Skip to main content

import - Assign the contents of a "Copy" operation programmatically to a variable?


While an earlier question Cut and paste data from a spreadsheet nominally addresses the issue, a year has past since its OP asked it and I hope that others have come across simpler or more direct ways to do this. I think a new solution might interest the community.



I often want to get values (numbers, dates, ...) from a spreadsheet someone has sent me into Mathematica.


Import[...] works fine, but seems a bit cumbersome, when I just want to grab something like a single column or row of values and work with them immediately.


"Copy/Paste" can give one odd results. Set up a short list of numbers in a spreadsheet ...


enter image description here


and copy them into Mathematica:


enter image description here


This seems a bit inconsistent and neither of these gives one a list of the data elements like Import[...].


Note: I use Mac OS X if it makes any difference.




Mr. Wizard supplies a pretty good solution in the original question, which references code for a palette that Szabolcs posted at see page, see Pasting tabular data into notebooks.



I've added a title and minor formatting to the palette:


CreatePalette[
Column[{Button["TSV",
Module[{data, strip},
data = NotebookGet[ClipboardNotebook[]][[1, 1, 1]];
strip[s_String] :=
StringReplace[s, RegularExpression["^\\s*(.*?)\\s*$"] -> "$1"];
strip[e_] := e;
If[Head[data] === String,
NotebookWrite[InputNotebook[],

ToBoxes@Map[strip, ImportString[data, "TSV"], {2}]]]],
ImageSize -> 85],
Button["CSV",
Module[{data, strip},
data = NotebookGet[ClipboardNotebook[]][[1, 1, 1]];
strip[s_String] :=
StringReplace[s, RegularExpression["^\\s*(.*?)\\s*$"] -> "$1"];
strip[e_] := e;
If[Head[data] === String,
NotebookWrite[InputNotebook[],

ToBoxes@Map[strip, ImportString[data, "CSV"], {2}]]]],
ImageSize -> 85],
Button["Table",
Module[{data},
data = NotebookGet[ClipboardNotebook[]][[1, 1, 1]];
If[Head[data] === String,
NotebookWrite[InputNotebook[],
ToBoxes@ImportString[data, "Table"]]]], ImageSize -> 85]},
Alignment -> {Center, Center}], WindowTitle -> "Paste Data",
WindowSize -> {85, Fit}]


One can install this by choosing Palettes > Install Palette (from the file menu) then provide the source file, palette name, and installation directory to the Install Mathematica Item dialog, and click Finish.




This works, but it does not have the immediacy that one gets used to when working with Mathematica, so...


Does a simple way exist to assign the contents of a "Copy" operation (from a spread sheet or csv file) programmatically to a variable as a list of elements, which one could immediately press into service?


Given that a clipboard appears to hold copies of content in different formats, an elegant solution might involve a right click to select how one wanted it pasted. Maybe a feature request to Wolfram would do it (eventually:-).



Answer



For Windows users something could perhaps be created using NETLink. For a basic demonstration:


Needs["NETLink`"];
InstallNET[];

LoadNETType["System.Windows.Forms.Clipboard"];

Now copy the data in Excel and evaluate:


ImportString[Clipboard`GetText[], "Table"]
(* {{1}, {2}, {3}, {4}, {5}} *)

Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...