Try the E8601DT21. informat. ... View more

Is this  &i*-1 Just a convoluted way to write -&i ?   ... View more

Alternatively with perl regular expression   data want; set have; name_nbr=prxchange('s/(.*\~)(.*)/$2/',-1,name); name=prxchange('s/(.*)(\~.*)/$1/',-1,name); run; ... View more

Any reason you didn't use arrays instead of a macro here? ... View more

Hi @harrylui    Does the following code could meet your expectations? data have; input test1 test2; datalines; 100 90 99 90 98 90 97 90 ; run; data want; set have; if _n_=1 then net=test1-test2; retain net; test1 = 100; test2 = 90; run; ... View more

A sum of elements divided by the number of elements is a MEAN. So this code: sum_checking=sum(checking1,checking2,checking3,checking4,checking5,checking6,checking7); count_word=countw(CLEANED_BY_WORD); percent=sum_checking/count_word; Assuming that "cleaned_by_word" matches the number of checking values then perhaps could be replaced with percent_sum = mean(checking1, checking2, checking3, checking4,checking5, checking6, checking7); I am a bit leery of this code: if percent=0.8 then modifed_name=clean_by_word2; if percent=0.75 because with 7 elements I would not expect to get 0.8 of 0.75 for any division by 7.  7* 0.8 = 5.6, 7*0.75 =5.25. There are similar issues with most of the possible counts. I suspect that you maybe should be checking a range of values for percent but without any actual data and desired outcome this is a bit difficult to verify. ... View more