I wish to extract the two words "blood" and "loss" within the closest proximity to a substring. I have the codes below, but the ID 4 didn't work. I wish to get the substring "blood loss", not "bloods but blood loss".
data test;
infile datalines truncover;
input id $2. string $80.;
datalines;
1 there is one blood something loss
2 no something else here
3 three blood loss again blood loss can not believe loss of blood
4 two bloods but blood loss
;
run;
data test1;
set test;
rx=prxparse("/blood.*?loss|loss.*?blood/i");
start=1;
stop =length(trim(string));
do until (p=0);
call prxnext(rx,start,stop,trim(string),p,l);
if p>0 then do;
sub=substr(string,p,l);
output;
end;
end;
run;
Very small change is needed, if bloods need to be ignored. Add a space between the blood and . in the first part of regex and it will try and match the word blood. Below is the replacement prxparse statement.
rx=prxparse("/blood .*?loss|loss.*?blood/i");
As per the updated comment, for matching string "blood loss" when string is like "blood something blood loss" then negative lookahead can help.
prxparse("/blood (.(?!blood))*?loss/i")
Related
I have a number of text entries (municipalities) from which I need to remove the s at the end.
Data test;
input city $;
datalines;
arjepogs
askers
Londons
;
run;
data cities;
set test;
if prxmatch("/^(.*?)s$/",city)
then city=prxchange("s/^(.*?)s$/$1/",-1,city);
run;
Strangely enough, my s's are only removed from my first entry.
What am I doing wrong?
You defined CITY as length $8. The s in Londons is in the 7th position of the string. Not the LAST position of the string. Use the TRIM() function to remove the trailing spaces from the value of the variable.
data have;
input city $20.;
datalines;
arjepogs
Kent
askers
Londons
;
data want;
set have;
length new_city $20 ;
new_city=prxchange("s/^(.*?)s$/$1/",-1,trim(city));
run;
Result
Obs city new_city
1 arjepogs arjepog
2 Kent Kent
3 askers asker
4 Londons London
You could also just change the REGEX to account for the trailing spaces.
new_city=prxchange("s/^(.*?)s\ *$/$1/",-1,city);
Here is another solution using only SAS string functions and no regex. Note that in this case there is no need to trim the variable:
data cities;
set test;
if substr(city,length(city)) eq "s" then
city=substr(city,1,length(city)-1);
run;
I have some data in the form of a column in a dataset (named Person_details), where each has an unknown number of names, with the name (split up by spaces), followed by an underscore, followed by that persons identifier (7 characters).
Is there a way to split these entries up automatically, rather than repeatedly finding the position of the underscore, and then taking the substring before and after?
Person_details:
Evan Davies_123F323 Adam John Smith_342D427 Karl Marx_903C943
There are an unknown number of names in each cell, e.g. some have just one name and some have 20. Also complicated by the fact that some entries have middle name(s).
The ideal output would be in the form
Name Code
Evan Davies 123F323
Adam John Smith 342D427
Karl Marx 903C943
You could just use SCAN() instead.
data have;
string='Evan Davies_123F323 Adam Smith_342D427 Karl Marx_903C943';
length name $50 code $7 ;
do index=1 to countw(string,' ');
name = catx(' ',name,scan(string,index,' '));
if index(name,'_') then do;
code = scan(name,-1,'_');
name = substr(name,1,length(name)-length(code)-1);
output;
name=' ';
end;
end;
run;
Result
You can use a Perl regular expression (regex) to detect and extract pieces from patterned text. SAS routine PRXNEXT iterates through matches, and function PRXPOSN extracts pieces.
Example:
data have;
text = 'Evan Davies_123F323 Adam John Smith_342D427 Karl Marx_903C943';
run;
data want(keep=name code);
rx = prxparse('/(.+?)_(.{7}( |$))/');
set have;
start = 1;
stop = length(text);
do seq = 1 by 1;
call prxnext(rx,start,stop,text,position,length);
if position=0 then leave;
name = prxposn(rx,1,text);
code = prxposn(rx,2,text);
output;
end;
run;
I have a SAS string that always starts with a date. I want to remove the date from the substring.
Example of data is below (data does not have bullets, included bullets to increase readability)
10/01/2016|test_num15
11/15/2016|recom_1_test1
03/04/2017|test_0_8_i0|vacc_previous0
I want the data to look like this (data does not have bullets, included bullets to increase readability)
test_num15
recom_1_test1
test_0_8_i0|vacc_previous0
Index find '|' position in the string, then substr substring; or use regular expression.
data have;
input x $50.;
x1=substr(x,index(x,'|')+1);
x2=prxchange('s/([^_]+\|)(?=\w+)//',1,x);
cards;
10/01/2016|test_num15
11/15/2016|recom_1_test1
03/04/2017|test_0_8_i0|vacc_previous0
;
run;
This is a great use case for call scan. If your length of date is constant (always 10), then you don't actually need this (start would be 12 then and skip to the substr, as user667489 noted in comments), but if it's not this would be helpful.
data have;
length textstr $100;
input textstr $;
datalines;
10/01/2016|test_num15
11/15/2016|recom_1_test1
03/04/2017|test_0_8_i0|vacc_previous0
;;;;
run;
data want;
set have;
call scan(textstr,2,start,length,'|');
new_textstr = substr(textstr,start);
run;
It would also let you grab the second word only if that's useful (using length third argument for substr).
Is it possible to use the number in this string:
'xx8xx'
by replacing the number with 8 spaces to get this string:
'xx xx'
I can identify the number between the xx but the replacement syntax does not work as intended:
PRXCHANGE(s/xx([\d]*)xx/' ' x $1/io, -1, 'xx8xx')
Is there a way to use the number being held in $1 to repeat the space character by that number i.e. something like ' ' x $1?
Any help much appreciated!
Tiaan
Supposed you need to replace with three blank.
data _null_;
x=prxchange('s/(xx)\d+(xx)/$1 $2/', -1, 'xx8xx');
_x=prxchange('s/(?=\w+)(\d+)/ /',1,'xx8xx');
put _all_;
run;
Edit:
I missed important information. Tranwrd and repeat could be used to get it.
data _null_;
x=tranwrd('xx8xx', prxchange('s/.*(\d+).*/$1/',1,'xx8xx'), repeat(' ',prxchange('s/.*(\d+).*/$1/',1,'xx8xx')));
put _all_;
run;
You'll need to extract first, then compile a new regex. This will be expensive since you have to compile once per line.
data have;
input xstr $;
datalines;
xx8xx
xx3xx
xx4xx
;;;;
run;
data want;
set have;
rx1 = prxparse('/xx([\d])*xx/io');
rc1 = prxmatch(Rx1,xstr);
num_x = prxposn(rx1,1,xstr);
rx2 = prxparse(cat('s/(xx)[\d]*(xx)/$1',repeat(" ",num_x-1),'$2/i'));
newstr = prxchange(rx2,-1,xstr);
run;
I came across this question today morning and I am still trying to figure out it can be done. the following dataset is present and has a character variable CAT.
CAT
A
AB
B
ABCD
CB
.
.
.
and so on.
We need to write a SAS program to introduce commas in-between each character of the string if the length of the string is more than 1. I used length() function and used a do loop to create different variables and it just got messy. How do i tackle this?
Regular expression solution:
data have;
input CAT $;
datalines;
A
AB
B
ABCD
CB
;;;;
run;
data want;
set have;
cat_c = prxchange('s/(?<=[[:alpha:]])([[:alpha:]])/,$1/io',-1,CAT);
put cat_c=;
run;
The first parenthetical group is a look-behind for an alpha character; then the captured alpha character. Then replace with comma and character. If you want something other than [[:alpha:]] (ie, A-Z) then supply that as a class.
The solution using length and do loop isn't bad, honestly, if you want something that is more readable to novice programmers. Just use SUBSTR left of the equal sign.
data want2;
set have;
if length(cat) > 1 then
do _t = 1 to length(cat)-1;
substr(cat_c,2*_t-1,2)=substr(cat,_t,1)||',';
end;
substr(cat_c,2*length(cat)-1,1)=substr(cat,length(cat),1);
put cat_c=;
run;