About Astounding

Astounding · ‎04-09-2012

mei, You are likely to need variable names to hold each new piece of information. Try this: data want; set have nobs=_nobs_; back_1 = lag1(stock_return); back_2 = lag2(stock_return); back_3 = lag3(stock_return); back_4 = lag4(stock_return); if _n_ < _nobs_ then set have (keep=stock_return rename=(stock_return=forward_1) firstobs=2); if _n_ < _nobs_-1 then set have (keep=stock_return rename=(stock_return=forward_2) firstobs=3); run; Good luck.

Astounding · ‎04-04-2012

You can also simulate the result, using %EVAL. This expression also returns a 1: %eval ( (4+5) ne 0) However, %SYSEVALF will also return 0 for calculated missing values. %EVAL returns an error. Also note that %EVAL is still limited to integer arithmetic. These expressions return different results: %sysevalf(4/5, boolean) %eval( (4/5) ne 0) Hope this helps, rather than confuses the issue!

Astounding · ‎04-04-2012

Hai.kuo, Yes, if you are willing to rely on removing all digits. You wouldn't want to encounter: 25.YOUR NAME IS 'R2D2.'

Astounding · ‎04-04-2012

No, no comma. The idea is that we'll specify just 2 arguments to SUBSTR. When the third argument is omitted, SUBSTR will take all remaining characters. So we really need to compute the second argument, telling SUBSTR where to begin.

Astounding · ‎04-04-2012

K, See my revised approach, for a simpler, better way. But note that the original approach should have generated the equivalent of: text = substr(text, length("2") +2) Good luck.

Astounding · ‎04-04-2012

It's a little messy, but it can be crammed into one statement: text = substr(text, length(scan(text, 1, '.'))+2); This assumes that every observation needs to remove something from the beginning of TEXT, so that IF/THEN is not required. Good luck. Edited: Actually, this is a little safer (as well as simpler) in case there is are spaces between the number and the dot: text = substr(text, index(text, '.') +1);

Astounding · ‎04-04-2012

Sister, If you're willing to go with the DATA step, here's a slight improvement: data have (rename=(&rename)); set have (firstobs=2); run; Good luck.

Astounding · ‎04-04-2012

Linlin, I think the simplest approach would be to create 2 data sets instead of 1 originally: data have1 have2; input ...; if _n_=1 then output have1; else output have2; cards;

Astounding · ‎04-04-2012

hdg, It is possible that Hai.kuo's solution worked. He is typically quite thoughtful and correct. My brain is fried from trying to follow it, though. Here's another approach. First, construct a shell data set, containing the code / dates you are hoping to get. Sorting is done in anticipation of merging later: proc sort data=have; by code date; run; proc summary data=have; by code; var date; output out=shell (drop=_type_ _freq_) min=min_date max=max_date; run; data shell; set shell; by code; do date=min_date to max_date; if 1 < weekday(date) < 7 then output; end; keep code date; run; Then merge the shell back with the data: data want; merge have (rename=(info=original_info)) shell; by code date; if original_info > . then info = original_info; retain info; * After checking the result: drop original_info; run; I'm sure somebody will look at my approach and find it just as unusual and brain-frying. But that's how I would do it. Good luck.

Astounding · ‎04-03-2012

Greg, Happy to help. As long as you are reading up on the details, here are a couple of items to consider. They may not be needed in this particular application. First, the length of Y will be 156. When SUBSTR creates a new variable, the new variable's length is the same as that of the incoming string. Second, SUBSTR can appear on the left-hand side of an assignment statement. This would be legal: substr(x, _n_, 2) = Y; Or simply eliminate the creation of Y: substr(x, _n_, 2) = substr(value, _n_, 2); This would become important if it were possible for VALUE to contain embedded blanks. (I know that won't happen here where you have hex codes.) In that case, the CATS function would remove them. Good luck.

Astounding · ‎04-03-2012

Greg, The culprit is the RETAIN statement. Just get rid of it, and X will start out blank for each new observation. Once you do that, you should also remove the 4 statements you added, to try to reset X. Also, a small short-cut ... Replace this: put "_n_ = " _n_; Use this instead: put _n_=;

Astounding · ‎04-03-2012

Linlin, In that case, I vote for option #2. My main consideration is that the definition of "efficiency" should include more than CPU time. It should include maintenance time. If you ever need to revise the code, or even review it 6 months later, it will take a little of your time to compare the two SELECT statements to verify that they retrieve the same records. That maintenance time is more expensive than the extra CPU time it might take to run one variation or the other. In fact, I would even consider making the program run a little longer by retrieving: trim(memname) That will make long values for &NAMES easier to read if you ever need to examine it down the road. And if you switch to: distinct(trim(memname)) it will take even longer, but will alphabetize the list of MEMNAMEs. That would make it easier for you to find a particular data set name if you are visually searching for it within &NAMES. Good luck.

Astounding · ‎04-03-2012

Linlin, If your intent here is to capture &SQLOBS, this alternative might do it: proc sql; select count(memname) into : n from dictionary.tables where libname='WORK'; quit; But if you need the list of names as well, this variation won't do that. Is efficiency really a factor when reading from DICTIONARY.TABLES? Good luck.

Astounding · ‎04-03-2012

Greg, In your original %LET statement, you added quotes. But in Linlin's example, his %LET statement omitted the quotes. So the DATA step added them later (using double quotes instead of single, because single quotes would prevent resolution of &VAR.).

Astounding · ‎04-03-2012

Greg, You've got a lot of the right pieces in place. A few tweaks should do it. First, assign a length to X. In place of x='';, use: length x $ 26; At least 26 looks like the right length based on your description. Then inside the loop, change the assignment to: x = trim(x) || substr(&var., i, 2); That should be all it takes. Good luck. Ooppss ... upon rereading the problem, You might need to add to the DO loop: BY 2; And the length of 14 might be sufficient.

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