Hi,
I have a csv file:
01/02/2009,146,394 |
02/02/2009,278,791 |
03/02/2009,224,637 |
04/02/2009,275,725 |
05/02/2009,276,1Â 101 |
06/02/2009,247,1Â 103 |
07/02/2009,154,412 |
08/02/2009,192,609 |
09/02/2009,308,992 |
10/02/2009,333,1Â 287 |
11/02/2009,342,1Â 250 |
12/02/2009,375,1Â 975 |
13/02/2009,275,1Â 019 |
14/02/2009,120,487 |
15/02/2009,133,567 |
16/02/2009,210,770 |
I remove  with the command 'compress', no problem.
But to remove the space, there is a problem, the command 'compress', does nothing.
Why?
My goal is not to change the csv file but to transform eg:
1Â 056 -- compress --> 1 056 --> compress --> 1056
data toto;
infile 'C:\.........................\sessionsetpagesvues.csv' dlm=',' firstobs=8 truncover;
input Index_des_jours :ddmmyy. Sessions :numx. x3 $;
x4=compress(x3,'Â ','a');
x5=compress(x4,'','s');
Pages_vues= input(x5, numx.);
run;
proc print data=toto(obs=350);run;
proc contents data=toto;run;
x5=compress(x4,'','s');
This command does not work why?
Thank you for your help.
It's not a space '20'x it just looks like one. It is non-breaking space 'A0'x.
It's not a space '20'x it just looks like one. It is non-breaking space 'A0'x.
I'm not sure why that doesn't work, as that was my first answer.
Anyway, you can use Perl regular expressions like so:
data test;
variable = "1 045";
rid = prxparse('s/\s*//');
call prxchange(rid, -1, variable);
drop rid;
run;
Just to add, the compress function has various modifiers you can apply to it which should simplfy your code:
x4=compress(x3,,'kd');
Note I put no characters in the second paramter, the 'KD' or keep decimals, does everything needed here. For example:
data a; abcd="1Â 101"; def=compress(abcd,,"kd"); run;
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