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souji
Obsidian | Level 7

Hi,

Can you please help me to split single column to multiple column, here below city1 has full address with city,state and zip code. I need to split it city1 column into city , state, Zip1, zip2 columns

 

city1

BRONX NY  10467-6021

LAS VEGAS NV  89119

BATTLE MOUNTAIN NV  89820-1998

NORTH LAS VEGAS NV  89031-3432

 

Thanks

Soujanya

1 ACCEPTED SOLUTION

Accepted Solutions
Jagadishkatam
Amethyst | Level 16
Could you please mark the response as answered if you agree you got the correct answer. This helps others searching for a similar issue.
Thanks,
Jag

View solution in original post

10 REPLIES 10
PaigeMiller
Diamond | Level 26

Can we assume that your data always has a 2 digit state? Or can sometimes NY be spelled as N.Y. or New York? Are the locations always in the United States, or can the data contain locations in Canada or Uruguay?


Will there always be a five or 9 digit zip code? Does the 9 digit zip code always have a hyphen? 

--
Paige Miller
souji
Obsidian | Level 7

Thanks for quick response ,

State always has a 2 digit state and in the United States only

zip code has 9 digits with hyphen ( Ex: BRONX NY  10467-6021 , BATTLE MOUNTAIN NV  89820-1998)

And some rows has 5 digit zip code only ( Ex: LAS VEGAS NV  89119), no hyphen and pin code

PaigeMiller
Diamond | Level 26

Okay, here's a start. You can determine where the two character state value ends and the zip code begins. Example:

 

data a;
	string='BATTLE MOUNTAIN NV  89820-1998';
	str1=anyalpha(string,-30);
run;

the -30 indicates you go right to left, starting at position 30 (this is the max length of the string), and you get STR1 = 18, meaning NV ends in position 18 of the string. ANything to the right of position 18 is the zip. Position 17-18 is the state, and anything in positions 1 through 16 is the city.

--
Paige Miller
Jagadishkatam
Amethyst | Level 16

Hi Soujanya,

 

Please let me know if this helps.

 

data have;
input city1 :&$200.;
cards;
BRONX NY 10467-6021
LAS VEGAS NV 89119
BATTLE MOUNTAIN NV 89820-1998
NORTH LAS VEGAS NV 89031-3432
;


data want;
set have;
city=prxchange('s/(.*)(\s\w{2,2})(\s\d.*)/$1/oi',-1,city1);
state=prxchange('s/(.*)(\s\w{2,2})(\s\d.*)/$2/oi',-1,city1);
zip1=scan(prxchange('s/(.*)(\s\w{2,2})(\s\d.*)/$3/oi',-1,city1),1,'-');
zip2=scan(prxchange('s/(.*)(\s\w{2,2})(\s\d.*)/$3/oi',-1,city1),2,'-');
run;
Thanks,
Jag
souji
Obsidian | Level 7

Thank You Jagadish, Can you please explain or please refer any document that explain in detail

it was worked and trying to learn

 

soujanya

Jagadishkatam
Amethyst | Level 16

You may refer to my paper on perl regular expressions

 

https://www.phusewiki.org/docs/2019%20Amsterdam/Papers_presentations/CT/CT%20Final%20Papers/CT12.pdf 

Thanks,
Jag
Jagadishkatam
Amethyst | Level 16
Could you please mark the response as answered if you agree you got the correct answer. This helps others searching for a similar issue.
Thanks,
Jag
PaigeMiller
Diamond | Level 26

Here's my full code

 

data have;
input city1 :&$200.;
cards;
BRONX NY 10467-6021
LAS VEGAS NV 89119
BATTLE MOUNTAIN NV 89820-1998
NORTH LAS VEGAS NV 89031-3432
;
data want;
	set have;
	length zip $ 10 city $ 40 state $ 2;
	pos=anyalpha(city1,-200);
	zip=substr(city1,pos+1);
	state=substr(city1,pos-1,2);
	city=substr(city1,1,pos-2);
	drop pos;
run;
--
Paige Miller
souji
Obsidian | Level 7
Thank You
souji
Obsidian | Level 7

Hi Miller,

Can you please make me understand below, why you add pos+1  for zip variable(+1) and pos-1 for state (-1),pos-2 for city(-2)

pos=anyalpha(city1,-200);

zip=substr(city1,pos+1);

state=substr(city1,pos-1,2);
city=substr(city1,1,pos-2);

 

Thank you very much

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