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Evaluate two variable functions at given points?

Okkes Dulgerci
Posted 9 years ago

Hi All,

Assume I have list of two variable function and want to evaluate at some point. First func. in first list 4x take first point {1/12, 1/6} and second func. 4y takes second point {1/6, 1/12}. and second func list at second point list and so on. How can I do this? I tried Map, Thread, Through, Apply but no luck. Any suggestion? Thanks.. 

f={{4 x, 4 y, 1 - 4 x + 4 y, 2 - 4 x, 2 - 4 y,1 + 4 x - 4 y}, {-1 + 4 x, 4 y, 2 - 4 x + 4 y, 3 - 4 x, 2 - 4 y, 
  4 x - 4 y}, {-2 + 4 x, 4 y, 3 - 4 x + 4 y, 4 - 4 x,2 - 4 y, -1 + 4 x - 4 y}, {4 x, -1 + 4 y, -4 x + 4 y, 2 - 4 x, 
  3 - 4 y, 2 + 4 x - 4 y}, {-1 + 4 x, -1 + 4 y, 1 - 4 x + 4 y,  3 - 4 x, 3 - 4 y, 1 + 4 x - 4 y}, {-2 + 4 x, -1 + 4 y, 
  2 - 4 x + 4 y, 4 - 4 x, 3 - 4 y,  4 x - 4 y}, {4 x, -2 + 4 y, -1 - 4 x + 4 y, 2 - 4 x, 4 - 4 y, 
  3 + 4 x - 4 y}, {-1 + 4 x, -2 + 4 y, -4 x + 4 y, 3 - 4 x, 4 - 4 y,   2 + 4 x - 4 y}, {-2 + 4 x, -2 + 4 y, 1 - 4 x + 4 y, 4 - 4 x, 
  4 - 4 y, 1 + 4 x - 4 y}}

pts={{{1/12, 1/6}, {1/6, 1/12}, {1/3, 1/6}, {5/12, 1/3}, {1/3, 5/12}, {1/
   6, 1/3}}, {{1/3, 1/6}, {5/12, 1/12}, {7/12, 1/6}, {2/3, 1/3}, {7/12, 5/12}, {5/12, 1/3}}, {{7/12, 1/6}, {2/3, 1/12}, {5/6, 1/
   6}, {11/12, 1/3}, {5/6, 5/12}, {2/3, 1/3}}, {{1/12, 5/12}, {1/6, 1/ 3}, {1/3, 5/12}, {5/12, 7/12}, {1/3, 2/3}, {1/6, 7/12}}, {{1/3, 5/
   12}, {5/12, 1/3}, {7/12, 5/12}, {2/3, 7/12}, {7/12, 2/3}, {5/12, 7/ 12}}, {{7/12, 5/12}, {2/3, 1/3}, {5/6, 5/12}, {11/12, 7/12}, {5/6, 
   2/3}, {2/3, 7/12}}, {{1/12, 2/3}, {1/6, 7/12}, {1/3, 2/3}, {5/12, 5/6}, {1/3, 11/12}, {1/6, 5/6}}, {{1/3, 2/3}, {5/12, 7/12}, {7/12, 
   2/3}, {2/3, 5/6}, {7/12, 11/12}, {5/12, 5/6}}, {{7/12, 2/3}, {2/3,  7/12}, {5/6, 2/3}, {11/12, 5/6}, {5/6, 11/12}, {2/3, 5/6}}}
POSTED BY: Okkes Dulgerci
6 Replies
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Okkes Dulgerci
Okkes Dulgerci
Posted 9 years ago

It helped a lot, thank you all..
POSTED BY: Okkes Dulgerci

There is nothing wrong in generating lists of expressions and then use proper programming to calculate the results. The WL does not "insist" on a specific programming style and there are several ways of doing the required task. The best of all is to mix procedural, functional and rewrite rules programming styles so the program will be elegant, short and efficient. IMHO, the original poster used expressions such as

f[1,1][x_,y_]:=4x

just to clarify what he meant in his original post

best

yehuda
POSTED BY: Yehuda Ben-Shimol

A similar method to the one Gianluca proposed could be:

f = Map[Function[{pt}, # /. {x :> pt[[1]], y :> pt[[2]]}] &, f, {2}];

a function call would then look like this (I admit this is not exactly what you want ...):

f[[1, 3]][pts[[1, 3]]]

or likewise

dims = Dimensions[f];
Table[f[[n, m]][pts[[n, m]]], {n, 1, dims[[1]]}, {m, 1, dims[[2]]}]

Regards -- Henrik
POSTED BY: Henrik Schachner

The way you have defined f, it is a little awkward to make it behave like a function. This is a solution, that uses two new symbols myF and myPts, which behave the way you want:

Clear[myF, x, y];
Do[myF[n, m][{x_, y_}] = f[[n, m]], {n, Length[f]},
 {m, Length[f[[1]]]}]
myPts[n_, m_] := pts[[n, m]];
myF[1, 1][myPts[1, 1]]
POSTED BY: Gianluca Gorni

I'm nut sure you are clear enough

Dimensions[f]

returns {9,6} and

Dimensions[pts]

returns {9,6,2}

As I see it, for each expression in f (you have 9*6=54 of those) you need to replace x and y from the 54 pairs (in the same position) from its The simple solution for that is to use MapThread at level 2 as follows. Note that that the level is required to use each instance of you functions with the correct pair of x and y values. Than use Thread to get a list of replacement rules so they will be used by the MapThread function (I forgot to add the Thread in my earlier reply.

MapThread[(#1 /.Thread[ {x, y} -> #2]) &, {f, pts}, 2]

you can also do it as

MapThread[#1 /. #2 & , {f, pts /. {a_, b_} :> {x -> a, y -> b}}, 2]

Evaluating one of the above for the given f and pts lists results with

{{1/3, 1/3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3, 1/3, 1/3}, {1/3,
   1/3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/
  3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 
  1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3, 1/3, 1/3, 1/3}, {1/3, 1/3, 1/3,
   1/3, 1/3, 1/3}}

best yehuda
POSTED BY: Yehuda Ben-Shimol
Okkes Dulgerci
Okkes Dulgerci
Posted 9 years ago

Thanks for your reply. Sorry my unclear explanation. This is what I want

f[1,1][x_,y_]:=4x,  then f[1,1][pts[1,1]]=f[1,1][1/12, 1/6]=4*1/12=1/3
 f[1,2][x_,y_]:=4y, then f[1,2][pts[1,2]]=f[1,2][1/6, 1/12]=4*1/12=1/3
 f[1,3][x_,y_]:= 1 - 4 x + 4 y,  then f[1,3][pts[1,3]]=f[1,3][1/3, 1/6]=1-4*1/3+4*1/6=1/3
POSTED BY: Okkes Dulgerci
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