Convert a table to text in excel - each cell on a new line [closed] - excel

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I have a table in excel, in the example below each number represents a cell:
11 12 13 14 15
21 22 23 24 25
31 32 33 34 35
I would like to convert this table to a text file that looks like this:
11 12
13
14
15
21 22
23
24
25
31 32
33
34
35
I am aware of the transpose function and a table to text but I can't seem to achieve what I illustrated above.

Answer 1

Using a simple VBA is efficient to create the actual file.
This code dumps data from the range from A1 to the last used cell in column E of the activesheet, to a file C:\temp\dummy.txt
Please change your path to suit
Sub RipData()
Dim X
Dim lngRow As Long
Dim lngCol As Long
Dim objFSO As Object
Dim objTF As Object
X = Range([a1], Cells(Rows.Count, "E").End(xlUp))
Set objFSO = CreateObject("scripting.filesystemobject")
Set objTF = objFSO.createtextfile("C:\temp\dummy.txt")
For lngRow = 1 To UBound(X, 1)
objTF.writeline X(lngRow, 1) & vbTab & X(lngRow, 2)
For lngCol = 3 To UBound(X, 2)
objTF.writeline X(lngRow, lngCol)
Next
objTF.writeline
Next
objTF.Close
End Sub

Answer 2

I usually apply a combination of ROUND, MOD, and OFFSET to turn a tabular set of data into a single column. This problem has the added wrinkle of wanting to put the second number in a group in a column to the right while skipping that number in the first column. And you want to put place a spacer row between each group of numbers. Both of these requirements make for more complicated than usual formulas.
For the first column, beginning in cell A5, or in another cell in column A the row of which is an even multiple of 5, use the following formula,
= IFERROR(
OFFSET(
$A$1,
ROUNDDOWN( ROW(A5) / 5 - 1, 0),
MOD( ROW(A5), 5) + (MOD( ROW(A5), 5) <> 0)
) /
(MOD( ROW(A5), 5) <> 4),
""
)
and copying down the column. This assumes that the data rows begin in cell A1.
For the second column, beginning with the cell in column B to the right of the starting cell in column A, enter this formula,
= IFERROR(
OFFSET($A$1,
ROUNDDOWN( ROW(B5) / 5 - 1, 0),
MOD( ROW(B5), 5) + 1
) /
NOT( MOD( ROW(B5), 5) > 0),
""
)
again copying it down.
How the formulas work
Both the column A and the column B function are elaborations of OFFSET, which takes as its arguments a starting address, the number of rows down (or up) its result range will begin, and the number of columns to the right (or left) that the range will. (It also takes two more arguments, which are the width and the height of the range to return. Since we're only concerned with single cells, we can leave these two arguments out.)
For example, the row calculation for column A uses the expression
ROUNDDOWN(ROW(A5)/5-1,0).
In cell A5, this formula resolves to (5/5 - 1) or 0, with no rounding needed. So, the row offset is 0 from A1, which makes sense because the A5 formula is processing the first row of data.
In cell A6, the formula becomes 6/5 - 1, or 1.2 - 1, or 0.2, which rounds down to 0, again what we need since we're still grabbing numbers from row 1.
This continues until cell A10, when we get 10/5 - 1, or 1. We're finished with the first row of data (which had an offset of 0 rows from a1) and now move on to the second. The value of the row offset will continue as 1 up to cell A15, when it will go up by 1 again.
The calculation of the column offset is a bit trickier:
MOD(ROW(A5),5)+(MOD(ROW(A5),5)<>0))
The first term is MOD(ROW(A5),5). In cell A5, that becomes MOD(5, 5), with a result of 0 since the integer remainder of 5 / 5 is 0. Again, makes sense--the column offset of 0 from cell A1 means the we will pick up the value in column A. In cell A6, we've got MOD(6, 5), for a column offset of 1. This means the value in A6 will come from column B.
But we don't want that: the value in column B of each data row is supposed to be shown in column B of the result range. We need to skip from column A to column C to get the next value for column A of the results.
Hence the second term (MOD(ROW(A5),5)<>0)). This evaluates to TRUE for every row that is not an even multiple of 5--these are the rows that show a (result) value in column A and column B. TRUE evaluates to 1 when used in an arithmetic expression. So, we are adding 1 to the column offset when the formula is in rows 6, 7, 8, 9, 11, 12, etc., thereby skipping over column B of the data row.
Finally, the divisor (MOD(ROW(A5),5)<>4). This expression will evaluate to TRUE (or 1) when the row the formula is in does not have a remainder of 4 when divided by 5. That means it evaluates to FALSE or 0 only when the formula is in rows 9, 14, 19, etc.
These in fact are the rows for which we want a space between the groups of numbers. The purpose of this divisor is to produce a formula error when the formula is in those rows. We then catch that error in the IFERROR function that encloses the entire formula, and the output of the formula becomes "" -- the empty string.
I won't go through the column B expression, which uses the same kind of maneuvers to pick up the second value in each data row and show it beside the first values.
For convenience in copying and pasting to your worksheet, here are the unformatted versions of the formulas:
Cell A5 Formula: =IFERROR(OFFSET($A$1,ROUNDDOWN(ROW(A5)/5-1,0),MOD(ROW(A5),5)+(MOD(ROW(A5),5)<>0))/(MOD(ROW(A5),5)<>4),"")
Cell B5 Formula: =IFERROR(OFFSET($A$1,ROUNDDOWN(ROW(B5)/5-1,0),MOD(ROW(B5),5)+1)/NOT(MOD(ROW(B5),5)>0),"")