I am writing my first macro for Libre Office right now and I have come into a bit of a problem: My code throws the error: BASIC Runtime error; Sub- or Function procedure not defined.
The line with the "If Cells (RowCnt,ChkCol......) throws the error.
I've looked through other entries on here, but I could not find the error... can anyone help me?
REM ***** BASIC *****
Sub Zeilennausblenden_Nullsummen
BeginRow=4
EndRow = 46
ChkCol= D
For RowCnt = BeginRow To EndRow step 1
If Cells(RowCnt,ChkCol).Value > 1 Then
Cells(RowCnt,ChkCol).EntireRow.Hidden = True
End if
Next
End Sub
PS: The function should hide all rows in which an integer higher than "1" appears in column "D"
Thanks in advance
Here is what the code looks like in LibreOffice Basic (aka StarBasic):
Sub Zeilennausblenden_Nullsummen
BeginRow=4
EndRow = 46
ChkCol= 3
oSheet = ThisComponent.Sheets(0)
For RowCnt = BeginRow To EndRow step 1
oCell = oSheet.getCellByPosition(ChkCol,RowCnt)
If oCell.Value > 1 Then
oRow = oSheet.getRows().getByIndex(RowCnt)
oRow.IsVisible = False
End if
Next
End Sub
I wasn't sure if BeginRow should be 3 or 4, because it's zero-based. You can test it and decide.
Note that a macro is not necessary in order to accomplish this task. The easiest way is to go to Data -> More Filters -> Standard Filter.
That's because CELLS isn't a StarBasic function.
It's VBA (different programming language). Some versions of OpenOffice support the use of it if a statement (Option VBASupport 1) is put in first line of source code.
Check the net for Andrew Pitonyak's "OpenOffice Macros Explained" document - very good for learning and available in German translation, too.
Related
I am trying to split a string variable into multiple dummy coded variables. I used these sources to get an idea of how one would achieve this task in SPSS:
https://www.ibm.com/support/pages/making-multiple-string-variables-single-multiply-coded-field
https://www.spss-tutorials.com/spss-split-string-variable-into-separate-variables/
But when I try to adapt the first one to my needs or when I try to convert the second one to a macro, I fail.
In my dataset I have (multiple) variables that contain a comma seperated string that represents different combinations of selected items (as well as missing values). For each item of a specific variable I want to create a dummy variable. If the item was selected, it should be represented with a 1 in the new dummy variable. If it was not selected, that case should be represented with a 0.
Different input variables can contain different numbers of items.
For example:
ID
VAR1
VAR2
DMMY1_1
DMMY1_2
DMMY1_3
1
1, 2
8
1
1
0
2
1
1, 3
1
0
0
3
3, 1
2, 3, 1
1
0
1
4
2, 8
0
0
0
Here is what I came up with so far ...
* DEFINE DATA.
DATA LIST /ID 1 (F) VAR1 2-5 (A) VAR2 6-12 (A).
BEGIN DATA
11, 28
21 1, 3
33, 12, 3, 1
4 2, 8
END DATA.
* MACRO SYNTAX.
* DEFINE VARIABLES (in the long run these should/will be inside the macro function, but for now I will leave them outside).
NUMERIC v1 TO v3 (F1).
VECTOR v = v1 TO v3.
STRING #char (A1).
DEFINE split_var(vr = TOKENS(1)).
!DO !#pos=1 !TO char.length(!vr).
COMPUTE #char = char.substr(!vr, !#pos, 1).
!IF (!#char !NE "," !AND !#char !NE " ") !THEN
COMPUTE v(NUMBER(!#char, F1)) = 1.
!IFEND.
!DOEND.
!ENDDEFINE.
split_var vr=VAR1.
EXECUTE.
As I got more errors than I can count, it's hard to narrow down my problem. But I think the problem has something to do with the way I use the char.length() function (and I am a bit confused when to use the bang operator).
If anyone has some insights, I would really appreciate some help :)
There is a fundamental issue to understand about SPSS macro - the macro does not read or interact in any way with the data. All the macro does is manipulate text to write syntax. The syntax created will later work on the actual data when you run it.
So, for example, Your first error is using char.length(!vr) within the syntax. You are trying to get the macro to read the data, calculate the length and use, but that simply can't be done - the macro can only work with what you gave it.
Another example in your code: you calculate #char and then try to use it in the macro as !#char. So that obviously won't work. ! precedes only macro functions or arguments. #char, in your code, is neither, and it can't become one - can't read the data into the macro...
To give you a litte push forward: I understand you want the macro loop to run a different number of times for each variable, but you can't use char.length(!vr). I suggest instead have the macro loop as many times as necessary to be sure you can deal with the longest variable you'll need to work with.
And another general strategy hint - first, create syntax to deal with one specific variable and one specific delimiter. Once this works, start working on a macro, keeping in mind that the only purpose of the macro is to recreate the same working syntax, only changing the parameters of variable name and delimiter.
With my new understanding of the SPSS macro logic (thanks to #eli-k) the problem was quite easy to solve. Here is the working solution.
* DEFINE DATA.
DATA LIST /ID 1 (F) VAR1 2-5 (A) VAR2 6-12 (A).
BEGIN DATA
11, 28
21 1, 3
33, 12, 3, 1
4 2, 8
END DATA.
* DEFINE MACRO.
DEFINE #split_var(src_var = !TOKENS(1)
/dmmy_var_label = !DEFAULT(dmmy) !TOKENS(1)
/dmmy_var_lvls = !TOKENS(1))
NUMERIC !CONCAT(!dmmy_var_label,1) TO !CONCAT(!dmmy_var_label, !dmmy_var_lvls) (F1).
VECTOR #dmmy_vec = !CONCAT(!dmmy_var_label,1) TO !CONCAT(!dmmy_var_label, !dmmy_var_lvls).
STRING #char (A1).
LOOP #pos=1 TO char.length(!src_var).
COMPUTE #char = char.substr(!src_var, #pos, 1).
DO IF (#char NE "," AND #char NE " ").
COMPUTE #index = NUMBER(#char, F1).
COMPUTE #dmmy_vec(#index) = 1.
END IF.
END LOOP.
RECODE !CONCAT(!dmmy_var_label,1) TO !CONCAT(!dmmy_var_label, !dmmy_var_lvls) (SYSMIS=0) (ELSE=COPY).
EXECUTE.
!ENDDEFINE.
* CALL MACRO.
#split_var src_var=VAR2 dmmy_var_lvls=8.
I am having a couple of issues to put this in a functional format.
select from tableName where i=fby[(last;i);([]column_one;column_two)]
This is what I got:
?[tableName;fby;enlist(=;`i;(enlist;last;`i);(+:;(!;enlist`column_one`column_two;(enlist;`column_one;`column_two))));0b;()]
but I get a type error.
Any suggestions?
Consider using the following function, adjust from the buildQuery function given in the whitepaper on Parse Trees. This is a pretty useful tool for quickly developing in q, this version is an improvement on that given in the linked whitepaper, having been extended to handle updates by reference (i.e., update x:3 from `tab)
\c 30 200
tidy:{ssr/[;("\"~~";"~~\"");("";"")] $[","=first x;1_x;x]};
strBrk:{y,(";" sv x),z};
//replace k representation with equivalent q keyword
kreplace:{[x] $[`=qval:.q?x;x;"~~",string[qval],"~~"]};
funcK:{$[0=t:type x;.z.s each x;t<100h;x;kreplace x]};
//replace eg ,`FD`ABC`DEF with "enlist`FD`ABC`DEF"
ereplace:{"~~enlist",(.Q.s1 first x),"~~"};
ereptest:{((0=type x) & (1=count x) & (11=type first x)) | ((11=type x)&(1=count x))};
funcEn:{$[ereptest x;ereplace x;0=type x;.z.s each x;x]};
basic:{tidy .Q.s1 funcK funcEn x};
addbraks:{"(",x,")"};
//where clause needs to be a list of where clauses, so if only one whereclause need to enlist.
stringify:{$[(0=type x) & 1=count x;"enlist ";""],basic x};
//if a dictionary apply to both, keys and values
ab:{$[(0=count x) | -1=type x;.Q.s1 x;99=type x;(addbraks stringify key x),"!",stringify value x;stringify x]};
inner:{[x]
idxs:2 3 4 5 6 inter ainds:til count x;
x:#[x;idxs;'[ab;eval]];
if[6 in idxs;x[6]:ssr/[;("hopen";"hclose");("iasc";"idesc")] x[6]];
//for select statements within select statements
//This line has been adjusted
x[1]:$[-11=type x 1;x 1;$[11h=type x 1;[idxs,:1;"`",string first x 1];[idxs,:1;.z.s x 1]]];
x:#[x;ainds except idxs;string];
x[0],strBrk[1_x;"[";"]"]
};
buildSelect:{[x]
inner parse x
};
We can use this to create the functional query that will work
q)n:1000
q)tab:([]sym:n?`3;col1:n?100.0;col2:n?10.0)
q)buildSelect "select from tab where i=fby[(last;i);([]col1;col2)]"
"?[tab;enlist (=;`i;(fby;(enlist;last;`i);(flip;(lsq;enlist`col1`col2;(enlist;`col1;`col2)))));0b;()]"
So we have the following as the functional form
?[tab;enlist (=;`i;(fby;(enlist;last;`i);(flip;(lsq;enlist`col1`col2;(enlist;`col1;`col2)))));0b;()]
// Applying this
q)?[tab;enlist (=;`i;(fby;(enlist;last;`i);(flip;(lsq;enlist`col1`col2;(enlist;`col1;`col2)))));0b;()]
sym col1 col2
----------------------
bah 18.70281 3.927524
jjb 35.95293 5.170911
ihm 48.09078 5.159796
...
Glad you were able to fix your problem with converting your query to functional form.
Generally it is the case that when you use parse with a fby in your statement, q will convert this function into its k definition. Usually you should just be able to replace this k code with the q function itself (i.e. change (k){stuff} to fby) and this should run properly when turning the query into functional form.
Additionally, if you check out https://code.kx.com/v2/wp/parse-trees/ it goes into more detail about parse trees and functional form. Additionally, it contains a script called buildQuery which will return the functional form of the query of interest as a string which can be quite handy and save time when a functional form is complex.
I actually got it myself ->
?[tableName;((=;`i;(fby;(enlist;last;`i);(+:;(!;enlist`column_one`column_two;(enlist;`column_one;`column_two)))));(in;`venue;enlist`venueone`venuetwo));0b;()]
The issues was a () missing from the statement. Works fine now.
**if someone wants to add a more detailed explanation on how manual parse trees are built and how the generic (k){} function can be replaced with the actual function in q feel free to add your answer and I'll accept and upvote it
I am trying to read in a whole set of environmental input/output data from the WIOD (World Input output database) through a nested loop over the countries and years. Now I have done something similar for the base/data accounts before. Now I try to load in the environmental data. My previously working code looked as follows:
for yr = 95:99
V(:,:,yr-94) = xlsread(['wiot' num2str(yr) '_row_apr12.xlsx'],['WIOT_19'
num2str(yr)],'E1443:BCI1448');
end
Now my code which is not working with the error message "File name must be a character vector." looks as follows:
country =
{'AUS','AUT','BEL','BGR','BRA','CAN','CHN','CYP','CZE','DEU','DNK','ESP',...
'EST','FIN','FRA','GBR','GRC','HUN','IDN','IND','IRL','ITA','JPN','KOR',...
'LTU','LUX','LVA','MEX','MLT','NLD','POL','PRT','ROU','RUS','SVK','SVN',...
'SWE','TUR','TWN','USA','ROW'}
for c = 1:41
for year = 1995:1995
F_NRG(:,(c*35)-34:(c*35),year-1994) = transpose(xlsread([country(c)
'_EU_May12.xlsm'],[num2str(year)],'AD2:AD36'));
end
end
I do not get it because the filename should be a string if I select the country c via country(c)? The xlsread is nested in a transpose command and the cells where I want to save the read in data are computed slightly more complicated but principally it should be the same? The following code also does render a string for each c.
for c = 1:41
country(c)
end
Can you help me find my coding mistakes? Why doesn't Matlab recognize the file name as a string?
Thank you for your help.
Try this,
for c = 1:41
for year = 1995:1995
F_NRG(:,(c*35)-34:(c*35),year-1994) = transpose(xlsread([country{c}
'_EU_May12.xlsm'],[num2str(year)],'AD2:AD36'));
end
end
I wrote a macro in LibreOffice Calc and it is able to run correctly. But if I close the file and reopen, it always show #NULL! instead of the correct value. What am I missing here?
My macro code
Rem Attribute VBA_ModuleType=VBAModule
Option VBASupport 1
Function Calculate(CalType As String) As Double
'
' Calculate Macro
'
Dim i As Integer
Calc = 0
i = 1
Do While Not IsEmpty(Cells(i, 2))
If (Cells(i, 3).Value = CalType And (Cells(i,2) = "A" Or Cells(i,2) = "B")) Then
Calculate = Calculate + Cells(i, 4).Value
ElseIf (Cells(i, 3).Value = CalType And Cells(i,2) = "C") Then
Calculate = Calculate - Cells(i, 4).Value
End If
i = i + 1
Loop
'
End Function
The calling function will be something like =Calculate(J6)
The file is saved as .ods format.
The Cells call did not work at all for me. It is from VBA, not LO Basic. However I do not think that is the main problem.
LibreOffice expects that user-defined functions will be simple, only accessing the cell that contains the formula. Since the spreadsheet has not been fully loaded yet when the function is called, it is not possible to read other cells.
The workaround is to ignore errors and wait until the document is fully loaded before running the function. Take the following code as an example:
Function ReadOtherCell(row, col)
On Error GoTo ErrorHandler
oSheet = ThisComponent.CurrentController.ActiveSheet()
oCell = oSheet.getCellByPosition(row, col)
ReadOtherCell = "value is '" & oCell.getString() & "'"
Exit Function
ErrorHandler:
Reset
End Function
Sub RecalculateAll
' This is for user-defined functions that need to read the spreadsheet.
' Assign it to the "View created" event,
' because before that, the spreadsheet is not fully loaded.
ThisComponent.calculateAll
End Sub
Enter foo in A1, and =ReadOtherCell(0,0) in A2. So far, this has the same problem -- It will fail when the document is first opened.
Now, go to Tools -> Customize. In the Events tab, highlight View created. Press Macro... and find the RecalculateAll function. Then press OK.
Now when the document is closed and reopened, cell A2 should show the result value is 'foo'.
This is derived from B. Marcelly's answer at http://ooo-forums.apache.org/en/forum/viewtopic.php?f=20&t=73090&sid=f92a89d676058ab597b4b4494833b2a0.
I had the same problem.
I noticed that in the module, i had an empty Sub Main
After i 've erased it, the functions started working again
I want to iterate through all classes and packages in a special path.
After that, I want to get the MethodList.
In the command window I can use following and it’s working fine:
a = ?ClassName;
a.MethodList(X);
Now I separate this into a function:
function s = befehlsreferenz(path)
s = what(path); % list MATLAB files in folder
for idx = 1:numel(s.classes)
c = s.classes(idx);
b = ?c;
b.MethodList(0);
end
end
I get an error:
Too many outputs requested. Most likely cause is missing [] around left hand side that has a comma separated list
expansion. Error in (line 7) b.MethodList(0);
While debugging I can see:
c: 1x1 cell = ‘Chapter’
b: empty 0x0 meta.class
Why is b empty? How can I get the methodlist?
1 Edit:
Here is an example class, also not working with it.
classdef TestClass
%TESTCLASS Summary of this class goes here
% Detailed explanation goes here
properties
end
methods
function [c] = hallo(a)
c = 1;
end
end
end
When struggling with operators in Matlab, it's typically the best choice to use the underlying function instead, which is meta.class.fromname(c)
Relevant documentation: http://de.mathworks.com/help/matlab/ref/metaclass.html
Further it seems s.classes(idx); is a cell, use cell indexing: s.classes{idx} ;