"There's a way to do it better - find it." - Thomas A. Edison

When it comes to SAS coding, this quote by Thomas A. Edison is my best advisor. Time permitting, I love finding better ways of implementing SAS code.

But what code feature means “better” – brevity, clarity or efficiency? It all depends on the purpose of your code. When code is to illustrate a coding concept or technique, clarity is a paramount. However, when processing large data volumes in near real-time, code efficiency becomes critical, not just a luxury or convenience. And brevity won’t hurt in either case. Ideally, your code should be a combination of all three features - brevity, clarity and efficiency.

Parsing a character string

In this blog post we will solve a problem of parsing a character string to find a position of n-th occurrence of a group of characters (substring) in that string.

The closest out-of-the-box solution to this problem is SAS’ FIND() function. Except this function searches only for a single/first instance of specified substring of characters within a character string. Close enough, and with some do-looping we can easily construct what we want.

After some internet and soul searching to find the Nth occurrence of a substring within a string, I came up with the following DATA STEP code snippet:

   p = 0;
   do i=1 to n until(p=0); 
      p = find(s, x, p+1);
   end;

Here, s is a text string (character variable) to be parsed; x is a character variable holding a group of characters that we are searching for within s; p is a position of x value found within s; n is an instance number.

If there is no n-th instance of x within s found, then the code returns p=0.

In this code, each do-loop iteration searches for x within s starting from position p+1 where p is position found in prior iteration: p = find(s,x,p+1);.

Notice, if there is no prior-to n instance of x within s, the do-loop ends prematurely, based on until(p=0) condition, thus cutting the number of loops to the minimal necessary.

Reverse string search

Since find() function allows for a string search in a reverse direction (from right to left) by making the third augment negative, the above code snippet can be easily modified to do just that: find Nth instance (from right to left) of a group of characters within a string. Here is how you can do that:

   p = length(s) + 1;
   do i=1 to n until(p=0); 
      p = find(s, x, -p+1);
   end;

The difference here is that we start from position length(s)+1 instead of 0, and each iteration searches substring x within string s starting from position –(p-1)=-p+1 from right to left.

Testing SAS code

You can run the following SAS code to test and see how these searches work:

data a;
   s='AB bhdf +BA s Ab fs ABC Nfm AB ';
   x='AB';
   n=3;
 
   /* from left to right */
   p = 0;
   do i=1 to n until(p=0); 
      p = find(s, x, p+1);
   end;
   put p=;
 
   /* from right to left */
   p = length(s) + 1;
   do i=1 to n until(p=0); 
      p = find(s, x, -p+1);
   end;
   put p=;
run;

FINDNTH() function

We can also combine the above left-to-right and right-to-left searches into a single user-defined SAS function by means of SAS Function Compiler (PROC FCMP) procedure:

proc fcmp outlib=sasuser.functions.findnth;
   function findnth(str $, sub $, n);
      p = ifn(n>=0,0,length(str)+1);
      do i=1 to abs(n) until(p=0);
         p = find(str,sub,sign(n)*p+1);
      end;
      return (p);
   endsub;
run;

We conveniently named it findnth() to match the Tableau FINDNTH(string, substring, occurrence) function that returns the position of the nth occurrence of substring within the specified string, where the occurrence argument defines n.

Except our findnth() function allows for both, positive (for left-to-right searches) as well as negative (for right-to-left searches) third argument while Tableau’s function only allows for left-to-right searches.

Here is an example of the findnth() function usage:

options cmplib=sasuser.functions;
data a;
   s='AB bhdf +BA s Ab fs ABC Nfm AB ';
   x='AB';
   n=3;
 
   /* from left to right */
   p=findnth(s,x,n);
   put p=;
 
   /* from right to left */
   p=findnth(s,x,-n);
   put p=;
run;

Using Perl regular expression

As an alternative solution I also implemented SAS code for finding n-th occurrence of a substring within a string using Perl regular expression (regex or prx):

data a;
   s='AB bhdf +BA s Ab fs ABC Nfm AB ';
   x='AB';
   n=3;
 
   /* using regex */
   xid = prxparse('/'||x||'/o');
   p = 0;
   do i=1 to n until(p=0);
      from = p + 1;
      call prxnext(xid, p + 1, length(s), s, p, len);
   end;
   put p=;
run;

However, efficiency benchmarking tests demonstrated that the above solutions using FIND() function or FINDNTH() SAS user-written function run roughly twice faster than this regex solution.

Overlapping and performance considerations

The above solutions with FIND() and FINDNTH() functions assume that the searched substring can overlap with its prior instance. For example, if we search for a substring ‘AAA’ within a string ‘ABAAAA’, then the first instance of the ‘AAA’ will be found in position 3, and the second instance – in position 4. That is, the first and second instances are overlapping. For that reason, when we find an instance we increment position p by 1 (p+1) to start the next iteration (instance) of the search.
However, if such overlapping is not a valid case in your searches, and you want to continue search after the end of the previous substring instance, then we should increment p not by 1, but by length of the substring x. That will speed up our search (the more the longer our substring x is) as we will be skipping more characters as we go through the string s. In this case, in our search code we should replace p+1 to p+w, where w=length(x), and also replace p=0 to p=1-w for left-to-right search initial p value:

data a; 
   s='AB bhdf +BA s Ab fs ABC Nfm AB '; 
   x='AB'; 
   n=3; 
   w=length(x); 
 
   /* from left to right */ 
   p = 1 - w; 
   do i=1 to n until(p=0);  
      p = find(s, x, p+w); 
   end; 
   put p=;
 
   /* from right to left */ 
   p = length(s) + 1; 
   do i=1 to n until(p=0);  
      p = find(s, x, -p+w); 
   end; 
   put p=; 
run;

Just make sure, you calculate your w=length(x) before (not within) the do-loop to cut out recalculation of the same over and over.

Challenge

Can you come up with an even better solution to the problem of finding Nth instance of a sub-string within a string? Please share your thoughts and solutions with us. Thomas A. Edison would have been proud of you!