Hello SAS community,
I came across those strange issues as I was testing mnemonics used to assess the outcome of the MODIFY statement within a DATA step in SAS Studio. Below is my code used to create two identical master datasets and one transaction dataset:
data master1 master2;
input a b;
datalines;
0 1
2 3
4 5
6 7
8 9
;
run;
data transaction;
input a b;
datalines;
0 9
1 8
1 7
1 6
2 5
3 4
3 3
3 2
4 1
;
run;
Starting with master data1, I intended to update the values in this dataset with matching records as well as append the 1st non-matching records from the transaction dataset:
data master1;
modify master1 transaction;
by a;
if _iorc_=%sysrc(_sok) then replace;
else if _iorc_=%sysrc(_dsenmr) then do;
put "First non-matching observation found!";
output;
_error_=0;
end;
else if _iorc_=%sysrc(_dsemtr) then do;
put "Subsequent non-matching observations found!";
_error_=0;
end;
run;
proc print data=master1;
run;
However, here is the message I get from the SAS log when I ran the code above:
1 OPTIONS NONOTES NOSTIMER NOSOURCE NOSYNTAXCHECK;
55
56 data master1;
57 modify master1 transaction;
58 by a;
59 if _iorc_=%sysrc(_sok) then replace;
60 else if _iorc_=%sysrc(_dsenmr) then do;
61 put "First non-matching observation found!";
62 output;
63 _error_=0;
64 end;
65 else if _iorc_=%sysrc(_dsemtr) then do;
66 put "Subsequent non-matching observations found!";
67 _error_=0;
68 end;
69 run;
First non-matching observation found!
First non-matching observation found!
NOTE: There were 1 observations read from the data set WORK.MASTER1.
NOTE: The data set WORK.MASTER1 has been updated. There were 7 observations rewritten, 2 observations added and 0 observations
deleted.
NOTE: There were 9 observations read from the data set WORK.TRANSACTION.
NOTE: DATA statement used (Total process time):
real time 0.00 seconds
user cpu time 0.01 seconds
system cpu time 0.00 seconds
memory 1317.43k
OS Memory 28844.00k
Timestamp 10/20/2016 11:29:58 PM
Step Count 60 Switch Count 82
Page Faults 0
Page Reclaims 452
Page Swaps 0
Voluntary Context Switches 273
Involuntary Context Switches 0
Block Input Operations 0
Block Output Operations 280
70
71 proc print data=master1;
72 run;
NOTE: There were 7 observations read from the data set WORK.MASTER1.
NOTE: PROCEDURE PRINT used (Total process time):
real time 0.02 seconds
user cpu time 0.03 seconds
system cpu time 0.00 seconds
memory 1608.50k
OS Memory 28836.00k
Timestamp 10/20/2016 11:29:58 PM
Step Count 61 Switch Count 24
Page Faults 0
Page Reclaims 124
Page Swaps 0
Voluntary Context Switches 42
Involuntary Context Switches 0
Block Input Operations 0
Block Output Operations 8
73
74 OPTIONS NONOTES NOSTIMER NOSOURCE NOSYNTAXCHECK;
86
As you can see, only the first PUT statement displays its message to the log, but not the second one. Why is that? Isn't this weird?
Moreover, this isn't the only strange thing that happened. In fact, here is the output:
Obs a b
1 0 9
2 2 5
3 4 1
4 6 7
5 8 9
6 1 6
7 3 2
Since I chose to get the first non-matching record from the transaction dataset (i.e. the "_dsenmr" mnemonic code), aren't the last two lines from the output supposed to be "6 1 8" and "7 3 4"??
Now, with master data 2, I used the same approach except that I wanted to append subsequent non-matching records instead of the first one:
data master2;
modify master2 transaction;
by a;
if _iorc_=%sysrc(_sok) then replace;
else if _iorc_=%sysrc(_dsenmr) then do;
put "First non-matching observation found!";
_error_=0;
end;
else if _iorc_=%sysrc(_dsemtr) then do;
put "Subsequent non-matching observations found!";
output;
_error_=0;
end;
run;
proc print data=master2;
run;
After submitting the code above, I got the correct message in the log. However, the output still worries me:
Obs a b
1 0 9
2 2 5
3 4 1
4 6 7
5 8 9
6 1 6
7 3 2
This is also strange because the "_dsemtr" mnemonic code implies getting ALL non-matching observations other than the first one, which means that I'm missing rows where "a=1 and b=7" and "a=3 and b=3".
I'm really stuck on these issues and hopefully someone could help me out on this.
Thanks a bunch!!
I guess that in the end only the _dsenmr mnemonic operator and not _dsemtr needs to be used for this purpose, right? I'm saying this because this is a variation of your code where _dsemtr shows up first, but it does not work:
