Transwrd multiple words in a string

[vickys]

I have a long string like this

"fruits: mangovegetables: carrotdrinks: cokeseafood: fishrestaurant: jasmineplace: London"

But I am trying to add "." (dot) so that I can differentiate which category they are and the expected string is as below

"fruits: mango.vegetables: carrot.drinks: coke.seafood: fish.restaurant: jasmine.place: London"

I tried like something

data want;

set have;

_list = o_list;

length word $200 ;

do

word='vegetables:', 'drinks:', 'seafood:', 'restaurant:', 'place:' ;

n_list = tranwrd(n_list, strip(word), .word);

end;

n_list = compbl(n_list);

drop word ;

run;

but they are doing nothing to the string. same string with out transwrd.

Appreciate your help Guru's

Thanks in advance

[ed_sas_member]

Hi @vickys 

Please try this:

data have;
	length n_list $ 200;
	n_list="fruits: mangovegetables: carrotdrinks: cokeseafood: fishrestaurant: jasmineplace: London";
run;

data want;
	set have;
	array word (5) $12 _temporary_ ('vegetables:', 'drinks:', 'seafood:', 'restaurant:', 'place:');
	do i=1 to dim(word);
		n_list = prxchange(cats('s/(.*)(',word(i),')(.*)/$1.$2$3/'),1,n_list);
	end;
	drop i;
run;

Best,

[ballardw]

One thing that you want to consider is that you are adding characters to a string. So the result should be longer that what you start with. Depending on the defined length of a variable you may get truncation because the longer result does not fit into the original variable.

So you either need to create a new variable or do something to address the length issue. Here's an example:

data example;
   n_list="fruits: mangovegetables: carrotdrinks: cokeseafood: fishrestaurant: jasmineplace: London";
   do word='vegetables:', 'drinks:', 'seafood:', 'restaurant:', 'place:' ;
      n_list = tranwrd(n_list, strip(word), cats('.',strip(word)));
      put n_list=;
   end;
run;

Look in the log and you will see the result start losing characters at the end.

Since you have 5 words that you are replacing with an added character you need to have a result variable that is at least 5 characters longer than your source variable or define the variable to be longer.

data example;
   length n_list $ 100;
   n_list="fruits: mangovegetables: carrotdrinks: cokeseafood: fishrestaurant: jasmineplace: London";
   do word='vegetables:', 'drinks:', 'seafood:', 'restaurant:', 'place:' ;
      n_list = tranwrd(n_list, strip(word), cats('.',strip(word)));
      put n_list=;
   end;
run;

Also you should always read your log. Your program had an error because of the way you called tranwrd:

28      n_list = tranwrd(n_list, strip(word), .word);
                                                ---
                                                22
                 -------
                 72
ERROR 22-322: Syntax error, expecting one of the following: !, !!, &, *, **, +, ',', -, /, <, <=,
              <>, =, >, ><, >=, AND, EQ, GE, GT, IN, LE, LT, MAX, MIN, NE, NG, NL, NOTIN, OR, ^=,
              |, ||, ~=.

ERROR 72-185: The TRANWRD function call has too many arguments.

If you want to add a character you need to use a function or operator to do so. Using .word is neither.  The use of dot plus a letter or an underscore references a special missing value in SAS. So the .w is treated as a numeric missing generating the "Numeric values have been converted …" warning. Then the characters following the ORD of word are treated as another variable which is undefined so also treated as numeric. Both of those mean too many arguments for Tranwrd.