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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] ...


5

Since c is common to all the entries, I would do this: MapThread[f[##, c] &, {{a1, a2}, {b1, b2}}] (* ==> {f[a1, b1, c] , f[a2, b2, c]} *) Here the ## stands for SlotSequence and accepts the pair of arguments fed into it by Inner, taken from the two Lists. This is based on constructing a pure function (identified by the & at the end) that is ...


5

Map[f, Unevaluated@{x, y, z}] (* {True, True, True} *) Update: ClearAll[f, x, y, z, q, w, e, r, t] f[vars : {(_) ..}] := Map[ValueQ, Unevaluated@vars]; SetAttributes[f, HoldAllComplete]; f[{x}] (* {False} *) x = 10; f[{x}] (* {True} *) f[{x, y}] (* {True, False} *) f[{x, y, z, q, w, e, r, t}] (* {True, False, False, False, False, False, False, False} ...


4

Uuuh, that's like a breeze.... ace= {1, 14, 27, 40}; uno = Partition[RandomSample[Range[52], 52], 13]; Intersection[#, ace] & /@ uno This returns the Aces returned in each hand let's go gambling....hi,hi,hi


4

Or set another attribute, Listable and be done with Map: SetAttributes[f,{HoldAll,Listable}]; x=10; y=.; z=.; f[{x,y,z}] (* {True, False, False} *)


3

Just write simpler code! Here's an example... first, the setup: cards = 52; uno = Partition[RandomSample[Range[cards], cards], cards/4]; aces = 1 + {0, 1, 2, 3} cards/4; (it's a bit easier to test if you set cards = 12). FreeQ[ Intersection[aces, #] & /@ uno, {}] Enjoy.


3

Thread[f[{a1, a2}, {b1, b2}, c]] (* {f[a1, b1, c], f[a2, b2, c]} *) Thread[f[{a1, a2}, {b1, b2}, c1 -> c11]] (* {f[a1, b1, c1 -> c11], f[a2, b2, c1 -> c11]} *)


3

lst = {a1, a2, a3, a4, a5}; Transpose[{Cos[lst], Sin[lst]}]


3

lst = {a1, a2, a3, a4, a5}; {Cos@#, Sin@#} & /@ lst (* {{Cos[a1], Sin[a1]}, {Cos[a2], Sin[a2]}, {Cos[a3], Sin[a3]}, {Cos[a4], Sin[a4]}, {Cos[a5], Sin[a5]}} *) Or Through@{Cos, Sin}@# & /@ lst (* same result *)


3

If you can't use Listable to your code is probably to use: ClearAll[f] (*first definition handles cases when you use list:={x,y};f[list]*) f[listof_Symbol] := f @@ (Hold[listof] /. OwnValues[listof]); (*general definition handles list of symbols*) f[vars_ : {(_) ..}] := ReleaseHold @ Map[ValueQ, Hold[vars], {2}]; (*as in kugler's answer the line ...


2

The two standard approaches to this problem, Unevaluated and Listable, have already been posted. If you need a different kind of evaluation control please consider my original method:(1),(2),(3),(4) Load step How do I evaluate only one step of an expression? then: SetAttributes[f2, HoldAll] f2[s_Symbol] := step[s] /. {_[x_List] :> f2 /@ ...


2

Your operation using the v10 operator form for MemberQ, along with Alternatives: uno = RandomSample[Range @ 52] ~Partition~ 13; And @@ MemberQ[1 | 14 | 27 | 40] /@ uno Or using modular arithmetic as already provided by others: x = RandomSample @ Range @ 52; Sign[x ~Mod~ 13] ~Partition~ 4 // Total {12, 11, 13, 12} This is the number of non-ace ...


2

Just for fun: You can count number of "aces" (obviously arbitrary choice) 4 sets of 13 cards: Total /@ Partition[ Boole[Mod[#, 13] == 1] & /@ RandomSample[Range@52, 52], 13] For simulation: tab[n_] := Table[Total /@ Partition[Boole[Mod[#, 13] == 1] & /@ RandomSample[Range@52, 52], 13], {n}]; Now cheating by just testing ...


2

Use this instead: FindInstance[#, {a, b}]& /@ {a + b > 1 && a + b < 2, a + b > 2 && a + b < 3} or, in fact, this: FindInstance[#, {a, b}]& /@ {{a + b > 1, a + b < 2}, {a + b > 2, a + b < 3}}


1

Here's one way... g[1][x_] := 1; g[2][x_] := x; g[3][x_] := x^2; norm[r_, s_] := Integrate[r[x]*s[x] + r'[x]*s'[x], {x, 0, 1}]; gram=Array[norm[g[#1], g[#2]] &, {3, 3}] or... gram=Table[norm[a, b], {a, {g1, g2, g3}}, {b, {g1, g2, g3}}]; or... gram=Outer[norm, {g1, g2, g3}, {g1, g2, g3}]


1

Your answers are correct. The first two solutions occur "at infinity" but your last two (after Chop) are {x->-4, y->0} and {x->0, y->0}, as is correct.



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