Skip to main content

installation - How can I install packages distributed as .paclet files?


I downloaded a Mathematica package. It came as a file with the .paclet extension. How can I install or uninstall it?




Answer



This answer is for people who are not interested in package development, and just need to install a Mathematica package which they obtained as a .paclet file.


Note: Before using the functions in this guide, it may be necessary to evaluate Needs["PacletManager`"]. Specifically, this is needed when your notebooks's default context is set to "Unique to This Notebook".


Installing


Packages distributed as paclets can be installed using the PacletInstall function. Evaluate this function with its argument the full path to the downloaded .paclet file.


For example, if SomePackage-1.0.0.paclet was downloaded into the ~/Downloads directory, then evaluate


PacletInstall["~/Downloads/SomePackage-1.0.0.paclet"]

You may use Insert → File Path... to easily insert the path to this file.


Getting information



.paclet files are typically named as SomePackage-1.0.0.paclet. Here SomePackage is the paclet name and 1.0.0 is its version. To get information about the active version of a currently installed paclet, use PacletInformation with its name:


PacletInformation["SomePackage"]

To list all installed versions of a paclet, use


PacletFind["SomePackage"]

Typically the latest version will be the active one. This is listed first by PacletFind.


Uninstalling


To uninstall all versions of a paclet named SomePackage, use


PacletUninstall["SomePackage"]


Warning: Mathematica uses many system paclets. Do not attempt to uninstall these. Only uninstall paclets that you have installed yourself.


Paclets with documentation


Packages that come with documentation may show up in the Documentation Center's Add-ons browser. Open the Documentation through the Help menu and look for the Add-ons and Packages link in the lower right corner:



Clicking it will show a list of those installed packages (both paclets and traditional ones) that come with documentation. The list may have buttons for opening the package's documentation or for uninstalling it.


Information for developers


If you are looking to distribute your own packages as paclets, read the following threads:



Comments

Post a Comment

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[ ...
Post a Comment
Read more

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...
Post a Comment
Read more

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...
Post a Comment
Read more