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22

We can take advantage of the fact that IntegerDigits is very fast when the base is large. But not too large: no bigger than $2^{63}-1$ on a 64-bit system or $2^{31}-1$ on a 32-bit one, because Mathematica's machine integers are signed. Additionally, non-power-of-two bases require more work to get the result than just partitioning a bit-string, and are ...

21

matOP = {{0, 1, 0, 0, 1, 0, 1, 0}, {0, 1, 0, 0, 1, 0, 1, 0}, {0, 0, 1, 0, 1, 0, 1, 0}, {0, 0, 0, 1, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0}}; $\left( \begin{array}{cccccccc} 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & ... 21 The file appears to be a Unified Summary File from the Schlumberger Eclipse Reservoir Simulator. This file format uses Compaq Visual Fortran variable length record encoding. Mathematica does not offer any built-in functionality to read this file format, so we will have to parse it ourselves. We start by defining a convenience function to read big-endian ... 19 Leonid has shown how to get a big improvement in speed with Java and has mentioned that by using a library function written e.g. in C you could probably get further improvement as you can then avoid additional copies/data transfer. I just wanted to add an all Mathematica solution as it probably indicates how similar problems can be tackled where for ... 16 I was able to get 50x speedup w.r.t. your fastest code by using highly optimized Java buffered read functionality. The idea The idea is quite simple: use buffered read to reduce the IO overhead, and use Java to reduce the symbolic Mathematica overhead. Implementation You will have to run the Java reloader. Then, you call JCompileLoad@" import java.... 16 There is in fact an easy test to determine if an integer is a power of$2$, thanks to bit twiddling: hadamardMatrix := {{1}} hadamardMatrix := {{1, 1}, {1, -1}} hadamardMatrix[n_Integer /; Positive[n] && BitAnd[n, n - 1] == 0] := KroneckerProduct[hadamardMatrix, hadamardMatrix[n/2]] 15 This should be reasonably efficient. BitRotateRight[n_Integer, m_Integer] /; n > 0 && m >= 0 := BitShiftRight[n, m] + BitShiftLeft[Mod[n, 2^m], (BitLength[n] - m)] Will need to modify for left rotations if you want to support negative m. I added the restriction that n be positive. Not exactly sure what should be the behavior for negative n. ... 15 Solution It appears this bug is the result of attempted parallelism gone wrong. I believe it is corrected in all cases by setting this System Option: SetSystemOptions[ "ParallelOptions" -> {"MachineFunctionParallelThreshold2" -> Infinity} ] This appears to be an out and out bug and I tagged the question accordingly. Original observations: Compact ... 14 mtrx = {{0, 1, 0, 0, 1, 0, 1, 0}, {0, 1, 0, 0, 1, 0, 1, 0}, {0, 0, 1, 0, 1, 0, 1, 0}, {0, 0, 0, 1, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 1, 0, 0}, {0, 0, 0, 0, 1, 0, 0, 0}}; Once we re-number the elements of mtrx using a function like renumber = Module[{i = 1}, # /. 1 :> i++] &; (* thanks: Mr.W *) mtrx2 =renumber@mtrx ... 14 ClearAll[countZ] countZ = 1 ## & @@ Dimensions @ # - Total[#, 2] &; countZ@binarym 17 Timings: SeedRandom binm = RandomInteger[{0, 1}, {10000, 10000}]; countZ @ binm // AbsoluteTiming {0.23863, 49996224} WaveletsCountZeros@binm // AbsoluteTiming {0.520381, 49996224} (Length @ # - Total @ #) & @ Flatten @ binm // AbsoluteTiming (*... 13 I imported the data this way: data = Import["ftp://software.seg.org/pub/datasets/2D/Hess_VTI/timodel_vp.segy.gz", "Binary"]; nByteFile = Length@data nByteTrace = (nByteFile - 3600)/3617 nTraceSample = (nByteTrace - 240)/4 22573680 6240 1500 You have to convert bytes to IBM 32-bit floating-point yourself. This takes a list of bytes, ... 13 An interesting question which I've never specifically considered before. Some observations: Log/Log // FullSimplify Log2 Log[2, 8] 3 3 3 And @@ IntegerQ /@ Log2[2^Range] And @@ Table[IntegerQ@Log2[2^RandomInteger[5*^8]], {500}] True True Mathematica documentation explicitly states: Log2 gives exact integer or rational number ... 12 To decipher the header structure is not really a question that we should solve here, but I can show you how to read this file in a structured way. In order to assist you in deciphering and to demonstrate the necessary tools I have build the below file and data browser. Everything you need can be found in this short program. (*Use this out-commented part if ... 12 How about this? nextpow2[a_] := Ceiling @ Log[2, Abs @ a]; or nextpow2[a_] := Ceiling[RealExponent[a, 2]] The same thing in a different style: f1 = Ceiling @* Log2 @* Abs; (* v10 syntax *) Or: f2 = ⌈Log2 @ Abs @ #⌉ &; A plot: Plot[f2[x], {x, -10, 10}, Filling -> 0] 12 You can input numbers in any base up to 36 using the notation base^^digits. Digits over 9 are represented using a, b, c, ... You can print numbers in any base up to 36 using BaseForm. Thus, In:= a=2^^0.10101 Out= 0.65625 In:= BaseForm[a^2,2] Out//BaseForm= Subscript[0.0110111001, 2] Note that the internal representation of numbers doesn't ... 10 There isn't really such a thing as binary arithmetic (at least in Mathematica). Numbers can be represented in any base, and this user-visible representation is completely independent from how arithmetic is done. Try this: BaseForm[(2^^1010101011)*(2^^1111101110), 2] Things to look up: BaseForm Digits in numbers 10 You could try Tuples: Tuples[{0, 1}, 4] (* {{0,0,0,0},{0,0,0,1},{0,0,1,0},{0,0,1,1}, {0,1,0,0},{0,1,0,1},{0,1,1,0},{0,1,1,1}, {1,0,0,0},{1,0,0,1},{1,0,1,0},{1,0,1,1}, {1,1,0,0},{1,1,0,1},{1,1,1,0},{1,1,1,1}} *) 10 bin = 1011011; words = (bin // IntegerDigits) /. {1 -> "one", 0 -> "zero"}; (str = StringJoin[Riffle[words, " "]]) // InputForm (* "one zero one one zero one one" *) Speak[str] 10 Updated to include suggestions from comments My original idea, using BitAnd, switching$0\leftrightarrow1\$, and multiplying. This idea uses 3 vectorized binary operations: erode1[list_] := Times[ list, Subtract[ 1, BitAnd[list, PadRight[list, Length@list, 0, 1]] ] ] Here is @Shadowray's improvement, which uses 2 vectorized ...

10

twosComplement[x_, n_] := UnitBox@IntegerDigits[x, 2, n] twosComplement[35, 8] {1, 1, 0, 1, 1, 1, 0, 1}

9

LengthWhile[#, Positive] & /@ binarym {3, 4, 2, 4, 6} Or binarym /. {x : Longest[1 ..], __} :> Length[{x}]

9

I will turn Kuba's comment into an answer. The EncryptedObject you create using Encrypt contains the encrypted data and the initialization vector of the encryption method as a ByteArray. These can be turned into an array of digits using Normal and IntegerDigits. key = GenerateSymmetricKey[]; message = "Hello world"; eobj = Encrypt[key, message]; bin = eobj[...

9

I don't think there is a built-in function to generate the two's complement representation. Easy to implement though. twosComplement[x_, n_] := IntegerDigits[2^x - n, 2, n] twosComplement[35, 8] (* {1, 1, 0, 1, 1, 1, 0, 1} *)

9

The problem is the following: Mod[Plus @@ IntegerDigits[n, 2], 2] Mod[2 + n, 2] This happens because IntegerDigits stays unevaluated for non integer inputs: IntegerDigits[n, 2] IntegerDigits[n, 2] Note that this sum remains unevaluated: Sum[ThueMorse[n]/(n + 1)^s, {n, 0, Infinity}] Sum[ThueMorse[n]/(n + 1)^s, {n, 0, Infinity}] Ideally Alpha should ...

8

The following function is based on the Ramp and Differences functions, as suggested in a comment by @garej . Its speed and low memory are surprising. rampDiff[list_] := Ramp@Prepend[Differences[list], First[list]] It was tested against the following functions from previous answers and comments: ClearAll["Global*"] erode1[list_] := Times[list, ...

8

Update: ff works as is in version 9. For versions 10+, as commented by @Mr.Wizard, it needs to be modified to prevent the index i from exceeding the length of the input list: ffX[m_] := With[{n = Length@First@m}, Module[{i = 1}, While[i <= n && Positive[#[[i]]], ++i]; i - 1] & /@ m] (* thanks: @Mr.Wizard *) Original answer: ClearAll[f1] ...

7

The short answer is, you can't. The only function that can really consume these objects is not exposed at the language level but only in the kernel's own C code. To be honest, am not sure why it is even in the system. My best guess is that there a few obscure case where it might be generated by internal code. Given that we don't even document it anymore (...

7

Using old-school GraphPlot: Edit: the comments that broke the code have been removed. input = {{0,1,0,0,1,0,1,0}, {0,1,0,0,1,0,1,0}, {0,0,1,0,1,0,1,0}, {0,0,0,1,0,0,1,0}, {0,0,0,0,1,1,0,0}, {0,0,0,0,1,0,0,0}}; new = Module[{i = 1}, input /. 1 :> i++]; Developer`PartitionMap[Union @@ # &, %, {2, 2}, 1]; Union @@ DeleteCases[%, 0, {-1}]; ...

7

If I am not mistaken the 212 format is storing two signals and each 3-byte group provides a 12-bit reading of that pair. Importing with Byte isn't too bad for this set: data = Import["http://physionet.org/physiobank/database/mitdb/100.dat", "Byte"] ~ Partition ~ 3 Then two helpers to parse it: twosComplement[a_, n_] := If[a < 2^(n - 1), a, a - 2^n] ...

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