About lizzy28

lizzy28 · ‎04-15-2015

I figured out what the problem was. After I recoded mhi_ctg into four dummy variables mhi_grp1-4, mhi_grp1 was excluded from the regression, and thus the observations with missing values in mhi_grp1 were treated the same way as the ones taking 1 in the variable. Thank you, Ballard and Paige!

lizzy28 · ‎04-15-2015

Sorry for the confusion. mhi_grp1-4 were derived from mhi_ctg as mutually exclusive dummy variables. As mhi_grp2-4 were included in both regressions, I believe that they were supposed to run the same data set. Thanks.

lizzy28 · ‎04-15-2015

Thanks, Ballard. But when I run the data with missing values excluded, the result was different from that given by not excluding missing values on purpose. The key difference is that coefficient values were totally different. I'm attaching my data in my thread. The way I used was: proc reg data=temp; model lcost_all_adj=age_diag_gp1 age_diag_gp2 age_diag_gp4 age_diag_gp5 mhi_grp2 mhi_grp3 mhi_grp4 days_debr_grp2 days_debr_grp3 days_debr_grp4 flap negpthp days_dswi_grp2 days_dswi_grp3 days_dswi_grp4 los_cs_grp2 los_cs_grp3 los_cs_grp4 los_cs_grp5 comorbid_grp2 comorbid_grp3 sepsis transf_bleedcomp; run; This gave me corrected total of 1198, as shown below: Number of Observations Read 1198 Number of Observations Used 1198 Analysis of Variance Source DF Sum of Squares Mean Square F Value Pr>F Model 23 441.70924 19.20475 24.58 <.0001 Error 1174 917.16903 0.78123 Corrected Total 1197 1358.87827 However, when I particularly excluded the missing values, as below proc reg data=temp; model lcost_all_adj=age_diag_gp1 age_diag_gp2 age_diag_gp4 age_diag_gp5 mhi_grp2 mhi_grp3 mhi_grp4 days_debr_grp2 days_debr_grp3 days_debr_grp4 flap negpthp days_dswi_grp2 days_dswi_grp3 days_dswi_grp4 los_cs_grp2 los_cs_grp3 los_cs_grp4 los_cs_grp5 comorbid_grp2 comorbid_grp3 sepsis transf_bleedcomp; where mhi_ctg^=.; run; I have Number of Observations Read 992 Number of Observations Used 992 Analysis of Variance Source DF Sum of Squares Mean Square F Value Pr > F Model 23 381.081 16.56874 21.28 <.0001 Error 968 753.8316 0.77875 Corrected Total 991 1134.913

lizzy28 · ‎04-15-2015

I use both SAS and STATA to run a log-linear regression with the same dataset. The coefficient magnitudes were somehow different. One of the variables in my dataset had 18% missing values. I was wondering whether it was because SAS applied imputation when running regression. Anyone knows the difference between SAS and STATA in running regression for data with missing values? Thanks a lot.

lizzy28 · ‎02-25-2015

I think I overthought. I did "04"--"06"="adult" at the beginning. Thanks a lot.

lizzy28 · ‎02-25-2015

Hi all, I've got lots of help here. Here is another question I come across. I have a character variable agegrp with values of "01", "02", "03", "04", "05", and "06". The value labels are 01: younger than 8, 02: age 9-12, 03: age 13-17, 04-06: adults. To do proc format for the variable, is there a quick way to do 04-06? Here is how I started: proc format; value $agegrp "01"="younger than 8" "02"="age 9-12" "03"="age 13-17" "04"="adult" "05"="adult" "06"="adult"; run; Thanks a lot.

lizzy28 · ‎02-11-2015

Thanks, Reeza. It does work now -- don't know why it didn't previously. Thanks a lot!

lizzy28 · ‎02-11-2015

I have windows 7 and SAS 9.3. I am learning macro, and this is my first time to do array within a macro program. I know that there is a function for it, but cannot really understand the function. There are 15 variables (proc1-proc15) with the same attributes in 10 datasets (data1-data10). Here is my program: %macro create_var(); %do i=1 to 10; data mylib.data&i; set mylib.data&i; tx=0; array proc[15] proc1-proc15; do j=1 to 15; if proc(j) in (8540, 8541, 8542) then tx=1; end; run; %end; %mend; %create_var(); run; The red part is my array in regular data step. Anyone can teach me how to modify it to make the macro program work? Thanks a lot.

lizzy28 · ‎02-10-2015

Thanks, Ballard. I see your point.

lizzy28 · ‎02-10-2015

Hi all, I've asked a similar question previously at . But the answer does not seem to address my current case. I have 10 STATA files which I need to convert to SAS files. Their files names are like "us_ny_YEAR_core.dta"; here, "YEAR" stands for each specific year from 2001 to 2010. As I tried to do the following macro to convert the files, it does not work. %macro convert_data(); %do i=2001 %to 2010; proc import out=ny&i datafile="C:\data\us_ny_&i_core.dta";run; %end; %mend; %convert_data(); run; Anyone has any suggestion? Thanks so much! Lizi

lizzy28 · ‎02-05-2015

Thanks, Sanjay. I have one more question: is it possible to change the font of x-axis(Desks, Chairs, Lamps) /y-axis (0, ...,2000) labels?

lizzy28 · ‎02-05-2015

Thanks, Dirisks! I tried including ERRORBARATTRS = (color = red) at the end of the SCATTERPLOT statement. It looks there is no error message, but I do not see the bar color change.

lizzy28 · ‎02-05-2015

Sorry, I'll keep that in my mind. I have SAS 9.3. I also just added some sample data in my question. Thanks a lot.

lizzy28 · ‎02-05-2015

Hi all, This is the very first time I am using SAS to do a customized chart. I have SAS 9.3 I tried to draw confidence intervals in gplot and find the thread So I bascially followed Sanjay's codes like. proc template; define statgraph errorbars; begingraph; entrytitle 'Product Revenue'; layout overlay / yaxisopts=(offsetmin=0) xaxisopts=(display=(ticks tickvalues line)); barchart x=product y=revenue; scatterplot x=product y=revenue / yerrorupper=upper yerrorlower=lower markerattrs=(size=0); endlayout; endgraph; end; run; But, I cannot figure out how to change the color of the bars. Can anyone tell me how to do that? Here is my data: days Mean Upper Lower 0 day 34.36 31.97 36.76 1-3 days 37.75 33.87 41.64 4-7 days 42.69 38.43 46.96 >7 days 48.60 43.07 54.12 "days" is my x-axis, and "Mean" is the variable I want to plot for a bar chart, with 95% CI of upper and lower values. Thanks so much! Lizi

lizzy28 · ‎01-28-2015

Thanks a lot, Ballard!

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