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31

Preamble This is a very good question, because answering it will make it very clear what immutability means, both in general and in the context of Associations. General A few general words on immutability Associations are immutable data structures. This means that they carry no state, and a copy of an Association is another completely independent ...


20

I hesitate to add anything after @Leonid's comprehensive answer, but I'd like to point out that an easy way to achieve the stated goal is to define f like this: f[x_] := <| x, "isFirstValueTrue" -> x@"firstValue" |> ... which yields the desired result when mapped across the associations in x: f /@ x (* { <|"firstValue" -> True, ...


16

Please see the Utility function section for a concise summary. An arbitrary density plot for the example: den = DensityPlot[Sin[x] Sin[y], {x, -180, 180}, {y, -90, 90}] : Extract the graphics primitives from the density plot: prim = First @ Cases[den, Graphics[a_, ___] :> a, {0, -1}, 1]; Plot them directly with GeoGraphics while setting the ...


13

This is because of the compilation that kicks in automatically if the list in Map exceeds a certain number of elements. "MapCompileLength" /. ("CompileOptions" /. SystemOptions["CompileOptions"]) (* Out: 100 *) shows that the default setting is that if the list contains more than 100 elements then Map will be compiled. MapThread on the other hand does not ...


12

Use Map with a levelspec of {-1}: Map[g, {a, b, {c, d}, {{e}}}, {-1}] {g[a],g[b],{g[c],g[d]},{{g[e]}}}


11

This perhaps: Function[{a, b}, a[#]/b[#] &] @@@ {{a, b}, {c, d}, {e, f}} (* Out: {a[#1]/b[#1] &, c[#1]/d[#1] &, e[#1]/f[#1] &} *) Mr.Wizard's way of writing it (see comment) looks like this in the frontend:


11

Map is automatically compiled. Yes, even with RandomChoice. Try it: f = Compile[{{p, _Real, 0}, {t, _Integer, 1}}, Map[# RandomChoice[{p, 1 - p} -> {1, 0}] &, t] ]; f // InputForm (* -> clean bytecode *) Check its performance: p = 0.1; t = Table[1, {10^6}]; SeedRandom[1000]; AbsoluteTiming[a = Map[# RandomChoice[{p, 1 - p} -> {1, 0}] ...


10

This is really a natural fit for Outer: t = Table[{i, j}, {i, 1, 2}, {j, 1, 2}]; Outer[Apply, {Plus, Subtract, Times, Divide}, t, 2] (* ==> {{{2, 3}, {3, 4}}, {{0, -1}, {1, 0}}, {{1, 2}, {2, 4}}, {{1, 1/2}, {2, 1}}} *)


9

I would prefer to use the "GeoImage" styling, because you can use other projections when using it. Let's say you have data for the whole world in a matrix: data = Table[ Sin[x Degree] Sin[y Degree], {y, -90, 90}, {x, -180, 180}] Then you use ListDensityPlot: den1 = ListDensityPlot[data, AspectRatio -> 1/2, Frame -> None, PlotRangePadding ...


8

If you look at SystemOptions[], like so, Column[ OpenerView /@ (Replace[SystemOptions[], Rule[x_, y_] -> List[x, y], 1]) ] you see that under CompileOptions, if you click on the triangle to open it, there is an option "MapCompileLength" -> 100. Set it to eg 10 and see it it helps (do SetSystemOptions["CompileOptions" -> ...


8

"Reflected" padding works as desired but "Periodic" padding is missed. There is corresponding definition for "Reflected" RandomProcesses`TemporalDataUtilitiesDump`toCannonicalPadding[ RandomProcesses`TemporalDataUtilitiesDump`td_, "Reflected", RandomProcesses`TemporalDataUtilitiesDump`w_, RandomProcesses`TemporalDataUtilitiesDump`Caller_] := ...


7

Here is one way: movingMapCircular[f_, l_List] := MapThread[f@* List, {l, RotateLeft[l]}]; For example: movingMapCircular[f, {1, 2, 3, 4}] (* {f[{1, 2}], f[{2, 3}], f[{3, 4}], f[{4, 1}]} *) A generalization of this approach for arbitrary window size may look like: ClearAll[movingMapCircular]; movingMapCircular[f_, l_List, {n_Integer}] := MapThread[ ...


7

I would use the following, Quiet[ Check[ PowerMod[#, -1, 126] + #, ##&[], PowerMod::ninv ]& /@ Range[125], PowerMod::ninv ] (* {2, 106, 34, 110, 106, 92, 34, 146, 142, 92, 146, 124, 128, 106, 160, 110, 106, 92, 160, 146, 142, 92, 146, 124, 128, 106, 160, 110, 106, 218, 160, 146, 142, 218, 146, 250} *) where ##&[] inserts a ...


7

This duplicates the behavior of yours (no effect on zeroes at ends): smoothee=ReplacePart[#, i_ /; i > 1 && i < Length@# && #[[i]] == 0 :> Mean[{#[[i - 1]], #[[i + 1]]}]] &; smoothee[{0, 1, 3, 4, 6, 8, 0, 11, 12, 0, 13, 0}] (* {0, 1, 3, 4, 6, 8, 19/2, 11, 12, 25/2, 13, 0} *) Here's a goofy ...


7

MovingMap is doing, what it is supposed to do. Evaluating AbsoluteTime /@ (Data[[1 ;; 5]][[All, 1]]) {3439843200, 3440102400, 3440188800, 3440275200, 3440361600} gives the timestamps for the first five data points in absolute time. The output of MovingMap[foo[#] &, Data, 5] or simpler MovingMap[foo[#] &, Data[[1 ;; 5]], 5] ...


7

nPoints = 9; lst = RandomReal[{-1, 1}, {nPoints, 3}]; lst = Partition[lst, {2}, 1] EuclideanDistance[First[#], Last[#]] & /@ lst ps. after posting saw Guesswhoitis above. I think using @@@ is nicer there. I used simple Mapping to do the job. So for completion, a less typing solution is as shown above by Guesswhoitis which is EuclideanDistance @@@ ...


7

You very nearly had it. What you need, instead of Map[], is Apply[]. This can then be combined with Map[], like so: mat = Table[{i, j}, {i, 2}, {j, 2}]; Apply[#, mat, {2}] & /@ {Plus, Subtract, Times, Divide}


6

First, your construction of f is a bit malformed. You are using the operator form of Select, therefore the Select expression itself acts as a function; you do not need to add @# & to it. Use instead: f = Select[#[[2]] >= 10 &]; The reason that your operations are not the same can be seen by mapping a dummy function foo: foo /@ a <|1 ...


6

molekyla777's answer can be very helpful but it is not technically correct. The question specifies "every element of a list" but using a levelspec of {-1} will apply the function to every atomic element regardless of its head: Map[f, 1 + 5 x + 10 x^2 + 10 x^3, {-1}] f[1] + f[5] f[x] + f[10] f[x]^f[2] + f[10] f[x]^f[3] Of course this can be very ...


6

You can almost always turn to replacement patterns when you need to transform expressions: Cases[ {{a, b}, {c, d}, {e, f}}, {x_, y_} :> (x[#]/y[#] &) ] {a[#1]/b[#1] &, c[#1]/d[#1] &, e[#1]/f[#1] &} Cases defaults to levelspec {1} so this is safer than using /..


6

Here is a way to use MapThread: MapThread[Function[{r, s}, r + # & /@ s], {tensorR, tensorS}, 2] For numeric tensors, it can be compiled to squeeze out a little more performance: Compile[{{tensorR, _Real, 3}, {tensorS, _Real, 4}} , MapThread[Function[{r, s}, r + # & /@ s], {tensorR, tensorS}, 2] ] Performance Measurements @kguler's solution ...


6

Since version 9 you do not need to do anything extra. tab1 = {{a1, a1 + a2}, {b1, b2*b2}} tab2 = {{2, 5}, {5, 2}} Solve[tab1 == tab2] {{a2 -> 3, b1 -> 5, a1 -> 2, b2 -> -Sqrt[2]}, {a2 -> 3, b1 -> 5, a1 -> 2, b2 -> Sqrt[2]}}


6

ds[MapIndexed[Append[#1, "age" -> ages[[First@#2]]] &]] {<|"name" -> "bob", "age" -> 1|>, <|"name" -> "joe", "age" -> 2|>} Note 10.0.2 throws a warning - who knows what's going on in that private type system: First::normal: Nonatomic expression expected at position 1 in \ First[TypeSystem`ZSignatures`PackagePrivate`i] ...


6

Here are two possibilities. First, use MovingMap: ClearAll[av]; av[{l_, 0, r_}] := (l + r)/2; av[{_, m_, _}] := m; and then smoothMM[list_] := Join[{First@list}, MovingMap[av, list, 3], {Last@list}] or, you can use in-place assignments: smooth2[list_] := Module[{copy = list, pos = Flatten[Position[list[[2 ;; -2]], 0]] + 1 }, copy[[pos]] = ...


6

Update: ClearAll[imtF] imtF[foo_] := Module[{i = 1}, foo[#, i++] & /@ Transpose@#] & Examples: imtF[#2 (Plus @@ #) &][{{a, b, c}, {e, f, g}}] (* {a + e, 2 (b + f), 3 (c + g)} *) xx = {{a, b, c}, {e, f, g}, {x, y, z}}; imtF[#2 (Plus @@ #) &][xx] (* {a + e + x, 2 (b + f + y), 3 (c + g + z)} *) imtF[Plus @@ Times@## &][xx] (* {a + e + ...


6

While I'm not sure why the error you see is generated, you can fix your sumT function by taking the fF call out of the Map form: sumT = Compile[ {{tab, _Real, 2}}, Total[fF /@ Map[enn[#[[1]], #[[2]]] &, tab]], CompilationTarget -> "C"]; This worked fine for me in version 10.1 of Mathematica: sumT[tab] // AbsoluteTiming (* ==> {0.000667, ...


6

It happens because Compile cannot infer the types returned by the subsidiary functions enn and fF. In the second approach, you partially solved this by specifying it manually for enn. In principle you could have done that in the first approach as well if you had specified it for fF too, but in practice getting the type inference to work out correctly is not ...


5

Update Well I guess I should retire for the evening to a less brain-intensive activity as apparently I can't think clearly. One could of course use Outer: Outer[Compose, {f, g}, {a, b, c}] {{f[a], f[b], f[c]}, {g[a], g[b], g[c]}} However I recommend that you do not do this as you will not gain the auto-compilation of Map, meaning this method will ...


5

Another method with a single plot: ClearAll[myFunc2]; yvalues = {0.1`, 0.5`, 1.`, 2.`, 3.`}; myFunc2[x_?NumericQ, y_?NumericQ] := ConditionalExpression[Sin[x y], IntervalMemberQ[Interval[{0, Cos[y]}], x]]; {min, max} = Through[{Min, Max}[Cos[yvalues]]]; Plot[myFunc2[x, #] & /@ yvalues, {x, min, max}, Evaluated -> True, BaseStyle -> ...


5

A method using a single Plot expression: myFunc[x_, y_] := Sin[x y] yvalues = {0.1, 0.5, 1.0, 2.0, 3.0}; Block[{x}, Plot[#, {x, -5, 5}, PlotRange -> All, PlotLegends -> Automatic] &[ If[Cos[#] <= x <= 0 || 0 <= x <= Cos[#], myFunc[x, #]] & /@ yvalues ] ] Notes: I use Block to keep x localized. I did not programmatically ...



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