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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
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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