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Re: st: RE: twoway scatter returns "unmatched quote"


From   David Airey <[email protected]>
To   [email protected]
Subject   Re: st: RE: twoway scatter returns "unmatched quote"
Date   Mon, 16 Mar 2009 12:31:09 -0500

.

You have an unmatched quote in note().

-Dave

On Mar 16, 2009, at 12:25 PM, Rike wrote:

with -set race on- this is where it stops...

------------------------------------------------------------------
end _fr_area_parse_and_log ---
- _fr_area_parse_and_log `log' plotregion1 PLOTRegion , `r(rest)'
       = _fr_area_parse_and_log __000001 plotregion1 PLOTRegion ,
l1title(`""') b1title(`"birth_month"') n
ote("Graphs by dummy variable equal to one if urban, zero if rural)
----------------------------------------------------------------
begin _fr_area_parse_and_log ---
         - gettoken log 0 : 0
         - gettoken object 0 : 0
         - gettoken parseopt 0 : 0
         - local opt = lower("`parseopt'")
         = local opt = lower("PLOTRegion")
         - syntax [ , `parseopt'(string) * ]
         = syntax [ , PLOTRegion(string) * ]
unmatched quote
------------------------------------------------------------------
end _fr_area_parse_and_log ---
----------------------------------------------------------------------
end bygraph_g.parse_make ---
     - if (_rc == 4023) InvalidFamily `do'
     = if (_rc == 4023) InvalidFamily twoway
     - else if (_rc) exit _rc
       }
       if "`.`name'._scheme.isa'" != "" {
       set curscm `.`name'._scheme.objkey'
       }
       }
---------------------------------------------------------------------------------
end graph.Graph ---
   - local rc = _rc
   - gdi record = yes
   - gdi maybedraw = yes
   - gdi end
   - exit `rc'
   = exit 198
-----------------------------------------------------------------------------------------
end graph ---
------------------------------------------------------------------------------------------
end twoway ---
r(198);
r(198);





On Mon, Mar 16, 2009 at 12:14 PM, Rike <[email protected]> wrote:
Yes-

ta status
0 rural
1 urban

but I still can't get the graph work..



On Mon, Mar 16, 2009 at 11:28 AM, Richard Ochmann <[email protected]> wrote:
does this do what you want?

***
label define status 0 "rural", modify
label values status status
tab status
***

assuming the variable "status" exists.

best, rich




Rike <[email protected]>
Gesendet von: [email protected]
16.03.2009 16:16
Bitte antworten an
[email protected]


An
[email protected]
Kopie

Thema
Re: st: RE: twoway scatter returns "unmatched quote"






I added "modify"

label define status 0 "rural", modify

is what I did (sorry). But

ta status

returns

0

1 urban





On Mon, Mar 16, 2009 at 11:10 AM, Friedrich Huebler <[email protected] >
wrote:
Rike,

What do you mean by "it doesn't add a label for 0"? Stata can assign
labels to zeros. Did you forget the -modify- option?

. sysuse auto
. d foreign
             storage  display     value
variable name   type   format      label      variable label
------------------------------------------------------------
foreign         byte   %8.0g       origin     Car type

. lab list origin
origin:
          0 Domestic
          1 Foreign

. lab def origin 0 "Rural", modify
. lab list origin
origin:
          0 Rural
          1 Foreign

. tab foreign
  Car type |      Freq.     Percent        Cum.
------------+-----------------------------------
     Rural |         52       70.27       70.27
   Foreign |         22       29.73      100.00
------------+-----------------------------------
     Total |         74      100.00

Friedrich

On Mon, Mar 16, 2009 at 10:54 AM, Rike <[email protected]> wrote:
Hi,

I think the problem lies in the label. The variable "status"" takes 0
for rural and 1 for urban, but I'm not able to assign a label to 0
(and "urban" is already assigned to 1). I ran

la def status 0 "rural"

it doesn't add a label for 0 which is odd...


On Mon, Mar 16, 2009 at 7:21 AM, Nick Cox <[email protected]> wrote:
No quotes are included in what you report here as typed.

So, check out in turn

1. Any unmatched quotes in the variable label of -status-.

2. Any unmatched quotes in the value labels of -status- (if any).

3. Any unmatched quotes in the values of -status- (if it is a string
variable).

4. Repeat 1 for your other variables.

My guess is #1.

Then remove the unmatched quote _or_ insert a matching quote and
reissue
the -graph- command.

Nick
[email protected]

Rike

I tried it without the space (I mistakenly put it back there when I
copied the code for the list) but I get the same error. Any other
ideas?

Kieran McCaul

You have a space in the -by- option.
-by (status)- should be -by(status)-

Rike

I'm running the following code:

twoway scatter ratio month, by(status) c(l) xlabel(1(1)12)

twoway scatter ratio1 ratio2 ratio3 ratio4 month, by (status)
connect(l l l l) xlabel(1(1)12)

where status is a dummy. Stata returns an r(198) error, and it
happened when I added the by-option. I would also like to use
by(status, total) later on. Any suggestions? Thanks in advance.
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