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I have here a complicated bit of code that is not pretty nor easy to follow, but it represents a simplification of a larger body of code I am working with. I am a Mathematica novice and have already received some help on this issue from stackoverflow but it is still not solving my problem. Here is the code for which I hope you can follow along and assume what I am trying to get it to do. Thanks to you programming whizzes for the help.

a[b_, c_] = -3*b + 2*c + d + e + f;

g[b_, c_] := If[a[b, c] < 0, -3*a[b, c], a[b, c]];

h[T_, b_, c_] = (T/g[b, c]);

i[h_, T_, b_, c_] := If[h[T, b, c] > 0, 4*h[T, b, c], -5*h[T, b, c]];

j[b_, c_] := If[a[b, c] < 0, 5*a[b, c], 20*a[b, c]];

XYZ[h_, T_, i_, g_, j_, b_, c_] = T*i[h, T, b, c]*g[b, c] + j[b, c]

rules = {a -> 1, b -> 2, c -> 3, d -> 4, e -> 5, f -> 6, T -> 10};

XYZ[h, T, i, g, j, b, c] //. rules
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  • What exactly is the question? I think you should explain what you are trying to do and where you believe the error lies. Commented May 24, 2011 at 22:45
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    Also, as you refer to the previous question, it is wise to refer to it. Commented May 24, 2011 at 22:45
  • The goal is to get a numeric output for XYZ after replacing all the variables, however the above code does not work. Commented May 24, 2011 at 22:50
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    Last entry of the rules list, did you mean to have T -> 10? Commented May 24, 2011 at 22:56
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    Also the first rule a->1 doesn't make sense, since you also defined a function with that name, whose head will now be replaced by 1. After getting rid of that rule the expression evaluates to a numeric value. Commented May 24, 2011 at 23:07

1 Answer 1

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Preserving as much of your code as possible, it will work with just a few changes:

a[b_, c_] := -3*b + 2*c + d + e + f;

g[b_, c_] := If[# < 0, -3 #, #] & @ a[b, c]

h[T_, b_, c_] := T / g[b, c]

i[h_, T_, b_, c_] := If[# > 0, 4 #, -5 #] & @ h[T, b, c]

j[b_, c_] := If[# < 0, 5 #, 20 #] & @ a[b, c]

XYZ[h_, T_, i_, g_, j_, b_, c_] := T*i[h, T, b, c]*g[b, c] + j[b, c]

rules = {a -> 1, b -> 2, c -> 3, d -> 4, e -> 5, f -> 6, T -> 10};

XYZ[h, T, i, g, j, b, c] /. rules

(* Out= 700 *)

  1. If statements are again externalized, as in the last problem.

  2. all definitions are made with SetDelayed (:=), as a matter of good practice.

  3. The presumed error T - 10 in your rules is corrected to T -> 10

Notice that again ReplaceRepeated (//.) is not needed, and is changed to /.

We still have a nonsensical rule a -> 1 but it does not cause a failure.

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6 Comments

Thanks Mr. Wizard! This helped immensely! I now know how to use the # symbol.
@ZB18749 I am glad that helps. I recommend skimming Leonid Shifrin's online book to learn about Mathematica. Also, if are not already doing it, you can select a name or symbol and press F1 to get more information about. For example, if you select & in the code and press F1, one of the results is Function.
@ZB18749, something else, as a new Mathematica user, even if your code works, it is unlikely to be optimal. At some point you may want to post a working application along with a description of what it does to this sister site: codereview.stackexchange.com to get feedback on your methods.
Mr. Wizard, after reading the first chapter of Leonid's book, I am convinced that Mathematica is a far more powerful tool than I initially realized, while maintaining a fairly simplistic coding language.
I have one last and final question in regards to the above piece of code you wrote that works. I want to substitute b->1-c into rules = {b -> 1 - c, c -> 3, d -> 4, e -> 5, f -> 6, T -> 10}; but when I evaluate XYZ[h, T, i, g, j, b, c] /. rules the resulting output is not numeric. Is there an easy way to make the variable b in XYZ[h, T, i, g, j, b, c] equal to '1-c' and still have it evaluate for a given value of 'c' ? Thanks again
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