I'm trying to scan a column for character size. If the alphanumerical character size is met (qty 12), then the first two characters will be deleted. They will always be specific numbers (10). See below.
H063088955
F243066424
10G403085387
F253066457
E473057375
G503087343
10H303098124
G093075912
G433084322
10G403085388
Select the cells you wish to process and run:
Sub qwerty()
For Each r In Selection
v = r.Text
If Len(v) = 12 Then
r.Value = Mid(v, 3)
End If
Next r
End Sub
Related
I have a sheet in libreoffice Calc which has an Id Column with incremental value from 1 to N.
I need to create a Macro in VBA (linked to a button i will create later) where i can select the last ID (which is the MAX id also) and delete the entire row relating to this ID.
i tried this so far
Sub suppression
dim maxId as Integer
my_range = ThisComponent.Sheets(0).getCellRangebyName("B19:B1048576")
maxId = Application.WorksheetFunction.Max(range("Dépôts!B19:B1048576"))
MsgBox maxId
End Sub
Thanks a lot for your help.
In libreoffice BASIC you first need to get the data array of the cell range. This is an array of arrays each representing a row of the cell range. It is indexed from zero irrespective of the location of the cell range within the sheet. Because your cell range is one column wide, each member array has only one member, which is at index zero.
As Jim K says, 'Application.WorksheetFunction' is from VBA. It is possible to use worksheet functions in LibreOffice BASIC, but these act on ordinary arrays rather than cell arrays, and the MAX function takes a one-dimensional array so it would be necessary to first reshape the data array using a loop. Furthermore, if you want to delete the row corresponding to the maximum value you are then faced with the problem of finding the index of that row using only the value itself.
It is much simpler to find the index by looping over the data array as shown in the snippet below.
Also, rather than traversing over a million rows, it would save computational effort to obtain the last used row of the spreadsheet via the BASIC function 'GetLastUsedRow(oSheet as Object)', which is supplied with LibreOffice. This is located in the 'Tools' library in 'LibreOffice Macros & Dialogs'. To use it you have to put the statement: 'Globalscope.BasicLibraries.LoadLibrary("Tools")' somewhere before you call the function.
To delete the identified row, get the XTableRows interface of the spreadsheet and call its removeByIndex() function.
The following snippet assumes that the header row of your table is in row 18 of the sheet, as suggested by your example code, which is in row 17 when numbered from zero.
Sub suppression()
' Specify the position of the index range
''''''''''''''''''''''''''''''''''''
Dim nIndexColumn As Long '
nIndexColumn = 1 '
'
Dim nHeaderRow As Long '
nHeaderRow = 17 '
'
''''''''''''''''''''''''''''''''''''
Dim oSheet as Object
oSheet = ThisComponent.getSheets().getByIndex(0)
' Instead of .getCellRangebyName("B19:B1048576") use:
Globalscope.BasicLibraries.LoadLibrary("Tools")
Dim nLastUsedRow As Long
nLastUsedRow = GetLastUsedRow(oSheet)
Dim oCellRange As Object
' Left Top Right Bottom
oCellRange = oSheet.getCellRangeByPosition(nIndexColumn, nHeaderRow, nIndexColumn, nLastUsedRow)
' getDataArray() returns an array of arrays, each repressenting a row.
' It is indexed from zero, irrespective of where oCellRange is located
' in the sheet
Dim data() as Variant
data = oCellRange.getDataArray()
Dim max as Double
max = data(1)(0)
' First ID number is in row 1 (row 0 contains the header).
Dim rowOfMaxInArray As Long
rowOfMaxInArray = 1
Dim i As Long, x As Double
For i = 2 To UBound(data)
x = data(i)(0)
If x > max Then
max = x
rowOfMaxInArray = i
End If
Next i
' if nHeaderRow = 0, i.e. the first row in the sheet, you could save a
' couple of lines by leaving the next statement out
Dim rowOfMaxInSheet As long
rowOfMaxInSheet = rowOfMaxInArray + nHeaderRow
oSheet.getRows().removeByIndex(rowOfMaxInSheet, 1)
End Sub
I'm trying to find the amount of sentences in this table:
Download Table here: http://www.mediafire.com/file/m81vtdo6bdd7bw8/Table_RandomInfoMiddle.mat/file
As you can see by the full-stops, there is one sentence in column one, and 2 sentences in column 3.
At the end of the day I desire to have a table with nothing but punctuation marks(with the exception of place holders like "", to keep the table rows the same length) that indicate the end of a sentence(Like "." or "?" or "!"), in order to calculate the total number of punctuation marks of each column.
This is my code(Yet unsuccessful):
EqualCoumns = [2:2:max(width(Table_RandomInfoMiddle))];
for t=EqualCoumns
MiddleOnlySentenceIndicators = Table_RandomInfoMiddle((Table_RandomInfoMiddle{:, t}=='punctuation'),:);
%Reomve all but "!.?" = Which is the only sentence enders
MiddleOnlySentenceIndicators(MiddleOnlySentenceIndicators{:, t} == ',', :) = [];
MiddleOnlySentenceIndicators(MiddleOnlySentenceIndicators{:, t} == ';', :) = [];
MiddleOnlySentenceIndicators(MiddleOnlySentenceIndicators{:, t} == ':', :) = [];
MiddleOnlySentenceIndicators(MiddleOnlySentenceIndicators{:, t} == '-', :) = [];
MiddleSentence_Nr(t) = height(MiddleOnlySentenceIndicators);
end
Right now this is almost giving good results, there is a little mistake somewhere.
(In the answer I would like to request only one thing, that I might have access to the results in the same table like form, it should look something like this(edited):
Any help will be appreciated.
Thank you!
If we use the table from my previous answer, t, we can use the following solution:
punctuation_table = table();
for col=1:size(t,2)
column_name = sprintf('Punctuation count for column %d',col);
punctuation_table.(column_name) = nnz(ismember(t(:,col).Variables,{'?',',','.','!'}));
end
which will create a table like this:
punctuation_table =
1×4 table
Punctuation count for column 1 Punctuation count for column 2 Punctuation count for column 3 Punctuation count for column 4
______________________________ ______________________________ ______________________________ ______________________________
2 0 2 0
This Macro is used to cut, insert and delete a cell range section of a workbook.
The problem I was trying to solve and gave up with the lack of response in another thread is why copying multiple non-adjacent rows to the MS clipboard often loses their row line-breaks when pasting.
E.g. Since trying to paste 3 non-adjacent rows into row 10, 11 and 12, often puts all 3 rows into row 10 with one row in fields A10-P10, the next row in Q10-AF10 and the last row into AG10-AV10...
I edited the Macro below to fix this mistake when this happens.
So, for example, I can now highlight row 10 and run the macro to cut/insert the fields Q10-AF10 to A11-P11 and delete/shift left the blank fields now in Q10-AF10.
I'm hoping for help to loop this process until there's no data outside Column A-P. In this case, no data outside cell P10.
Sub FixAllOnLine1OneRowAtATimeInsertToNextRow()
Application.ScreenUpdating = False
Dim copySheet As Worksheet
Dim pasteSheet As Worksheet
Set copySheet = ActiveSheet
Set pasteSheet = ActiveSheet
copySheet.Range("Q" & ActiveCell.Row & ":AF" & ActiveCell.Row).Copy
Range("Q" & ActiveCell.Row & ":AF" & ActiveCell.Row).Offset(1).Select
pasteSheet.Cells(ActiveCell.Row, 1).End(xlUp).Offset(1, 0).PasteSpecial xlPasteValues
Application.CutCopyMode = False
Application.ScreenUpdating = True
Columns("Q:AF").Select
Selection.Delete Shift:=xlToLeft
End Sub
Ok, I made some headway. I just have one super easy issue and then I need to loop it.
The first issue is that it cuts Column Q:AF correct of the row I've highlighted and shifts the entire Column Q:AF to the left, but it INSERTS the cut cells into the fixed range, A2:P2. I want to INSERT the cut cells down ONE row from my selection. I KNOW this is a couple characters in the Offset, I just can't get it.
Then, once it's working properly...say I highlight row 10, it cuts Q10:AF10 and instead INSERTS the cells into A11:P11 and shifts "Q:AF" to the left, then I need to figure out how to get it to loop until there's no more data to right of Column P. When this problem occurs pasting multiple rows from the clipboard all into the first row losing the row line-breaks, it's always quite a few rows.
Any ideas?
Thanks so much!
Mark
Sub FixAllOnLine1OneRowAtATimeInsertToNextRow()
Dim ws As Worksheet
Dim lNextRow As Long
Application.ScreenUpdating = False
Set ws = ActiveSheet
ws.Range("Q" & ActiveCell.Row & ":AF" & ActiveCell.Row).Copy 'Copy the row of the selected cell from Q:AF
ws.Range("Q" & ActiveCell.Row & ":AF" & ActiveCell.Row).Offset(1).Select 'Select the cells you have just copied. Not needed
ws.Cells(ActiveCell.Row, 1).End(xlUp).Offset(1, 0).Insert xlShiftDown ' Paste the copied values in to column "A" on next row?
'lNextRow = ws.Range("A" & Rows.count).End(xlUp).Row + 1 'Get Next Row number
'Range("A" & lNextRow).PasteSpecial xlPasteValues
Application.CutCopyMode = False
Range("Q:AF").Delete Shift:=xlToLeft
'Columns("Q:AF").Select
'Selection.Delete Shift:=xlToLeft
Application.ScreenUpdating = True
ActiveCell.Offset(RowOffset:=-1, columnOffset:=0).Activate 'Added to move active cell up one row to run it again for multiple groups to apply fix.
End Sub
Here's a solution in another direction just in case someone from the engines needs it...
Sub ReduceNoOfColumns()
Dim iRow As Integer 'Row to be manipulated
Dim iRowToPasteTo 'Row number to paste the copied cells
Dim iCurCol As Integer 'Current Column number of first cell with a value to cut
Dim NoOfCols As Integer 'integer to hold max number of columns
Dim sAddress As String
iRow = ActiveCell.Row
iRowToPasteTo = iRow + 1
NoOfCols = 16 'Set this number to the total number of columns you wish to have (in your case 16)
iCurCol = NoOfCols + 1
Do Until Cells(iRow, iCurCol).Value = "" 'Keep looping until we get to an empty column
sAddress = ColNoToLetter(iCurCol) & iRow & ":" & ColNoToLetter(iCurCol + NoOfCols - 1) & iRow
Rows(iRowToPasteTo & ":" & iRowToPasteTo).Insert Shift:=xlDown
Range(sAddress).Copy
Range("A" & iRowToPasteTo).PasteSpecial xlPasteAll
Range(sAddress).Clear
iCurCol = iCurCol + NoOfCols
iRowToPasteTo = iRowToPasteTo + 1
Loop
End Sub
Function ColNoToLetter(iCol As Integer) As String
Dim vArr
vArr = Split(Cells(1, iCol).Address(True, False), "$")
ColNoToLetter = vArr(0)
End Function
This question already has answers here:
how to count unique elements of a cell in matlab?
(2 answers)
Closed 7 years ago.
I want to determine the number of times a character appears in a character array, excluding the time it appears at the last position.
How would I do this?
In Matlab computing environment, all variables are arrays, and strings are of type char (character arrays). So your Character Array is actually a string (Or in reality the other way around). Which means you can apply string methods on it to achieve your results. To find total count of occurrence of a character except on last place in a String/Character Array named yourStringVar you can do this:
YourSubString = yourStringVar(1:end-1)
//Now you have substring of main string in variable named YourSubString without the last character because you wanted to ignore it
numberOfOccurrence = length(find(YourSubString=='Character you want to count'))
It has been pointed out by Ray that length(find()) is not a good approach due to various reasons. Alternatively you could do:
numberOfOccurrence = nnz(YourSubString == 'Character you want to count')
numberOfOccurrence will give you your desired result.
What you can do is map each character into a unique integer ID, then determine the count of each character through histcounts. Use unique to complete the first step. The first output of unique will give you a list of all possible unique characters in your string. If you want to exclude the last time each character occurs in the string, just subtract 1 from the total count. Assuming S is your character array:
%// Get all unique characters and assign them to a unique ID
[unq,~,id] = unique(S);
%// Count up how many times we see each character and subtract by 1
counts = histcounts(id) - 1;
%// Show table of occurrences with characters
T = table(cellstr(unq(:)), counts.', 'VariableNames', {'Character', 'Counts'});
The last piece of code displays everything in a nice table. We ensure that the unique characters are placed as individual cells in a cell array.
Example:
>> S = 'ABCDABABCDEFFGACEG';
Running the above code, we get:
>> T
T =
Character Counts
_________ ______
'A' 3
'B' 2
'C' 2
'D' 1
'E' 1
'F' 1
'G' 1
I need to search a text file that is about 30 lines. I need it to search row by row and grab numbers based on their position in the text file that will remain constant throughout the text file. Currently I need to get the first 2 numbers and then the last 4 numbers of each row. My code now:
FileToOpen = fopen(textfile.txt)
if FileToOpen == -1
disp('Error')
return;
end
while true
msg = fgetl(FileToOpen)
if msg == -1
break;
end
end
I would like to use the fgetl command if possible as I somewhat know that command, but if their is an easier way that will be more than welcome.
This looks like a good start. You should be able to use msg - '0' to get the value of the numbers. For ascii code the digits are placed next to each other in the right order (0,1,2,3,4,5,6,7,8,9). What you do when you subtract with '0' is that you subtract msg with the ascii code of '0'. You will then get the digits as
tmp = msg - '0';
idx = find(tmp>=0 & tmp < 10); % Get the position in the row
val = tmp(idx); % Or tmp(tmp>=0 & tmp < 10) with logical indexing.
I agree that fgetl is probably the best to use for text without specific structure. However, in case you have a special structure of the text you can use that and thus be able to use more effective algorithms.
In case you was actually after finding the absolute position of the digits in the text, you can save the msgLength = msgLength + length(msg) for every iteration and use that to calculate the absolute position of the digits.