Is there an easy way to copy a TableForm column and convert it into a vector?

Question

Is there an easy way to take a column output from a TableForm command, like this:

6.3121375196805859436
6.9003005747830988641
8.9768182443850834342
8.9138378561826490430
7.9892637649923361149
5.7310834764460301485
7.8127176986072821866
7.2290808062999679496
7.4923081278967455513

And convert it back into a "Table" or "List" or "Vector", i.e.,

    {6.3121375196805859436,6.9003005747830988641,8.9768182443850834342,8.9138378561826490430,7.9892637649923361149,5.7310834764460301485,7.8127176986072821866,7.2290808062999679496,7.4923081278967455513}

It is not so hard to do manually if there are only a dozen items, but when there are hundreds, it gets a bit tedious. What I've been doing is copy as "plain text", paste into Microsoft Word, replace "paragraph" marks with commas (Find ^p; Replace ,), put in the curly brackets, and copy it back into Mathematica. Surely there is an easier way?

Well, here is one way, and recall that I want to just translate within Mathematica itself:

StringReplace["{0.27053645617075607328
0.29665173751384943109
0.25127005359909123280
0.23465443792124858652
0.25447200323727824111
0.26802978565269905254
0.26622025533442188507
0.31359242691527905846
0.26810428054117807246
0.29408129579900499019
0.29001875144558125364
0.24823119448471783028
0.27626029400322289905
0.27347835282168678727
0.26924305884480351606
0.25995335230286428315
0.25585822835794553252
0.26573081058995874804
0.28819488915098074127
0.27186805052334818319
0.31654336155657666993}", "\n" -> ","]

That is, copy it as text, and paste it into the command

StringReplace["{paste here}", "\n" -> ","]

The output is then formatted correctly. I would have thought it would be a builtin command, and well, sort of, it is. Anyone else have an answer, please?

Answer 1

Clear["Global`*"]

TableForm should only be used for display. Assign a name to the input to TableForm and take the column that you want from the input.

SeedRandom[1234];

(mat = RandomReal[1, {9, 3}]) // TableForm

For the second column

mat[[All, 2]]

{* {0.521964, 0.0116446, 0.479332, 0.984993, 0.884729, 0.91956, 0.587943, \
0.696159, 0.632741} *}

Or, for the first column

mat[[All, 1]]

(* {0.876608, 0.377913, 0.543757, 0.759896, 0.459017, 0.263973, 0.98729, \
0.790215, 0.401549} *)

Answer 2

Using sample data

TableForm[RandomReal[{0, 1}, {5, 3}]]

Select middle column, Copy & Paste.

Add ; (optional), and evaluate, followed by

a = Flatten@%

This should set the middle column to a with the original precision.

Answer 3

<...> when the current context is dumped, all one may have is the printout of the table left of the original calculation, and if that takes 24 h to calculate, one would have the problem of translating a table into a vector whose origin is difficult to resurrect, in order to use it for further processing. That is the context I am most interested in.

If you have an output cell with TableForm output, you can create under it an input cell with the following code which will extract the contents of the TableForm from the output cell and assign it to a variable table:

table = First@
  Cases[NotebookRead[PreviousCell[]], 
   grid : TagBox[GridBox[__], _] :> ToExpression@grid, -1]

Here is how it can be used:

---

Another method is to right-click on the output cell bracket and select Convert To > InputForm. The result will look as follows:

Now you can easily select the contents of the recovered original TableForm input by double-clicking on the first curly bracket in the TableForm expression, after what you can simply copy the input matrix:

Answer 4

You could take the column with arrayName[[3]] or arrayName[[;;,3]] (if it is column three; the ;; is Span and captures all of that dimension) and export it to a file with Export[pathAndName.xlsx,arrayname[[appropriate designation]] to open it in Excel and copy from there.