How to pad out character fields in SAS? - sas

I am creating a SAS dataset from a database that includes a VARCHAR(5) key field.
This field includes some entries that use all 5 characters and some that use fewer.
When I import this data, I would prefer to pad all the shorter entries out to use all five characters. For this example, I want to pad on the left with 0, the character zero. So, 114 would become 00114, ABCD would become 0ABCD, and EA222 would stay as it is.
I've attempted this with a simple data statement, but of course the following does not work:
data test;
set databaseinput;
format key $5.;
run;
I've tried to do this with a user-defined informat, but I don't think it's possible to specify the ranges correctly on character fields, per this SAS KB answer. Plus, I'm fairly sure proc format won't let me define the result dynamically in terms of the incoming variable.
I'm sure there's an obvious solution here, but I'm just missing it.

Here is an alternative:
data padded_data_dsn; length key $5;
drop raw_data;
set raw_data_dsn(rename=(key=raw_data));
key = translate(right(raw_data),'0',' ');
run;

Data raw_data_dsn;
format key $5.;
key = '4'; key1 = CATT(REPEAT('0',5-length(key)),key);output;
key = 'A114'; key1 = CATT(REPEAT('0',5-length(key)),key);output;
key = 'A1140'; key1 = CATT(REPEAT('0',5-length(key)),key);output;
run;

I'm sure someone will have a more elegant solution, but the following code works. Essentially it is padding the variable with five leading zeros, then reversing the order of this text string so that the zeros are to the right, then reversing this text string again and limiting the size to five characters, in the original order but left-padded with zeros.
data raw_data_dsn;
format key $varying5.;
key = '114'; output;
key = 'ABCD'; output;
key = 'EA222'; output;
run;
data padded_data_dsn;
format key $5.;
drop raw_data;
set raw_data_dsn(rename=(key=raw_data));
key = put(put('00000' || raw_data ,$revers10.),$revers5.);
run;

Here's what worked for me.
data b (keep = str2);
format str2 $5. ;
set a;
catlength = 4 - length(str);
cat = repeat('0', catlength);
str2 = catt(cat, str);
run;
It works by counting the length of the existing string, and then creating a cat string of length 4 - that, and then appending the cat value and the original string together.
Notice that it screws up if the original string is length 5.
Also - it won't work if the input string has a $5. format on it.
data a; /*input dataset*/
input str $;
datalines;
a
aa
aaa
aaaa
aaaaa
;
run;
data b (keep = str2);
format str2 $5. ;
set a;
catlength = 4 - length(str);
cat = repeat('0', catlength);
str2 = catt(cat, str);
run;
input:
a
aa
aaa
aaaa
aaaaa
output:
0000a
000aa
00aaa
0aaaa
0aaaa

I use this, but only works with numeric values :S. Try with another formats in the INPUT
data work.prueba;
format xx $5.;
xx='1234';
vv=PUT(INPUT(xx,best5.),z5.);
run;

Related

SAS Retain not working for 1 string variable

The below code doesn't seem to be working for the variable all_s when there is more than 1 record with the same urn. Var1,2,3 work fine but that one doesn't and I cant figure out why. I am trying to have all_s equal to single_var1,2,3 concatenated with no spaces if it's first.urn but I want it to be
all_s = all_s + ',' + single_var1 + single_var2 + single_var3
when it's not the first instance of that urn.
data dataset_2;
set dataset_1;
by URN;
retain count var1 var2 var3 all_s;
format var1 $40. var2 $40. var3 $40. all_s $50.;
if first.urn then do;
count=0;
var1 = ' ';
var2 = ' ';
var3 = ' ';
all_s = ' ';
end;
var1 = catx(',',var1,single_var1);
var2 = catx(',',var2,single_var2);
var3 = catx(',',var3,single_var3);
all_s = cat(all_s,',',single_var1,single_var2,single_var3);
count = count+1;
if first.urn then do;
all_s = cat(single_var1,single_var2,single_var3);
end;
run;
all_s is not large enough to contain the concatenation if the total length of the var1-var3 values within the group exceeds $50. Such a scenario seems likely with var1-var3 being $40.
I recommend using the length function to specify variable lengths. format will create a variable of a certain length as a side effect.
catx removes blank arguments from the concatenation, so if you want spaces in the concatenation when you have blank single_varN you won't be able to use catx
A requirement that specifies a concatenation such that non-blank values are stripped and blank values are a single blank will likely have to fall back to the old school trim(left(… approach
Sample code
data have;
length group 8 v1-v3 $5;
input group (v1-v3) (&);
datalines;
1 111 222 333
1 . 444 555
1 . . 666
1 . . .
1 777 888 999
2 . . .
2 . b c
2 x . z
run;
data want(keep=group vlist: all_list);
length group 8 vlist1-vlist3 $40 all_list $50;
length comma1-comma3 comma $2;
do until (last.group);
set have;
by group;
vlist1 = trim(vlist1)||trim(comma1)||trim(left(v1));
vlist2 = trim(vlist2)||trim(comma2)||trim(left(v2));
vlist3 = trim(vlist3)||trim(comma3)||trim(left(v3));
comma1 = ifc(missing(v1), ' ,', ',');
comma2 = ifc(missing(v2), ' ,', ',');
comma3 = ifc(missing(v3), ' ,', ',');
all_list =
trim(all_list)
|| trim(comma)
|| trim(left(v1))
|| ','
|| trim(left(v2))
|| ','
|| trim(left(v3))
;
comma = ifc(missing(v3),' ,',',');
end;
run;
Reference
SAS has operators and multiple functions for string concatenation
|| concatenate
cat concatenate
catt concatenate, trimming (remove trailing spaces) of each argument
cats concatenate, stripping (remove leading and trailing spaces) of each argument
catx concatenate, stripping each argument and delimiting
catq concatenate with delimiter and quote arguments containing the delimiter
From SAS 9.2 documentation
Comparisons
The results of the CAT, CATS, CATT, and CATX functions are usually equivalent to results that are produced by certain combinations of the concatenation operator (||) and the TRIM and LEFT functions. However, the default length for the CAT, CATS, CATT, and CATX functions is different from the length that is obtained when you use the concatenation operator. For more information, see Length of Returned Variable.
Note: In the case of variables that have missing values, the concatenation produces different results. See Concatenating Strings That Have Missing Values.
Some example data would be helpful, but I'm going to give it a shot and ask you to try
all_s = cat(strip(All_s),',',single_var1,single_var2,single_var3);

Get string between two specific char positions

i have a long text string in SAS, and a value is within it of variable length but is always proceeded by a '#' and then ends with ' ,'
is there a way i can extract this and store as a new variable please?
e.g:
word word, word, #12.34, word, word
And i want to get the 12.34
Thanks!
Double scan should also work if you only have a single #:
data _null_;
var1 = 'word word, word, #12.34, word, word';
var2 = scan(scan(var1,2,'#'),1,',');
put var2=;
run;
You can make use of the substr and index functions to do this. The index function returns the first position of the character specified.
data _null_;
var1 = 'word word, word, #12.34, word, word';
pos1 = index(var1,'#'); *Get the position of the first # sign;
tmp = substr(var1,pos1+1); *Create a string that returns only characters after the # sign;
put tmp;
pos2 = index(tmp,','); *Get the position of the first "," in the tmp variable;
var2 = substr(tmp,1,pos2-1);
put var2;
run;
Note that this method only works if there is only one "#" in the string.
One way is to use index to locate the two 'sentinels' delimiting the value and retrieve the innards with substr. If the value is supposed to be numeric, an additional use of input function is needed.
A second way is to use a regular expression routines prxmatch and prxposn to locate and extract the embedded value.
data have;
input;
longtext = _infile_;
datalines;
some thing #12.34, wicked
#, oops
#5a64, oops
# oops
oops ,
oops #
ok #1234,
who wants be a #1e6,aire
space # , the final frontier
double #12, jeopardy #34, alex
run;
data want;
set have;
* locate with index;
_p1 = index(longtext,'#');
if _p1 then _p2 = index(substr(longtext,_p1),',');
if _p2 > 2 then num_in_text = input (substr(longtext,_p1+1,_p2-2), ?? best.);
* locate with regular expression;
if _n_ = 1 then _rx = prxparse('/#(\d*\.?\d*)?,/'); retain _rx;
if prxmatch(_rx,longtext) then do;
call prxposn(_rx,1,_start,_length);
if _length > 0 then num_in_text_2 = input (substr(longtext,_start, _length), ?? best.);
end;
* drop _: ;
run;
The regex way looks for ##.## variants, the index way looks only for #...,. Then input function will decipher scientific notation values the regex (example pattern)way will not 'locate'. The ?? option in the input function prevents invalid arguments NOTE:s in the log when the enclosed value can not be parsed as a number.
Another way to do is by using Regex and code is given below
data have;
infile datalines truncover ;
input var $200.;
datalines;
word word, word, #12.34, word, word
word1 #12.34, hello hi hello hi
word1 #970000 hello hi hello hi #970022, hi
word1 123, hello hi hello hi #97.99
#99456, this is cool
;
A small note about below regular expression and functions
(?<=#) Zero-width positive look-behind assertion and looking for # before the pattern of interest
(\d+.?\d+) here means digit followed or not followed by . and other digits
(?=,) Zero-width positive look-ahead assertion and looking for , after the pattern of interest
call prxsubstr finds the position and length of pattern and substr extracts the required values.
data want( drop=pattern position length);
retain pattern;
IF _N_ = 1 THEN PATTERN = PRXPARSE("/(?<=#)(\d+\.?\d+)(?=,)/");
set have;
call prxsubstr(pattern, var, position, length);
if position then
match = substr(var, position, length);
run;
if you want to get really lazy you can just do
want = compress(have,".","kd");

How to scan the string and convert dynamically in SAS

Supposed I have two strings to convert from SAS program name to table number.
My goal is to convert the first "f-2-2-7-5-vcb" to "2.2.7.5".
And this should be done dynamically. Like for "f-2-2-12-1-2-hbd87q",
it needed to be "2.2.12.1.2" .
How to accomplish this?
data input;
input str $ 1-20;
datalines;
f-2-3-1-5-vcb
f-2-4-1-6-rtg
f-2-3-11-1-3-hb17
;
run;
data want;
set input;
Sub=compress(substr(str,3,length(str)),,'kd') ;
run;
Bit of a longer way, but this works fine for me.
Use FIND() to find the first '-'
Use REVERSE() and FIND() to find the
last '-'
Use SUBSTR() and metrics + math from above to remove the first and
last components
Use TRANSLATE() to convert the - to periods.
z=find(str, '-');
end=find(strip(reverse(str)), '-');
string = translate(substr(str, z+1, length(str) - z - end), ".", "-");
A regular expression can match the dash delimited digits only sequence. The match, when extracted, can be transformed using translate.
data input;
input str $ 1-20;
rx = prxparse ("/^.*?((\d+)(-\d+)*)/");
if prxmatch(rx,str) then do;
call prxposn (rx,1,s,e);
name = substr(str,s,e);
name = translate(name,'.','-');
end;
datalines;
f-2-3-1-5-vcb
f-2-4-1-6-rtg
f-2-3-11-1-3-hb17
funky2-2-1funky
f-2-hb17
a2bfunky
;
run;
A funky situation occurs if the digits only token sequence is preceded by a token ending with digits, or succeeded by a token starting with digits.
data input;
input str $ 1-20;
string=translate(prxchange('s/\w+?\-(.*)\-\w+/$1/',-1,strip(str)),'.','-');
datalines;
f-2-3-1-5-vcb
f-2-4-1-6-rtg
f-2-3-11-1-3-hb17
;
run;
You can do this in one line. Use subtr to keep the text between the second word and last word:
translate(substr(str,find(str,scan(str,2,'-')),find(str,scan(str,-1,'-'))-find(str,scan(str,2,'-'))-1),'.','-')
find(str,scan(str,2,'-') : finds the starting position of the second
word.
find(str,scan(str,-1,'-') : finds the starting position of the last
word.
step2 - find(str,scan(str,2,'-'))-1 : find ending position of second
last word (length of text to copy).
Translate function: replaces '-' with '.'
substr(str,step1,step3) : copy text between second word and second to last.
Code:
data want;
set input;
Sub=translate(substr(str,find(str,scan(str,2,'-')),find(str,scan(str,-1,'-'))-find(str,scan(str,2,'-'))-1),'.','-');
put _all_;
run;
Output:
str=f-2-3-1-5-vcb Sub=2.3.1.5
str=f-2-4-1-6-rtg Sub=2.4.1.6
str=f-2-3-11-1-3-hb17 Sub=2.3.11.1.3

How to read informats data: $1,000.1M to 1000.1

The datasets include a list of numbers:
$1,000.1M
$100.5M
$1,002.3M
$23.4M
$120.3M
I want to read the variable as a numeric in SAS
the result should be:
Money(millions)
1000.1
100.5
1002.3
23.4
120.3
I used COMMAw.d to read this data, but cannot run
The code is:
input Money(millions) COMMA9.1;
run;
How to modify it?
Thank you very much!
The COMMA informat does not expect letters like 'M', it removes only commas, blanks, dollar signs, percent signs, dashes, and close parentheses.
You can just convert your raw string to a string containing a number by removing all characters you do not need:
data input;
length moneyRaw $200;
infile datalines;
input moneyRaw $;
datalines;
$1,000.1M
$100.5M
$1,002.3M
$23.4M
$120.3M
;
run;
data result;
set input;
* "k" modifier inverts the removed characters;
money = input(compress(moneyRaw,"0123456789.","k"),best.);
run;
Or if you know regex, you can add some intrigue to the code for anyone who reads it in the future:
data resultPrx;
set input;
moneyUpdated = prxChange("s/^\$(\d+(,\d+)*(\.\d+)?)M$/$1/",1,strip(moneyRaw));
money = input(compress(moneyUpdated,','),best.);
run;
I think you're best off reading it as a character and then processing it as in Dmitry's answer. But if it was a single column you could read it if you set the delimiter to M. I suspect this will work in a demo, but not in your full process.
data input;
informat moneyRaw dollar8.;
infile datalines dlm='M';
input moneyRaw ;
*moneyRaw = moneyRaw * (1000000);
format moneyRaw dollar32.;
datalines;
$1,000.1M
$100.5M
$1,002.3M
$23.4M
$120.3M
;
run;

Regular Expression dot in SAS

I'm new in this field, and try to use prxmatch and rxmatch to match some strings.
The pattern is a., which matches a string with more than 2 characters and a isn't the last one.
I run prxmatch('/a./', 'a') and rxmatch('/a./', 'a'), the result should be 0. But the system returns me 1.
So how can I get 0 in this case?
If you write a MCVE for this, you do get no match.
data test;
x='a';
rc=prxmatch('~a.~',x);
put x= rc=;
run;
However, if x is not length 1, it will match!
data test;
length x $5;
x='a';
rc=prxmatch('~a.~',x);
put x= rc=;
run;
Why?
Because in SAS, strings are not varchar, they are char. They have spaces padding the rest of the string out to its full length. So you would need to do either
data test;
length x $5;
x='a';
rc=prxmatch('~a[^ ]~',x);
put x= rc=;
run;
or, better,
data test;
length x $5;
x='a';
rc=prxmatch('~a.~',trim(x));
put x= rc=;
run;
(Note, I use ~ for my regex delimiter - you're free to use slash, or any other character, for that, it makes no difference.)