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ovince
Yugoslavia
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 Posted - 09/21/2006 : 3:10:47 PM
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Hi all,
Is it posible to put column data into legend and how? In each worksheet I have a column with 6 row data that I would like to appear on graph as a legend but no idea how to do it
Thanks
oliver |
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Deanna
China
Posts |
Posted - 09/21/2006 : 10:13:00 PM
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Suppose the second column of the below worksheet has the legend texts
To use the texts for the legend like this:
The following code should work:
void test_legend()
{
GraphPage gp("Graph1");
if ( !gp.IsValid() ) return;
vector<string> vS;
Dataset ds("data2", 1);
ds.GetStringArray(vS, 0, 2);
gp.LT_execute("legend -s"); //Create a legend on the graph page
string str;
str.Format("legend.text$=\\l(1) %s\n\\l(2) %s\n\\l(3) %s", vS[0], vS[1], vS[2]);
gp.LT_execute(str);
}
Deanna
OriginLab GZ Office |
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ovince
Yugoslavia
Posts |
Posted - 09/22/2006 : 12:23:43 AM
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Hello
Thank you. What if I already have one legend? In this form (bellow) the new legen just overwrite the old one. Is there a command option to append a new legend? This is the form:
LT_execute("legend -s;");
LT_execute("legend.text$='\l(1) all poins
\l(2) mean-no weight
\l(3) mean-weight
\l(4) median';");
LT_execute("legend.left=4500;");
LT_execute("legend.top=3100;");
LT_execute("legend.background = 1;");
legend with data number
gpg.LT_execute("legend -s");
string str;
str.Format("legend.text$= \\l(1)%s\n \l(2)%s\n \\l(3)%s\n \\l(4)%s\n \\l(5)%s\n \\l(6)%s\n \\l(7)%s", vS[0], vS[1], vS[2], vS[3], vS[4], vS[5], vS[6]);
gpg.LT_execute(str);
oliver
P.S. I also have a problem to implement the program makeLayout2.c that you sent via email. It reports some errors. Not clear why. I have replyed to that email with more detail. thanks |
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Deanna
China
Posts |
Posted - 09/22/2006 : 01:18:27 AM
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In this case, we can add a label instead. Please try the following code
void test_label()
{
GraphPage gp("Graph1");
if ( !gp.IsValid() ) return;
vector<string> vS;
Dataset ds("data2", 1);
ds.GetStringArray(vS, 0, 2);
string str;
str.Format("label \\l(1) %s\n\\l(2) %s\n\\l(3) %s", vS[0], vS[1], vS[2]);
gp.LT_execute(str);
}
I have checked tech@originlab.com for several times today. But I did not see your email. I have no idea what happens. Would you please send me the email again? Thank you.
Deanna
OriginLab GZ Office
Edited by - Deanna on 09/22/2006 01:19:13 AM |
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ovince
Yugoslavia
Posts |
Posted - 09/22/2006 : 01:51:10 AM
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hi,
Label is an object no? I was trying to positione it and make a nice border but does not work. This was the idea:
;string str;
;str.Format("label %s\n %s\n %s", vS[0], vS[1], vS[2], vS[3]);
;gpg.LT_execute("label.background = 1;");
;gpg.LT_execute("label.left=4500;");
;gpg.LT_execute("label.top=3100;");
;gpg.LT_execute(str);
(note that I have corrected \\l(1) %s\n\\l(2) %s\n\\l(3) %s to %s\n %s\n %s; could be this the reason; no I hope)
I have sent the email again. Should be there already. Thanks many...very happy for all helps
Oliver
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Deanna
China
Posts |
Posted - 09/22/2006 : 02:17:56 AM
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Yes, a label is an object. But its name is text.
I suggest that you change your code like:
str.Format("label %s\n%s\n%s\n%s", vS[0], vS[1], vS[2], vS[3]);
gp.LT_execute(str);
gp.LT_execute("text.background = 1;");
gp.LT_execute("text.left=4500;");
gp.LT_execute("text.top=3100;");
Note that there should be 4 '%s' if your want to put 4 strings into the str.
There is a very useful Labtalk command you might want to know. Active a graph page, and type
list -o
in the command window. The names of the objects on the active graph layer will be listed. :)
Deanna
OriginLab GZ Office |
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ovince
Yugoslavia
Posts |
Posted - 09/22/2006 : 02:39:52 AM
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hi,
new tricky problem. With your suggestion the title of the graph (grpName) is moved not the text that I would like. How to distinguish between two. This is the part of routine:
//set X and Y titles
LT_set_str("%P", grpName);
LT_execute("lab -p 45 2 -s %P;");
LT_execute("lab -xb (T);");
LT_execute("lab -yl (index);");
LT_execute("lab -r Legend;");
LT_execute("legend -s;");
LT_execute("legend.text$='\l(1) all poins
\l(2) mean-no weight
\l(3) mean-weight
\l(4) median';");
LT_execute("legend.left=4500;");
LT_execute("legend.top=3100;");
LT_execute("legend.background = 1;");
//legend with data number (Deanna)
vector<string> vS;
Dataset ds(wks, 8);
ds.GetStringArray(vS, 0, 6);
string str;
str.Format("label %s\n %s\n %s\n %s\n %s\n %s\n %s", vS[0], vS[1], vS[2], vS[3], vS[4], vS[5], vS[6]);
gpg.LT_execute(str);
gpg.LT_execute("text.background = 1;");
gpg.LT_execute("text.left=45;");
gpg.LT_execute("text.top=31;");
so the grpName is moved not the label
oliver |
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Deanna
China
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Posted - 09/22/2006 : 05:06:27 AM
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It seems that there are more than one labels in your graph. The label we have just created was named after text1, or something like that. The OC function does not use the correct name.
Let's first check the name of the label. We can do it as follows:
1. Select the label with the mouse;
2. In the script window, type:
select -g a;
%a=
This should output the name of the label.
With the correct name, you can modify your OC function, and it should be okay (I hope :P ).
Deanna
OriginLab GZ Office
Edited by - Deanna on 09/22/2006 05:07:38 AM
Edited by - Deanna on 09/22/2006 05:09:28 AM |
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Topic |
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worksheet vs legend