Re: Apply and up/down value questions

*To*: mathgroup at smc.vnet.net*Subject*: [mg61641] Re: Apply and up/down value questions*From*: dh <dh at metrohm.ch>*Date*: Mon, 24 Oct 2005 21:06:58 -0400 (EDT)*References*: <djcfvq$6a8$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Matt, see below. sincerely, Daniel Matt wrote: > Hello, > Again, I'm working my way through some of the programming examples > in Michael Trott's Programming Guidebook, and there's behaviour that I > cannot figure out. I've been over and over section A.5.2 and 2.5.10 in > the Mathematica Book, but I still don't understand up values when > applied to functions with more than one argument. Also, I had a > question about level specification, which I couldn't figure out even > after looking at section A.3.6. Any help is appreciated. First the > upvalue questions: > > I don't understand the point of using TagSet as opposed to UpSet: > e.g. in Michael Trott's Programming Guidebook on page 318 he states: > "For functions with several arguments, the information can be > associated with a certain prescribed argument rather than with all > arguments at the first level." However, I don't see any difference in > functionality between this: > consider the following: Clear[a1, a2]; f[a1, a2] ^:= 3; ??a1 Global`a1 a1 /: f[a1, a2] := 3 ??a2 Global`a2 a2 /: f[a1, a2] := 3 Clear[a1, a2]; a1 /: f[a1, a2] := 3; ??a1 Global`a1 a1 /: f[a1, a2] := 3 ??a2 Global`a2 here the assignment is only done to the explicitely specified a1. Clear[a1, a2]; f[a1, a2_] ^:= 3; ??a1 Global`a1 a1 /: f[a1, a2_] := 3 ??a2 Global`a2 here a2 is is not a variable, but the name of a pattern. That is why nothing is assigned to a2. > > Also, I'm stumped on this particular wording in the TagSet > documentation: > > "If f appears several times in lhs, then f/: lhs = rhs associates the > assignment with each occurrence" > In my oppinion, this is missleading and confusing. I think it only means : assignment to f > What does this really mean? > > > As regards Apply: > > Is it true that Apply[newHead, expr, 1] and Apply[newHead, expr, {1}] > will always be the same result given that the other two arguments are > identical? > > By definition, a level specification of n applies to levels 1 through > n. I have found that if I do something like Apply[newHead, expr, -2], > that what I actually end up with seems to be equivalent to newExpr = > Apply[newHead, expr, {1}] followed by Apply[newHead, newExpr, {-2}]. > Here is an example: > > Cell[BoxData[{ > RowBox[{ > RowBox[{"a1", " ", "=", " ", > RowBox[{"Array", "[", > RowBox[{"\[DoubleStruckR]", ",", > RowBox[{"{", > RowBox[{"2", ",", "2"}], "}"}], ",", > RowBox[{"{", > RowBox[{"2", ",", "4"}], "}"}]}], "]"}]}], > ";"}], "\[IndentingNewLine]", > RowBox[{"MatrixForm", "[", "a1", "]"}], "\[IndentingNewLine]", > RowBox[{ > RowBox[{"a11", " ", "=", " ", > RowBox[{"Apply", "[", > RowBox[{"newHead", ",", "a1", ",", > RowBox[{"{", "1", "}"}]}], "]"}]}], > ";"}], "\[IndentingNewLine]", > RowBox[{"MatrixForm", "[", "a11", "]"}], "\[IndentingNewLine]", > RowBox[{ > RowBox[{"a12", " ", "=", " ", > RowBox[{"Apply", "[", > RowBox[{"newHead", ",", "a11", ",", > RowBox[{"{", > RowBox[{"-", "2"}], "}"}]}], "]"}]}], > ";"}], "\[IndentingNewLine]", > RowBox[{"MatrixForm", "[", "a12", "]"}], "\[IndentingNewLine]", > RowBox[{ > RowBox[{"a13", " ", "=", " ", > RowBox[{"Apply", "[", > RowBox[{"newHead", ",", " ", "a1", ",", " ", > RowBox[{"-", "2"}]}], "]"}]}], > ";"}], "\[IndentingNewLine]", > RowBox[{"MatrixForm", "[", "a13", "]"}]}], "Input"] > > What is the general rule, or am I just misusing the level specification > in Apply? from the Help: " A negative level number -n represents all parts of an expression that have depth n. The depth of an expression, Depth[expr], is the maximum number of indices needed to specify any part, plus one. Levels do not include heads of expressions, except with the option setting Heads -> True. Level 0 is the whole expression. Level -1 contains all symbols and other objects that have no subparts. " This is a bit confusing. If you consider an expression as a tree, -n counts from the leaves. -1 means all atoms at the leafs. -2 all expression starting 1 level below(or above, depending on how your tree grows, up or down) the leaves. e.t.c. The Help specifies the same a bit more confusing. It means "all expressions ending in a leaf that has depth n". -1 means depth 1, that is expressions that can take at most 0 indices, the atoms. -2 means depth 2, that is expressions that can take at most 1 indices, e.t.c. Examples: Clear[a1, a2, a3, a4]; t = {a1, {a2, {a3, {a4}}}}; Level[t, {-1}] {a1, a2, a3, a4} this are the atoms at the leaves. Clear[a1, a2, a3, a4]; t = {a1, {a2, {a3, {a4}}}}; Level[t, {-2}] {{a4}} {a4} is the only subexpression that can take at most 1 index Clear[a1, a2, a3, a4]; t = {a1, {a2, {a3, {a4}}}}; Level[t, {-3}] {{a3, {a4}}} expression {a3, {a4}} can take at most 2 indices Clear[a1, a2, a3, a4]; t = {a1, {a2, {a3, {a4}}}}; Level[t, -1] {a1, a2, a3, a4, {a4}, {a3, {a4}}, {a2, {a3, {a4}}}} these are all possible subexpressions Clear[a1, a2, a3, a4]; t = {a1, {a2, {a3, {a4}}}}; Level[t, -2] {{a4}, {a3, {a4}}, {a2, {a3, {a4}}}} here the atoms with level -1 are missing With other functions like: Apply, Cases, Count, FreeQ, Map, MapIndexed, Position, Replace and Scan it should work the same, but the effect is not as clearly seen as with Level > > And finally, a question on modified built in functions: > > In Michael Trott's Programming Guidebook, on page 311, he uses the > following construct to remove a downvalue that had been specified for > the Cos and Sin functions: > > Unprotect[Cos]; > Clear[Cos]; > Protect[Cos]; > {Cos} > > I've noticed that if I omit the last line (i.e. {Cos}), that it still > works. What is the point of the last line? > I agree there is not point. > Thanks very much, > > Matt >

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