Count characters in a string variable without blank

[huang jialin]

Hi Everyone,

I was trying to count the number of characters in a string variable, which includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the same results.

However, what I am looking for is how to only count characters without strings. The expected result should be 26 (including # sign).

Your advice will be very appreciated. Thank you very much.

Example:
Name of the variable : Response

#BADA CDACDA   D BDAD    BBACBCA   ABB

Sincerely,
Jialin Huang

[Maguin, Eugene]

I assume all character strings contain US keyboard printing characters.

Let x by the variable with format A50

Compute nchars=0.

Loop #i=1 to 50.

+  if (substr(x,i,1) ne ‘ ‘) nchar=nchar+1.

End loop.

 

Can it be done in fewer commands???

 

Gene Maguin

 

From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of huang jialin
Sent: Thursday, September 27, 2012 5:11 PM
To: [hidden email]
Subject: Count characters in a string variable without blank

 

Hi Everyone,

I was trying to count the number of characters in a string variable, which includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the same results.

However, what I am looking for is how to only count characters without strings. The expected result should be 26 (including # sign).

Your advice will be very appreciated. Thank you very much.

Example:
Name of the variable : Response

#BADA CDACDA   D BDAD    BBACBCA   ABB

Sincerely,
Jialin Huang

[Jon K Peck]

compute nchars = char.length(replace(x, " ", ""))

Jon Peck (no "h") aka Kim
Senior Software Engineer, IBM
[hidden email]
new phone: [phone]




From:        "Maguin, Eugene" <[hidden email]>
To:        [hidden email]
Date:        09/27/2012 03:24 PM
Subject:        Re: [SPSSX-L] Count characters in a string variable without blank
Sent by:        "SPSSX(r) Discussion" <[hidden email]>

---

I assume all character strings contain US keyboard printing characters.
Let x by the variable with format A50
Compute nchars=0.
Loop #i=1 to 50.
+  if (substr(x,i,1) ne ‘ ‘) nchar=nchar+1.
End loop.
 
Can it be done in fewer commands???
 
Gene Maguin
 
From: SPSSX(r) Discussion [[hidden email]] On Behalf Of huang jialin
Sent: Thursday, September 27, 2012 5:11 PM
To: [hidden email]
Subject: Count characters in a string variable without blank
 
Hi Everyone,

I was trying to count the number of characters in a string variable, which includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the same results.

However, what I am looking for is how to only count characters without strings. The expected result should be 26 (including # sign).

Your advice will be very appreciated. Thank you very much.

Example:
Name of the variable : Response

#BADA CDACDA   D BDAD    BBACBCA   ABB

Sincerely,
Jialin Huang

[David Marso]

"Can it be done in fewer commands???"
YEP!!!!! ;-))))))

COMPUTE L=LENGTH(RTRIM(REPLACE(X," ",""))).
===========

DATA LIST /X (A100).
BEGIN DATA
#BADA CDACDA   D BDAD    BBACBCA   ABB
END DATA.

COMPUTE L=LENGTH(RTRIM(REPLACE(X," ",""))).

LIST.

Maguin, Eugene wrote

I assume all character strings contain US keyboard printing characters.
Let x by the variable with format A50
Compute nchars=0.
Loop #i=1 to 50.
+  if (substr(x,i,1) ne ‘ ‘) nchar=nchar+1.
End loop.

Can it be done in fewer commands???

Gene Maguin

From: SPSSX(r) Discussion [mailto:[hidden email]] On Behalf Of huang jialin
Sent: Thursday, September 27, 2012 5:11 PM
To: [hidden email]
Subject: Count characters in a string variable without blank

Hi Everyone,

I was trying to count the number of characters in a string variable, which includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the same results.

However, what I am looking for is how to only count characters without strings. The expected result should be 26 (including # sign).

Your advice will be very appreciated. Thank you very much.

Example:
Name of the variable : Response

#BADA CDACDA   D BDAD    BBACBCA   ABB

Sincerely,
Jialin Huang

[David Marso]

RTRIM simply strips trailing blanks but does NOT do anything with embedded spaces.
See solution below.
COMPUTE L=LENGTH(RTRIM(REPLACE(X," ",""))).

huang jialin wrote

Hi Everyone,

I was trying to count the number of characters in a string variable, which
includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for
the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the
same results.

However, what I am looking for is how to only count characters without
strings. The expected result should be 26 (including # sign).

Your advice will be very appreciated. Thank you very much.

Example:
Name of the variable : Response

#BADA CDACDA   D BDAD    BBACBCA   ABB

Sincerely,
Jialin Huang

[huang jialin]

Hi Everyone,

Your solutions worked. Thank you very much for your helps.

But I have a question. When I used REPLACE(X," ", "") to remove the blanks. SPSS showed an error " Character strings may only be assigned to string variables." But the variable is actually string. However, David and Jon solutions involved REPLACE function. There is no error. What is going on there?

Thank you again.

Sincerely,
Jialin Huang

On Thu, Sep 27, 2012 at 4:55 PM, David Marso <[hidden email]> wrote:

RTRIM simply strips trailing blanks but does NOT do anything with embedded
spaces.
See solution below.
COMPUTE L=LENGTH(RTRIM(REPLACE(X," ",""))).


huang jialin wrote

> Hi Everyone,
>
> I was trying to count the number of characters in a string variable, which
> includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for
> the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the
> same results.
>
> However, what I am looking for is how to only count characters without
> strings. The expected result should be 26 (including # sign).
>
> Your advice will be very appreciated. Thank you very much.
>
> Example:
> Name of the variable : Response
>
> #BADA CDACDA   D BDAD    BBACBCA   ABB
>
> Sincerely,
> Jialin Huang

-----
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[Jon K Peck]

The error does not have to do with replace.  It is telling you that you are trying to assign a string value to a variable that has not been declared as string type.  New variables are by default numeric, so you need to declare a string variable and a length code for it.
E.g.,
string huang(a50).
compute huang = <some string expression>.



Jon Peck (no "h") aka Kim
Senior Software Engineer, IBM
[hidden email]
new phone: [phone]




From:        huang jialin <[hidden email]>
To:        [hidden email]
Date:        09/28/2012 07:33 AM
Subject:        Re: [SPSSX-L] Count characters in a string variable without blank
Sent by:        "SPSSX(r) Discussion" <[hidden email]>

---

Hi Everyone,

Your solutions worked. Thank you very much for your helps.

But I have a question. When I used REPLACE(X," ", "") to remove the blanks. SPSS showed an error " Character strings may only be assigned to string variables." But the variable is actually string. However, David and Jon solutions involved REPLACE function. There is no error. What is going on there?

Thank you again.

Sincerely,
Jialin Huang


On Thu, Sep 27, 2012 at 4:55 PM, David Marso <david.marso@...> wrote:
RTRIM simply strips trailing blanks but does NOT do anything with embedded
spaces.
See solution below.
COMPUTE L=LENGTH(RTRIM(REPLACE(X," ",""))).


huang jialin wrote
> Hi Everyone,
>
> I was trying to count the number of characters in a string variable, which
> includes blanks. I tried to use CHAR.LENGTH( Response ), it yields 38 for
> the example. Then, I tried LENGTH(RTRIM( Response , " ")),which gave the
> same results.
>
> However, what I am looking for is how to only count characters without
> strings. The expected result should be 26 (including # sign).
>
> Your advice will be very appreciated. Thank you very much.
>
> Example:
> Name of the variable : Response
>
> #BADA CDACDA   D BDAD    BBACBCA   ABB
>
> Sincerely,
> Jialin Huang





-----
Please reply to the list and not to my personal email.
Those desiring my consulting or training services please feel free to email me.
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View this message in context: http://spssx-discussion.1045642.n5.nabble.com/Count-characters-in-a-string-variable-without-blank-tp5715332p5715337.html
Sent from the SPSSX Discussion mailing list archive at Nabble.com.

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