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| `^^` isn't a function. It's a syntax. Just as `2.` means "the numeric version of 2" and not the integer, `16^^ab` means "the number 171". This is handled by the parser, not the interpreter which is why `FullForm[Hold[16^^A]]` is `Hold[10]`. ... |
| Mathematica 10 comes with, and should use, its own installation of Java (specifically a Java 7, I believe at `$InstallationDirectory/SystemFiles/Java`). There should be no need to install Java yourself, even on Mac. I'm suspicious of the space... |
| I wish I could provide some insight as to **why** this is happening -- unless Workbench was still running or did not exit cleanly? -- but I believe that you'll find that you have a file `\WolframWorkspaces\Base\.metadata\.lock` (and similar for the... |
| You can also use Ordering.In[205]:= Ordering[{0.1, 0.03, 0, 4}, {1}] Out[205]= {3}(because the Min is actually 0, because 0 |
| In[25]:= ToString[Unevaluated[foo /@ {1, 2}], InputForm, CharacterEncoding -> "ASCII"] Out[25]= "foo /@ {1, 2}" |
| The first 2 don't work because the Condition on the rhs is interpreted as a condition. The third one doesn't work because HoldPattern doesn't prevent the interpretation of Condition, merely evaluations. The last one doesn't work because Verbatim... |
| For those that assume the last line is terminated by a new line, here's another approach[mcode]newlines = 1 - Unitize[bytes - 10]; formfeeds = 1 - Unitize[bytes - 13]; both = Rest[newlines]*Most[formfeeds]; Total[newlines] + Total[formfeeds] -... |
| Or SetAlphaChannel on the one that's missing a transparency. |
| Oops; that's obviously a factor of 100, not 1000. And over 10 of that (down to ~.01s) is recovered if you wrap the Append[...] in Dispatch. |
| Well, an absolutely terrible way of solving the specific question is to use Trace: [mcode]In[72]:= getUsage2[s_String]:=Extract[Trace[Symbol[s]],2,Apply[Function[symb,MessageName[symb,"usage"],{HoldFirst}],#]&] In[73]:= getUsage2["Plot"] Out[73]=... |