# Fourier, numbers-as-text, and macros.

N

#### Neal Carron

The output of the Fourier Transform fills most cells as numbers (all those
that are complex numbers), but all those that are real it fills as text(!)
The output of the inverse Fourier Transform fills all cells with
numbers-as-text(!)
My output column is formatted to Number, General. But the inverse FT result
is still written as text (they are all real numbers).
I need them as bonafide numbers (to be plotted).
Converting them to numbers is done by selecting the output cells, opening
the error icon on the upper left, and selecting "convert to number".
But this last action is not accepted when recording the strokes in a macro.

The steps suggested in HELP on "Convert numbers stored as text to numbers"
don't work.
So:
Can the FT routine be set to enter results directly as numbers?
If not, how do I convert numbers-as-text to numbers in a Macro?
- Neal

M

#### meh2030

The output of the Fourier Transform fills most cells as numbers (all those
that are complex numbers), but all those that are real it fills as text(!)
The output of the inverse Fourier Transform fills all cells with
numbers-as-text(!)
My output column is formatted to Number, General. But the inverse FT result
is still written as text (they are all real numbers).
I need them as bonafide numbers (to be plotted).
Converting them to numbers is done by selecting the output cells, opening
the error icon on the upper left, and selecting "convert to number".
But this last action is not accepted when recording the strokes in a macro.

The steps suggested in HELP on "Convert numbers stored as text to numbers"
don't work.
So:
Can the FT routine be set to enter results directly as numbers?
If not, how do I convert numbers-as-text to numbers in a Macro?
- Neal

Neal,

Have you tried =VALUE(A1), =A1+1-1, or =A1*1?

Best,

Matthew Herbert

D

#### Dana DeLouis

Hi. The output is a complex number represented in Excel as a string.
=Complex(5,0) will show up as 5, but as a string, much like you are seeing.

=ImReal(B1) will return the Real part.

Due to 'rounding issues', you may prefer to use
IMABS(B1) is you believe the Complex part is suppose to be zero.

The advantage also is that it will convert those cells that do have very
small imaginary parts.

HTH
Dana DeLouis

N

#### Neal Carron

Matthew,
I don't see how to carry out your operations on a column that has already
been filled with numbers-as-text.
You mean insert a temporary column A for the FT, and carry out one of your
suggestions to define the numeric value in the final column B? Like B1=A1*1.

N

#### Neal Carron

Dana,
All true, but I don't see how to apply your method to a column.
It seems I'd have to create a temporary column B, evaluate the FT in it, and
then set another column =ImReal(B1) as the final answer.
Is that what you meant?
- Neal

D

#### Dana DeLouis

Hi. This is a shorter / modified version of something I use.
Maybe you can get some ideas here.

Take the inverse of the data, and output it beginning at E1.
Note that you are assuming the data will be integers. This may not be
the case. You may be able to adjust it to your situation.
(with integer output, are you doing convolution, or high-precision
multiplication ?? )

Sub YourRoutine()
Call FFTInverse([C1:C16], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

For Each Cell In OutCell.Resize(N).Cells
Cell.Value = CDbl(WorksheetFunction.ImAbs(Cell))
Next Cell
End Sub

= = = = =
HTH
Dana DeLouis

N

#### Neal Carron

Dana,
My original function is real. None of the values is integer.
Its FFT is complex.

The inverse FFT is real, and it comes out of the Fourier routine as
numbers-as-text.
..ImReal(Cell)]

The complex FFT remains a problem. Those values that are complex appear in
their cells as numbers. But those that have zero imaginary part are text.
I can't identify ahead of time which ones will be text.
I need a function applying to all cells, that will convert text to real
numbers, and leave complex numbers alone.
How do I do that?
- Neal

Dana DeLouis said:
Hi. This is a shorter / modified version of something I use.
Maybe you can get some ideas here.

Take the inverse of the data, and output it beginning at E1.
Note that you are assuming the data will be integers. This may not be
the case. You may be able to adjust it to your situation.
(with integer output, are you doing convolution, or high-precision
multiplication ?? )

Sub YourRoutine()
Call FFTInverse([C1:C16], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

For Each Cell In OutCell.Resize(N).Cells
Cell.Value = CDbl(WorksheetFunction.ImAbs(Cell))
Next Cell
End Sub

= = = = =
HTH
Dana DeLouis

Neal said:
Dana,
All true, but I don't see how to apply your method to a column.
It seems I'd have to create a temporary column B, evaluate the FT in it, and
then set another column =ImReal(B1) as the final answer.
Is that what you meant?
- Neal

D

#### Dana DeLouis

Hi. I don't think I understand the question.
I think you want if a cell has a 0 imaginary part, convert the real part
to a number, and not string. If the cell does have an imaginary part,
then leave it alone.

Sub YourRoutine()
Call FFTInverse([C1:C4], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell
Const k As Double = 0.0000000001

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

With WorksheetFunction
For Each Cell In OutCell.Resize(N).Cells
If Abs(.Imaginary(Cell)) < k Then
Cell.Value = CDbl(.ImAbs(Cell))
End If
Next Cell
End With
End Sub

Suppose your calculation was suppose to return a real number "5".
Due to rounding, the vector could be rotated slightly.
Your vector of '5 might now be "4.9999975+.0049999999i"
Your use of ImReal will return 4.9999975.
We can get rid of small rotation errors via Abs.

?worksheetfunction.Imabs("4.9999975+.0049999999i")
5.00000000000053

Again, just something to consider.
= = = =
HTH :>)
Dana DeLouis

Neal said:
Dana,
My original function is real. None of the values is integer.
Its FFT is complex.

The inverse FFT is real, and it comes out of the Fourier routine as
numbers-as-text.
.ImReal(Cell)]

The complex FFT remains a problem. Those values that are complex appear in
their cells as numbers. But those that have zero imaginary part are text.
I can't identify ahead of time which ones will be text.
I need a function applying to all cells, that will convert text to real
numbers, and leave complex numbers alone.
How do I do that?
- Neal

Dana DeLouis said:
Hi. This is a shorter / modified version of something I use.
Maybe you can get some ideas here.

Take the inverse of the data, and output it beginning at E1.
Note that you are assuming the data will be integers. This may not be
the case. You may be able to adjust it to your situation.
(with integer output, are you doing convolution, or high-precision
multiplication ?? )

Sub YourRoutine()
Call FFTInverse([C1:C16], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

For Each Cell In OutCell.Resize(N).Cells
Cell.Value = CDbl(WorksheetFunction.ImAbs(Cell))
Next Cell
End Sub

= = = = =
HTH
Dana DeLouis

Neal said:
Dana,
All true, but I don't see how to apply your method to a column.
It seems I'd have to create a temporary column B, evaluate the FT in it, and
then set another column =ImReal(B1) as the final answer.
Is that what you meant?
- Neal

:

Hi. The output is a complex number represented in Excel as a string.
=Complex(5,0) will show up as 5, but as a string, much like you are seeing.

=ImReal(B1) will return the Real part.

Due to 'rounding issues', you may prefer to use
IMABS(B1) is you believe the Complex part is suppose to be zero.

The advantage also is that it will convert those cells that do have very
small imaginary parts.

HTH
Dana DeLouis

Neal Carron wrote:
The output of the Fourier Transform fills most cells as numbers (all those
that are complex numbers), but all those that are real it fills as text(!)
The output of the inverse Fourier Transform fills all cells with
numbers-as-text(!)
My output column is formatted to Number, General. But the inverse FT result
is still written as text (they are all real numbers).
I need them as bonafide numbers (to be plotted).
Converting them to numbers is done by selecting the output cells, opening
the error icon on the upper left, and selecting "convert to number".
But this last action is not accepted when recording the strokes in a macro.

The steps suggested in HELP on "Convert numbers stored as text to numbers"
don't work.
So:
Can the FT routine be set to enter results directly as numbers?
If not, how do I convert numbers-as-text to numbers in a Macro?
- Neal

N

#### Neal Carron

Dana,
Thank you. That worked.
[In your last Cell.Value = CDbl(.ImAbs(Cell)), I changed it to
..ImReal(Cell)]

This seems like an awfully laborious way just to change a number-as-text
to a number.
Any idea why the Fourier routine doesn't just put out bonafide numbers?
- Neal

Dana DeLouis said:
Hi. I don't think I understand the question.
I think you want if a cell has a 0 imaginary part, convert the real part
to a number, and not string. If the cell does have an imaginary part,
then leave it alone.

Sub YourRoutine()
Call FFTInverse([C1:C4], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell
Const k As Double = 0.0000000001

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

With WorksheetFunction
For Each Cell In OutCell.Resize(N).Cells
If Abs(.Imaginary(Cell)) < k Then
Cell.Value = CDbl(.ImAbs(Cell))
End If
Next Cell
End With
End Sub

Suppose your calculation was suppose to return a real number "5".
Due to rounding, the vector could be rotated slightly.
Your vector of '5 might now be "4.9999975+.0049999999i"
Your use of ImReal will return 4.9999975.
We can get rid of small rotation errors via Abs.

?worksheetfunction.Imabs("4.9999975+.0049999999i")
5.00000000000053

Again, just something to consider.
= = = =
HTH :>)
Dana DeLouis

Neal said:
Dana,
My original function is real. None of the values is integer.
Its FFT is complex.

The inverse FFT is real, and it comes out of the Fourier routine as
numbers-as-text.
.ImReal(Cell)]

The complex FFT remains a problem. Those values that are complex appear in
their cells as numbers. But those that have zero imaginary part are text.
I can't identify ahead of time which ones will be text.
I need a function applying to all cells, that will convert text to real
numbers, and leave complex numbers alone.
How do I do that?
- Neal

Dana DeLouis said:
Hi. This is a shorter / modified version of something I use.
Maybe you can get some ideas here.

Take the inverse of the data, and output it beginning at E1.
Note that you are assuming the data will be integers. This may not be
the case. You may be able to adjust it to your situation.
(with integer output, are you doing convolution, or high-precision
multiplication ?? )

Sub YourRoutine()
Call FFTInverse([C1:C16], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

For Each Cell In OutCell.Resize(N).Cells
Cell.Value = CDbl(WorksheetFunction.ImAbs(Cell))
Next Cell
End Sub

= = = = =
HTH
Dana DeLouis

Neal Carron wrote:
Dana,
All true, but I don't see how to apply your method to a column.
It seems I'd have to create a temporary column B, evaluate the FT in it, and
then set another column =ImReal(B1) as the final answer.
Is that what you meant?
- Neal

:

Hi. The output is a complex number represented in Excel as a string.
=Complex(5,0) will show up as 5, but as a string, much like you are seeing.

=ImReal(B1) will return the Real part.

Due to 'rounding issues', you may prefer to use
IMABS(B1) is you believe the Complex part is suppose to be zero.

The advantage also is that it will convert those cells that do have very
small imaginary parts.

HTH
Dana DeLouis

Neal Carron wrote:
The output of the Fourier Transform fills most cells as numbers (all those
that are complex numbers), but all those that are real it fills as text(!)
The output of the inverse Fourier Transform fills all cells with
numbers-as-text(!)
My output column is formatted to Number, General. But the inverse FT result
is still written as text (they are all real numbers).
I need them as bonafide numbers (to be plotted).
Converting them to numbers is done by selecting the output cells, opening
the error icon on the upper left, and selecting "convert to number".
But this last action is not accepted when recording the strokes in a macro.

The steps suggested in HELP on "Convert numbers stored as text to numbers"
don't work.
So:
Can the FT routine be set to enter results directly as numbers?
If not, how do I convert numbers-as-text to numbers in a Macro?
- Neal

D

#### Dana DeLouis

Any idea why the Fourier routine doesn't just put out bonafide numbers?

Hi. Excel represents Complex numbers as strings, so I guess this is the
only way to properly output the data.
Even though a number like =Complex(3,0) has only a real part of '3,
Excel represents this as the string '3.
One may think that "Val" would be a possible solution.
Just note that when you take the Val of a complex number with a zero
real, you are getting the complex part, and is probably not what you want.

Sub Demo()
Dim x As Double
Dim c As String
With WorksheetFunction
c = .Complex(0, 5.001)
x = Val(c)
End With
End Sub

'returns: 5.001

As a side note: Custom vba Fourier routines usually do not use
strings. Instead, they keep the Real, and Imaginary numbers in separate
arrays. Maybe something to look at in the future ??

= = = =
HTH :>)
Dana DeLouis

Neal said:
Dana,
Thank you. That worked.
[In your last Cell.Value = CDbl(.ImAbs(Cell)), I changed it to
.ImReal(Cell)]

This seems like an awfully laborious way just to change a number-as-text
to a number.
Any idea why the Fourier routine doesn't just put out bonafide numbers?
- Neal

Dana DeLouis said:
Hi. I don't think I understand the question.
I think you want if a cell has a 0 imaginary part, convert the real part
to a number, and not string. If the cell does have an imaginary part,
then leave it alone.

Sub YourRoutine()
Call FFTInverse([C1:C4], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell
Const k As Double = 0.0000000001

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

With WorksheetFunction
For Each Cell In OutCell.Resize(N).Cells
If Abs(.Imaginary(Cell)) < k Then
Cell.Value = CDbl(.ImAbs(Cell))
End If
Next Cell
End With
End Sub

Suppose your calculation was suppose to return a real number "5".
Due to rounding, the vector could be rotated slightly.
Your vector of '5 might now be "4.9999975+.0049999999i"
Your use of ImReal will return 4.9999975.
We can get rid of small rotation errors via Abs.

?worksheetfunction.Imabs("4.9999975+.0049999999i")
5.00000000000053

Again, just something to consider.
= = = =
HTH :>)
Dana DeLouis

Neal said:
Dana,
My original function is real. None of the values is integer.
Its FFT is complex.

The inverse FFT is real, and it comes out of the Fourier routine as
numbers-as-text.
.ImReal(Cell)]

The complex FFT remains a problem. Those values that are complex appear in
their cells as numbers. But those that have zero imaginary part are text.
I can't identify ahead of time which ones will be text.
I need a function applying to all cells, that will convert text to real
numbers, and leave complex numbers alone.
How do I do that?
- Neal

:

Hi. This is a shorter / modified version of something I use.
Maybe you can get some ideas here.

Take the inverse of the data, and output it beginning at E1.
Note that you are assuming the data will be integers. This may not be
the case. You may be able to adjust it to your situation.
(with integer output, are you doing convolution, or high-precision
multiplication ?? )

Sub YourRoutine()
Call FFTInverse([C1:C16], [E1])
End Sub

Sub FFTInverse(InData, OutCell)
Dim N
Dim Cell

N = InData.Cells.Count
OutCell.Resize(N).Clear

Run "ATPVBAEN.XLAM!Fourier", InData, OutCell, True, False

For Each Cell In OutCell.Resize(N).Cells
Cell.Value = CDbl(WorksheetFunction.ImAbs(Cell))
Next Cell
End Sub

= = = = =
HTH
Dana DeLouis

Neal Carron wrote:
Dana,
All true, but I don't see how to apply your method to a column.
It seems I'd have to create a temporary column B, evaluate the FT in it, and
then set another column =ImReal(B1) as the final answer.
Is that what you meant?
- Neal

:

Hi. The output is a complex number represented in Excel as a string.
=Complex(5,0) will show up as 5, but as a string, much like you are seeing.

=ImReal(B1) will return the Real part.

Due to 'rounding issues', you may prefer to use
IMABS(B1) is you believe the Complex part is suppose to be zero.

The advantage also is that it will convert those cells that do have very
small imaginary parts.

HTH
Dana DeLouis

Neal Carron wrote:
The output of the Fourier Transform fills most cells as numbers (all those
that are complex numbers), but all those that are real it fills as text(!)
The output of the inverse Fourier Transform fills all cells with
numbers-as-text(!)
My output column is formatted to Number, General. But the inverse FT result
is still written as text (they are all real numbers).
I need them as bonafide numbers (to be plotted).
Converting them to numbers is done by selecting the output cells, opening
the error icon on the upper left, and selecting "convert to number".
But this last action is not accepted when recording the strokes in a macro.

The steps suggested in HELP on "Convert numbers stored as text to numbers"
don't work.
So:
Can the FT routine be set to enter results directly as numbers?
If not, how do I convert numbers-as-text to numbers in a Macro?
- Neal