# Tag Info

15

str = {"1,2,3,5,10,12,13,17,26,30,32,41,42,43,113,115,121,125"} Flatten@ToExpression@StringSplit[str, ","] Short explanation: After executing StringSplit you get a list of separated "StringNumbers" like {{"1", "2", ... "125"}} ToExpression converts these "StringNumbers" to Integers. Flatten removes the outermost brackets. You can even omit Flatten ...

13

The coolest way is to check the answer to this question by David Carraher. I am shamelessly stealing his code here to write a function that gives you rules for up to maxNumber: ordinalRule[maxNumber_Integer] /; maxNumber > 0 := Block[{p}, Thread[ Function[{x}, x -> StringSplit[SpokenString[p[[#]]]][[2]] &[x] // Quiet] /@ ...

12

TeXForm is indeed your friend. It even gives you nicely formatted code: Table[RandomInteger[10], {3}, {4}] // TeXForm gives (* \left( \begin{array}{cccc} 9 & 5 & 10 & 9 \\ 6 & 10 & 3 & 9 \\ 9 & 5 & 9 & 7 \\ \end{array} \right) *)

12

Nested WolframAlpha approach, showing the intermediate steps: numberString[a_, k_: 10] := FixedPointList[ StringReplace[#, b : (DigitCharacter ..) :> WolframAlpha["spell " <> b, {{"Result", 1}, "Plaintext"}]] &, a, k] numberString["123456"] (* ==> {"123456", "123 thousand and 456", "one hundred twenty-three \ thousand and ...

12

data = FinancialData["SPY", "Jan. 1, 2011"] /. {d_List, v_} :> {AbsoluteTime@d, v}; model = a x^4 + b x^3 + c x^2 + d x + e; fit = FindFit[data, model, {a, b, c, d, e}, x] modelf = Function[{x}, Evaluate[model /. fit]] Plot[modelf[x], {x, Min@data[[All, 1]], Max@data[[All, 1]]}, Epilog -> Map[Point, data]] Edit Better (tick labels showing dates) ...

10

While this is overkill, I'm just trying everything I can do with these new (in V10) and exciting Mesh and Region functions. So here we go: f[x_, y_] := -E^(-(1 + x)^2 - y^2)/3 + 3*E^(-x^2 - (1 + y)^2)*(1 - x)^2 - 10*E^(-x^2 - y^2)*(x/5 - x^3 - y^5); gr = Plot3D[f[x, y], {x, -3, 3}, {y, -3, 3}, PlotRange -> All, PlotPoints -> 100]; We ...

9

I've got my own package that I've used for a few years to generate LaTeX from Mathematica. All the labs on my Mathematica course page were produced with this package. Here's a handout on probability theory for Calc II students that was produced by the package. Unfortunately, it's not at all polished and really not usable by anyone but me. I can present ...

9

Messy but a working method inWords[n_] := Module[ {r, numNames = {"", " one", " two", " three", " four", " five", " six", " seven", " eight", " nine"}, teenNames = {" ten", " eleven", " twelve", " thirteen", " fourteen", " fifteen", " sixteen", " seventeen", " eighteen", " nineteen"}, tensNames = {"", " ten", " twenty", " ...

9

There are two parts to accessing the contents of a Java Map object. The first is to traverse the iterator interface to extract the map elements. The second is to use accessor methods on those elements to extract their properties. For the purpose of discussion, let's create a map from strings to Java date objects: Needs["JLink"] InstallJava[]; $map = ... 9 I noticed only afterwards that you were asking for a built-in method. I'm not certain if you would actually want to rely on an undocumented feature instead of writing your own, since it's not hard. This requires v10 for StringTemplate. It could be replaced with something else, of course... Range[120] /. { tens_Integer /; Floor[Mod[tens, 100], 10] == 10 ... 8 You can start with that. StringCases["8.0,1034(*g opx ksp pl ilm liq q *)", x__ ~~ y : ("(*" ~~ ___ ~~ "*)") :> {x, y}] {{"8.0,1034", "(*g opx ksp pl ilm liq q *)"}} And You can do with the first elements what You need. Edit I can not be passive after Jacob's remarks :) Following solution is based on his but simplified in order to avoid ... 8 You can use the Range function in this way: Range[1, 2, 0.1] to get {1.,1.1,1.2,1.3,1.4,1.5,1.6,1.7,1.8,1.9,2.} If the problem is how to pass the parameter a you can use: 1- Using Apply. a = {1, 2, 0.1} Range@@a The Apply operator (@@) "decapitates" the List head from list a and changes it by Range. 2- Using Sequence a = {1, 2, 0.1}; ... 8 In Mathematica 8 you can use free form input : = Spell 15 and you get "fifteen" Or just write = thirty and you obtain 30 Since for larger numbers this approach yields expressions like {number words number, _ } one could nest this arbitrarily to obtain expressions containing only words. In fact, it is sufficient to nest only two times. For ... 8 This seems to work on my system at least, but as Mr.Wizard said it might be system dependent lineHeight = 1.5; conversion = 10;(*magic number*) scrollToThis = 80; paneHeight = 200; pos = (scrollToThis - 1/2)*lineHeight*conversion - paneHeight/2; Framed[ Pane[ Grid[List /@ data, Frame -> All, ItemSize -> {5, lineHeight}, ... 8 The reason your original code fails is that the TreeFrom object is only formatted as Graphics object, meaning that it converted for display rather that as part of the normal evaluation sequence. You can convert to and from box form to recover your Graphics object: tf = TreeForm[a + b^2 + c^3 + d]; gr = tf // ToBoxes // ToExpression gr /. (x_Framed ... 7 It seems that every time you open a new cell with no style, that is "", it automatically puts the span tag to it. To to avoid this we can use this rule: "ConversionRules" -> { "" -> {"", ""} } So now, evaluating ExportString[ Cell[TextData[{"This is an equation: ", Cell[BoxData[ FormBox[RowBox[{RowBox[{"f", "(", "x", ... 7 Don't use BaseForm[number,base] Use IntegerDigits[number,base] : IntegerDigits[10, 2] (* ---> {1, 0, 1, 0} *) It returns a List of Integers, which is a very good thing for further processing 7 N A one-character answer is disallowed by SE, so I will expand. N is mostly what I use. If I have an expression like$2 x + 3\$, I sometimes write it 2. x + 3. in Mathematica; then if x is numeric, whether it happens to be an Integer or not, the expression will always be Real or Complex.

7

I would use ImportString myself: string = "1,2,3,5,10,12,13,17,26,30,32,41,42,43,113,115,121,125"; ImportString[string, "CSV"] {{1, 2, 3, 5, 10, 12, 13, 17, 26, 30, 32, 41, 42, 43, 113, 115, 121, 125}} Are you using Mathematica for the operation which I imported from a homepage? There may be a more direct method in that case. Regarding your ...

7

Assuming that your 3D plot is in Graphics3D format, you should be able to just extract the points on the graph and use ListContourPlot. f[x_, y_] := -E^(-(1 + x)^2 - y^2)/3 + 3*E^(-x^2 - (1 + y)^2)*(1 - x)^2 - 10*E^(-x^2 - y^2)*(x/5 - x^3 - y^5); cp = ContourPlot[f[x, y], {x, -3, 3}, {y, -3, 3}, PlotRange -> All, PlotLabel -> "Computed from ...

6

About the first section: This simulates the behavior of Mathematica a bit more closely than Kuba's answer, creating an additional RowBox. However, the choice of how general the different aspects of the solution are may be a bit odd. Kuba's answer feels cleaner. Use the OPs definition of str. The following will generate an input cell where the comments are ...

6

You can access many different font characteristics via CurrentValue. Here is an approximation to convert between ItemSize and ImageSize: itemSize = {10, 10}; Overlay[{ Grid[ {{"Sample", "Text"}}, Frame -> All, Spacings -> {0, 0}, ItemSize -> itemSize, Alignment -> {Left, Center}], Row[{ Framed["Sample", ImageSize -> ...

6

Update As @VCL pointed out in his comment, just exporting a list of graphics does not work since the braces and commas of the list a exported as well. Additionally, the pdf is one single page. Here is an updated approach, which takes all imported pdf-pages and inserts them into a new notebook where every page is separated by a pagebreak. The resulting pdf ...

6

num = 1234567891234567899; triples = Reverse /@ Reverse@Partition[Reverse@IntegerDigits[num], 3, 3, 1, 0]; threePowers = {"septillion", "sextillion", "quintillion", "quadrillion", "trillion", "billion", "million", "thousand", ""}; singleRules = {0 -> "", 1 -> "one", 2 -> "two", 3 -> "three", 4 -> "four", 5 -> "five", 6 -> ...

6

This solution is similar in spirit to Prashant's. Though not particularly elegant, I avoid any calls to W|A and any other form of internet connectivity. Further down the post I also provide a solution to the inverse problem of returning the number when given English words. numberform[n_]:=With[{id=IntegerDigits@n}, ...

6

This question feels familiar but I could not find a true duplicate. You can use the string replacement that Oleksandr proposed here and then use ToExpression to convert the numbers: string = "-5.100686209408900133332e+02 -1.294005398404007344443e+01 \ -2.59376479781563728887e-02 -1.3043629998334040122222e+02" ToExpression /@ ...

6

From this answer NetworkGraphPlot`ExprTreePlot[a+b^2+c^3+d] /. (x_Framed :> Print[x])

6

A brute force but very simple approach (no undocumented functions needed) that I think will work with any version of Mathematica. ordinalRule[n_Integer /; Mod[n, 100] == 11] := generalRule[n] ordinalRule[n_Integer /; Mod[n, 100] == 12] := generalRule[n] ordinalRule[n_Integer /; Mod[n, 100] == 13] := generalRule[n] ordinalRule[n_Integer /; Mod[n, 10] == 1] = ...

6

{1970, 1, 1, 0, 0, 0} is shorthand for a DateObject, namely this one: DateObject[{1970, 1, 1, 0, 0, 0}] This represents January 1, 1970 in your current time zone. The origin of UNIX time is January 1, 1970 UTC: DateObject[{1970, 1, 1, 0, 0, 0}, TimeZone -> 0] Note now that the returned DateObject displays with the time zone GMT: For example, in ...

5

It is due to precedence. Need to use () in this case. Times @@ (Subscript[x, #] & /@ {1, 3, 7}) See this when-is-fg-not-the-same-as-fg topic for table of precedence in Mathematica.

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