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Version 10 empowered!


(also known as Spartacus)


See those two pins? Well today I'm going to... stick them in you.


Ok, let me come clean: Mr.Wizard is actually an alias of mine... I'm glad I can finally confess. ― Sjoerd C. de Vries

No, I'm Spatac... I mean, Mr.Wizard! ― R.M

I'm Mr.W, and so's my wife! ― acl


@Spartacus, it was not fixable. ― rcollyer

check again; I just fixed it. :-p ― Spartacus


"This is the internet, where the men are men, the women are men and the children are FBI agents..." ― yoda Dec 18 '11 at 6:06


"My main purpose of being here is in collecting the votes. I'd even go further and say that this is my main purpose in life." ― Leonid Shifrin


"I am making some voodoo dance and invoking the rain God while we talk" ― belisarius


Some people are fortunate. Others get married. ― belisarius


It's not that easy to do the untangling even manually though. – Szabolcs

Maybe you could turn it into a puzzle and let other people do the untangling :-) – Heike

I could make a web based game, and feed all my graphs to the visitors.
SE is a big game, isn't it? :D – Szabolcs


"ListContourPlot crush on simple 5-points data" (Question title)

"Yes, this must be love at first sight."ruebenko


... but I am working with a large body of code written by an incompetent fool (i.e. myself, several years ago...) – Simon Woods


I was tempted to edit your title: "Do rules rule?" But my nose is still bleeding as a result of a limerick. – belisarius


Project Euler


2d
comment Proper EdgeForm setting so boundary appears entirely within shape
I have wanted this myself but I have never found a solution within EdgeForm itself. I too have ended up shrinking shapes to make them fit. As far as I know Mathematica lacks the "inside stroke" method that you describe. Once again I hope I am proven wrong. :-)
2d
comment How can I display the numbers on the ordinate axis as exponentiation?
I believe this question is a duplicate of: (5369). Please review that Q&A and tell me if your question is not answered there.
2d
revised How can I display the numbers on the ordinate axis as exponentiation?
added 25 characters in body
2d
comment difficulty abstracting the OptionValue pattern
@seismatica I guess it would, but ## & is shorter. :-)
2d
revised difficulty abstracting the OptionValue pattern
added 288 characters in body
2d
revised difficulty abstracting the OptionValue pattern
edited tags
2d
answered difficulty abstracting the OptionValue pattern
2d
revised Selectively updating individual graphics in a dynamic, multiple-plot Show
deleted 1 character in body
2d
comment Blurred frontend text in Mathematica 10, Windows, High-DPI Display
Also, this seems to be closely related: (18419). I don't know if Mathematica 10 for Windows supports high-PPI displays.
2d
comment Blurred frontend text in Mathematica 10, Windows, High-DPI Display
Which version of Windows are you running?
2d
answered Selectively updating individual graphics in a dynamic, multiple-plot Show
2d
comment When must I use the Return function?
@m_goldberg Okay. :-)
2d
comment Optimization: run this process multiple times
I am closing this question because "output: it doesn't output as a table" is not true and there is no other question asked, other than the one trivially solved above.
2d
comment Optimization: run this process multiple times
@kuku Your table is a series of lists, each with four values. If you wish to plot x versus b you will need to extract these values. tab[[All, {1, 2}]] does this. Look up Part is the documentation.
2d
comment When must I use the Return function?
@m_goldberg That's why I also included a link, but only a link, in my answer. Although I hope that my answer is complementary to yours I also wanted to make it stand on its own so I thought it important to reference some guides to the rather mysterious behavior of Return.
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awarded  version-10
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answered When must I use the Return function?
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awarded  Nice Answer
2d
comment How to get the mean color of an artwork?
Clearly your result is closer to the OP's original, apart from the 4x4 "pixel" thing.
2d
comment How to get the mean color of an artwork?
@MichaelE2 Would you say that (42825) explains this issue, at least pragmatically?