3
$\begingroup$

Here is the code that I have so far:

fun[{data_, p_, n_, min_, max_}] := 
 Module[{}, lstplt = LinearModelFit[data, Table[x^i, {i, n}], x];
  Print[lstplt[x]]
   Print["Fit plotted from min to max temperature in data set range"]
   Print[Plot[lstplt[x], {x, min, max}]];
  reslist = 
   Inner[List, {data}[[1, All, 1]], lstplt["FitResiduals"], List];
  Print["Fit and data curve"]
   Print[Show[ListPlot[data], Plot[lstplt[x], {x, min, max}]]];
  Print["Residual Plot"]
   Print[ListPlot[reslist]];
  bres = Select[reslist, Abs[#[[2]]] > p &];
  gres = DeleteCases[reslist, Alternatives @@ bres];
gpoints = gres[[All, 1]] \[Intersection] reslist[[All, 1]];
Print["New data curve excluding points that had poor residual \
values"]
 Print[ListPlot[
  dataset13 = Select[data, gpoints~MemberQ~First[#] &]]];
  Print["New data set"];
  Print["Number of points"];
 Print[Length[dataset13]];
 Return[dataset13];]
data = Table[{x, RandomReal[{-.1, .1}] + x^2}, {x, 0, 15}];

I want to be able to do the following line of code:

fun[fun[data, .04, 2, 0, 15], 1, 3, 0, 15]

without having to type in the argument over and over again. Would NestList be the best way to tackle this? It looks like a nest function problem to me, however I want to be able to change the n value by one after each iteration. How could I simplify this last line that I want to be able to do so that I can type it in one line and say continue to apply the function 'fun' until n has reached some value N?

$\endgroup$
  • $\begingroup$ do you want p changing in each iteration or is it a typo in the last line? $\endgroup$ – kglr Jul 19 at 20:36
3
$\begingroup$

If p is fixed (at, say, p0) in each iteration and only n is changing, you can use Nest:

k = 3;
i = 1; 
Nest[fun[#, p0, ++i, 0, 15] &, data, k]

fun[fun[fun[data, p0, 2, 0, 15], p0, 3, 0, 15], p0, 4, 0, 15]

Alternatively,

First @ Nest[{fun[#[[1]], p0, #[[2]], 0, 15], #[[2]] + 1} &, {data, 2},  k]

fun[fun[fun[data, p0, 2, 0, 15], p0, 3, 0, 15], p0, 4, 0, 15]

If p varies in each iteration taking values in a specified list:

plist = Array[p, 3];
nlist = Range[Length @ plist] + 1;

Fold[fun[#1, plist[[#2 - 1]], #2, 0, 15] &, data, nlist]

fun[fun[fun[data, p[1], 2, 0, 15], p[2], 3, 0, 15], p[3], 4, 0, 15]

Also

i = 1; Nest[fun[#, plist[[i]], ++i, 0, 15] &, data, 3]
Fold[fun[#1, #2[[1]], #2[[2]], 0, 15]&, data, Transpose[{nlist, plist}]]
Fold[fun[#1, ## & @@ #2, 0, 15] &, data, Transpose[{nlist, plist}]]

same result

$\endgroup$
  • $\begingroup$ p will vary with each iteration, however the value will not be known until the previous iteration is done. How could I manipulate the function to represent this? I am trying to use Manipulate now and it is not giving a correct output. $\endgroup$ – Carrson Baldwin Jul 30 at 16:16
  • $\begingroup$ how is the value of p determined given previous iteration? $\endgroup$ – kglr Jul 30 at 16:21
  • $\begingroup$ There is no true criteria, I want to exclude outliers while keeping a majority of the points. It is really done from looking at the previous iteration residual graph and making a determination that way, which is why I wanted some dynamic variable to change the p value. $\endgroup$ – Carrson Baldwin Jul 30 at 19:22
1
$\begingroup$
m = 2;
Fold[fun[#1, #2, m++, 0, 15] &, data, {0.04, 1}]
(*    fun[fun[data, 0.04, 2, 0, 15], 1, 3, 0, 15]    *)

This Fold takes as the last argument the list of p values to run through, and it stops when this list is used up.

$\endgroup$
  • $\begingroup$ Not exactly what I want the program to do. This only takes the output of the first iteration and takes it to a degree fit that is one value higher than before. I want to take the result of the first iteration, plug that into the function 'fun' again (for a second iteration) and have the third argument of 'fun' be increased by one for the second iteration. Then I want to continue this process until it has reached say 6 iterations for example. How could this be done? $\endgroup$ – Carrson Baldwin Jul 19 at 20:26
  • $\begingroup$ So what to do with the second argument p? In your example there is a different value for p in every iteration. $\endgroup$ – Roman Jul 19 at 21:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.