Re: How to construct pure expressions

*To*: mathgroup at christensen.cybernetics.net*Subject*: [mg733] Re: [mg631] How to construct pure expressions*From*: villegas (Robert Villegas)*Date*: Wed, 12 Apr 1995 01:53:17 -0500*Keywords*: MapIndexed, index, atom*Organization*: Wolfram Research, Inc.

> x[ > x1[ > x11[ > x111[ x1111, x1112, x1113], > x112[ x1121, x1122, x1123], > x113[ x1131, x1132, x1133] > ], .. > In general, the list will have following properties: > * All expressions has a head that starts with the letter x. > > * The number after x correspond to the position list of that element. For > example, x211 is an element with position {2,1,1}. > > * The number of digits following x correspond to the level of that element. > For example, x111, x211 are elements in level 3, x1 or x2 is in level 1, > x is in level 0 because it has zero digits following x. Hello Xah, If you first make a dummy array that has the right structure, and then replace all the entries and the heads with "x" of the indices, you can do it with a short command. Note that all the places where your example wants x... are atomic, in that they're either bottom-level entries or they are heads. In Mathematica, this means we want to target level {-1} and we want Heads->True. Example for a 2 x 2 matrix, but slightly modifying the form of the x's from your example, just as an illustration: In[1]:= MapIndexed[x @@ DeleteCases[#2, 0] &, Table[Null, {3}, {3}], {-1}, Heads->True] Out[1]= x[][x[1][x[1, 1], x[1, 2], x[1, 3]], x[2][x[2, 1], x[2, 2], x[2, 3]], > x[3][x[3, 1], x[3, 2], x[3, 3]]] That makes it clear what MapIndexed is doing, and we can then just embellish it by replacing the "x @@ DeleteCases" business with a pre-defined function that takes care of creating the "x" symbols: MakeSymbol[base_String, {indices___, 0} | {indices___}] := ToExpression @ StringJoin[base, ToString /@ {indices}] IndexedExpression[base_String, dims:{___Integer?Positive}] := MapIndexed[MakeSymbol["x", #2]&, Array[Null &, dims], {-1}, Heads->True] (* I switched to using Array to make the code simpler given the way your example inputs dimensions *) Here's a test: In[4]:= IndexedExpression["x", {3, 3, 3}] Out[4]= x[x1[x11[x111, x112, x113], x12[x121, x122, x123], > x13[x131, x132, x133]], x2[x21[x211, x212, x213], > x22[x221, x222, x223], x23[x231, x232, x233]], > x3[x31[x311, x312, x313], x32[x321, x322, x323], x33[x331, x332, x333]]] Robby Villegas