I have the following code which works well for a needle plot to which I added a Loess to connect certain points. The Loess worked for 4 of the variables but I get an error message stating, "
The data is below and I have looked at it and can't figure out why SAS has an issue with the PAUC03L and PAUC37L data.
They both decrease as they should.
title 'Confidence Interval Range Kfast';
proc sgplot data=boot2 ;
/*highlow x=ratio low=low high=high / type=line;*/
loess x=ratio y=LOWER_ci/interpolation=cubic group=block;
highlow x=ratio low=lower_ci high=upper_ci / type=bar group=block
lineattrs=(color=black) name='a' attrid=Mono;
yaxis label='CI range' grid;
xaxis label='Ratio KaT/KaR' grid values =(0.5 to 2.1 by 0.1);
keylegend 'a' / location=inside position=TOP;
run;
Can some one tell me what is wrong with the boot2 data for PAUC03L and PAUC37L for the Lower_CI?
Group | Block | Upper_CI | Lower_CI | Ratio |
1 | PAUC03L | 84 | 80 | 0.62 |
1 | PAUC03L | 85 | 81 | 0.63 |
1 | PAUC03L | 88 | 85 | 0.7 |
1 | PAUC03L | 100 | 99 | 0.8 |
1 | PAUC03U | 108 | 106 | 1.25 |
1 | PAUC03U | 109 | 106 | 1.3 |
1 | PAUC03U | 113 | 109 | 1.5 |
1 | PAUC03U | 122 | 115 | 2 |
1 | PAUC03U | 125 | 116 | 2.1 |
1 | PAUC37L | 104 | 102 | 0.62 |
1 | PAUC37L | 104 | 101 | 0.63 |
1 | PAUC37L | 104 | 100 | 0.7 |
1 | PAUC37L | 90 | 88 | 0.8 |
1 | PAUC37U | 98 | 96 | 1.25 |
1 | PAUC37U | 96 | 95 | 1.3 |
1 | PAUC37U | 96 | 93 | 1.5 |
1 | PAUC37U | 94 | 89 | 2 |
1 | PAUC37U | 94 | 89 | 2.1 |
I have to admit, I am totally mystified by your comments. Your original code does not use the DEGREE= option, it does use the INTERPOLATION= option, which is what I responded to. And then you ignore the INTERPOLATION option in your reply.
Your original message talks about the lower confidence interval, even though the error messages don't talk about lower confidence intervals at all.
But anyway, the problem, as I implied, is that you have a very small amount of data to do some of this smoothing. Playing with the options won't overcome the fact that you have a very small amount of data.
Just a guess, but maybe you can't do cubic interpolation with just 4 data points.
I have to admit, I am totally mystified by your comments. Your original code does not use the DEGREE= option, it does use the INTERPOLATION= option, which is what I responded to. And then you ignore the INTERPOLATION option in your reply.
Your original message talks about the lower confidence interval, even though the error messages don't talk about lower confidence intervals at all.
But anyway, the problem, as I implied, is that you have a very small amount of data to do some of this smoothing. Playing with the options won't overcome the fact that you have a very small amount of data.
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