure we think it's the best way to model our data. You assume the weights are fixed but they aren't. How do you know that's ok? Do you understand the ramifications?   SAS will output standard errors of your estimated parameter estimates; you can just square them to get their variances.   Using 1/var will increase the influence of observations with standard errors below 1 and, relatively, decrease the influence of observations with standards errors above 1.   For example, if you have two observations with standard errors of .5 and 2 then 1/se is 2 and .5, respectively. The first observation has a weight that is 2/.5= 4 times higher than the second. If you use their variances, .25 and 4, respectively, then 1/var is 4 and .25, respectively. Using these weights, observation 1 has a weight that is 4/.25=16 (4 squared) times higher than the second.   ... View more

@lgslm wrote: Thank for your response!   What do you exactly mean with 'model-fitting approach'? I want to find a model that is fitting the data, we suppose it's linear with interaction effects.   In the meanwhile, I adjusted the 'stder' in the dataset to 'w' and I changed the values to 1/stderror.  This is what I have so far:   data trial; input x1$ x2$ x3$ y w; datalines; ... ... ... ; proc glm data=trial; class x1 x3; weight w; model y=x1 x2 x3 x1*x2 x1*x3 x2*x3 / solution; output out=trialresults p=pk ; proc print data=trialresults; run;   Now SAS is giving me this error, although all the values of x2 are numeric:  ERROR: Variable x2 should be either numeric or specified in the CLASS statement. Do I also need to state that x2 is numeric?   It's also giving me this error, although I stated a model (or at least that's what I think): ERROR: A MODEL statement must be given.   What could be wrong?   Remove the $ sign from X2 in your input statement so SAS will treat X2 as continuous in your proc glm.     "What do you exactly mean with 'model-fitting approach'? I want to find a model that is fitting the data, we suppose it's linear with interaction effects."   Well, what I suggested is known as inverse weighting (except I suggest using standard errors but you could also use the variances.) One issue is that SAS will treat the weights as fixed, as if they were measured with no error. Is that appropriate for your case? I don't know. I certainly think that the variances of your regression parameters will be underestimated because SAS won't capture the fact that your weights aren't fixed.  You could appeal to authority or context; if you are trying to publish in a particular field and this sort of approach is used all the time then you can just go with it. If you are coming up with this approach on your own, you should have some understanding of the ramifications and justification for this approach other than using an ad hoc method that does reduce the influence of observations with large standard errors on parameter estimates. What about the variances of those parameter estimates?  Can you cite a paper or other reference for your approach? Even an example that SAS uses in its manuals would be something.     ... View more

@Prest wrote: Hello again. I want to run a logistic regression. My dependent variable is Mets and independent variable is Q (Quartiles of Dietary Inflammatory Index) (coded as 1,2,3,4) My covariates are age, Energy intake,BMI, education (coded 0, 1, 2), Alcohol(0,1), exercise (0,1),smoking(o,1) and physical exercise (0,1).   proc surveylogistic data = Project1.DII_M; weight pooled_wt; cluster psu; strata kstrata; class Q (reference = '1') educn(reference='2') smoking (reference='0') exercise (reference='1') Drinking (reference='0'); model Mets (desc) = age HE_BMI educn Drinking smoking exercise N_EN Q; Domain sex; run;   I then want to calculate p for trend. To do this, I assigned the median value (Q_median) of the variable Adj_DII in each quartile category to each participant and modeled this as a continuous variable and read off the pr>t as my p for trend.    I used the following code to determine median values, assigning them to participants and running a logistic regression. /*obtaining median in each quartile*/ proc report data=Project1.DII_Q out=dii_median; columns sex Q Adj_DII; define sex / group "Sexe"; define Q / group "Quartile"; define Adj_DII / median; run; /*Assigning median value to individuals*/ data Project1.DII_M; set Project1.DII_Q; if sex=1 and Q=1 then Q_median=-1.10; else if sex=1 and Q=2 then Q_median=0.13; else if sex=1 and Q=3 then Q_median=1.13; else if sex=1 and Q=4 then Q_median=2.27; else if sex=2 and Q=1 then Q_median=-1.22; else if sex=2 and Q=2 then Q_median=0.10; else if sex=2 and Q=3 then Q_median=1.06; else if sex=2 and Q=4 then Q_median=2.06; if HE_BMI LE 18.5 then BMI_cat=0; else if 18.5 LT HE_BMI LT 23 then BMI_cat=1; else if 23 LT HE_BMI LT 25 then BMI_cat=2; else if HE_BMI GE 25 then BMI_cat=3; Keep Age sex pooled_wt kstrata psu HE_BMI educn Drinking smoking exercise N_EN Q Mets Hy_glu Abd_obs TG_high Low_HDL HTN Q_median Adj_DII BMI_cat ; run; /*obtaining p for trend..(method adopted from some paper*/ proc surveylogistic data = Project1.DII_M; weight pooled_wt; cluster psu; strata kstrata; model Mets(desc) = age HE_BMI educn Drinking smoking exercise N_EN Q_median; Domain sex; run; The p for trend obtained in this paper was 0.008 whereas mine is 0.09 so far from this value. I would like to know where I went wrong. I have attached my working data set. Without seeing the paper, could be for any number of reasons. How do you expect anyone to know how your work difers from a paper when you don't provide the paper? ... View more

@lgslm wrote: Dear SAS users   I am trying to fit the model: Y= a + b*X1 + c*X2 + d*X3 + e*X1*X2 + f*X1*X3 + g*X2*X3 using SAS version 9.4 with - Y the response (= estimate with standard error) - X1 and X3 categorical variables - X2 continuous variable   To specify the categorical variables, I included the CLASS statement before the model statement.   The response variables Y are estimates with a standard error (resulting from other models). How can I include the standard errors of the response variables in this model so that the more precisely estimates of Y (small standard errors) are more important for fitting the model than the Y values with bigger standard errors?   This is what I have so far:   data trial; input X1 X2 X3 Y stder; datalines; V 7 K 72.03 0.67 V 9 L 45.12 0.88 V 10 M 80.27 1.05 V 19 K 53.32 0.73 V 19 L 22.52 0.91 V 20 M 80.63 0.73 V 20 N 45.39 1.22 V 29 K 26.05 0.59 V 31 L 53.36 1.12 V 31 M 43.99 0.86 V 32 N 69.06 1.26 M 5 K 53.59 0.78 M 7 L 44.73 0.72 M 9 M 76.36 0.82 M 18 K 80.68 0.50 M 19 L 70.96 0.67 M 20 M 44.62 0.99 M 20 N 76.6 1.21 M 28 K 53.08 0.51 M 29 L 82.71 0.62 M 29 M 79.75 0.89 M 30 N 44.3 1.62 T 7 K 65.61 1.45 T 9 L 76.42 1.15 T 11 M 68.34 1.36 T 22 K 55.34 1.15 T 23 L 62.48 0.95 T 23 M 53.29 0.94 T 23 N 66.9 1.40 T 30 K 74.94 0.91 T 31 L 66.4 0.93 T 31 M 49.77 1.11 T 32 N 86.76 1.27;   Proc print data=trial; run; proc glm data=trial; class X1 X3; model Y=a+b*X1 + c*X2 + d*X3 + e*X1*X2 + f*X1*X3 + g*X2*X3; output out=trialresults p=pk ; proc print data=trialresults; run;   How do I include the WEIGHT statement?   Thank you for your advice!   Before asking how to do something like this in SAS, state what type of model-fitting approach you are using and then ask if that model-fitting approach is supported by SAS. I have no idea from your question what model-fitting approach you are using.  Is there a name for the model-fitting approach you describe?  I mean, you could create a "weight" for each record defined as 1/(standard error of Y). If standard errors > 1 then weights will be smaller for Y values with large standard errors and larger for Y values with smaller standard errors. If a handful of standard errors can fall below 1 you could just set their weights to 1. You could then use a weight statement with this "weight" and you observations with large weights would have a larger influence on the resulting model parameter estimates.   The issue with this is that this is ad hoc. You should try and identify a model-fitting methodology that is anchored to to some theory/approach and then see if SAS can implement that methodology.   ... View more

@statistician13 wrote: I am trying to create Excel output with ODS Excel and gain control over the sheet names.  The difficulty I'm encountering is that I have a freq procedure (actually the surveyfreq procedure) with multiple variables on the tables statement.  This produces one sheet per variable (as I'd hope), but I can't seem to figure out how to control the name of each tab in this case since both tabs are created with the same procedure and I can't use a byvariable as others have suggested using with proc report.   Here's same sample code: ods excel file="D:\Output1.xls" style=printer; title "Survey Results"; ods excel options(sheet_name="q0001"); proc surveyfreq data=surveydata missing nosummary; stratum stratum; weight FinalWt; tables cohort*(q0001 q0002)/cl nostd; run; quit; ods excel close; As you can see from the tables statement, the output creates two sheets:  One for cohort*q0001 and the other for cohort*q0002.  Unfortunately, this produces an Excel file with two Sheets named "q0001" and "q0001 2."  I'm trying to name the names "q0001" and "q0002" corresponding to the names of the variables.  I realize I can do this by breaking out the tables statement into two separate proc freq calls preceded with the ods excel options(sheet_name=questionname) statement, but I'd like to avoid this if possible. Any suggestions on how to do this? Thanks.      Proc surveyfreq creates two separate internal data sets ; one for each table. You can see this if you do: ods trace on; before proc surveyfreq; ods trace off;   You can use ODS to grab those data sets and print each one seprately afer your survyefreq procedure. This way you can use sheet_name to name each tab in your excel file.     This may not be what you want because it would be equivalent to running two different proc surveyfreqs.   Dave   ... View more

@George4 wrote: Thanks for the suggestions, unfortunately they don't seem to have done anything to shift the error, adding an ODS LISTING line doesn't remove the error message from my log.  Note that ods listing is explicitly closed by the %stpbegin macro, anyway I reopened it and set the gpath to the pathname of the work directory.   With ods listing open I get the below written into my log   ERROR: Physical file does not exist, /app/sas/config/Lev3/SASApp/SGPlot.png. NOTE: Listing image output written to /data/saswork1/SAS_work3B8400007F74_sl07377.e-ssi.net/SAS_work4EDA00007F74_sl07377.e-ssi.net/SGPlot1.png   I also tried moving the ods listing gpath= line to before the %STPBEGIN in case having gpath set prior to %STPBEGIN being resolved is important, but that didn't help either. The KB article didn't help me, essentially it says to close the ods listing destination or to open the ods html destination using the path attribute (both of which appear to be being done by %STPBEGIN according to my log).   Any more ideas.        Turn off ODS listing and try specifying a specific directory (where you know you have write access) to store the PNG file using the GPATH option on the ODS HTML statement.   ... View more

If you want to save your plot do this:   ODS LISTING GPATH="directory you can write to";   at the beginning of your program   Other approaches:   http://support.sas.com/kb/49/421.html     ... View more

@D_Z_ wrote: The code is below.  When the data gets concatenated the way it works right now is that it puts a space in between each line its concentrating.  what i would like it to do is put a carriage return instead.  Can anyone help me with this modification.  I have not found anything (prob because i couldn't determine the right search words) that will do this for me.   data work.order_notes; length order_notes $ 10000 /* need to be the max */; set sorted; by odr_odr_number dt_crtd; retain order_notes; if first.odr_odr_number then order_notes = scratch_pad; else order_notes=trim(order_notes) || ' ' || trim(scratch_pad); DROP SCRATCH_PAD; if last.odr_odr_number then output; run; thanks to anyone that can help me with this one.   Dean Well, on windows:   0x0D is hexadecimal for carriage return 0x0A is hex for line feed 0x0D0A is hex for carriage return followed by line feed   You can used one of these hex values as the separator instead of a space but you are creating a character variable in a SAS data set so I'm not sure how this helps you.   order_notes=trim(order_notes) || '0x0D0A' || trim(scratch_pad);     Hrm, you might need to tell SAS that 0x0D0A is hex, maybe like this: order_notes=trim(order_notes) || '0x0D0A'x || trim(scratch_pad);   ... View more