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I aim to calculate and preserve the results from the maximization of a function with two arguments and one exogenous parameter, when the maximum can not be derived (in closed form) by maximize. For instance, let
f[x_,y_,a_]=Max[0,Min[a-y,1-x-y]
be the objective function where a is positive. The maximization shall take place over [0,1]^2, therefore I set
m[a_]=Maximize[{f[x, y, a], 0 <= x <= 1 && 0 <= y <= 1 && 0 <= a}, {x,y}]
Obviously m can be evaluated at any point a and it is therefore possible to plot the maximizing x by employing
Plot[x /. m[a][[2]], {a, 0.01, 1}]
As I need to do several plots and further derivations containing the optimal solutions x and y (which of course are functions of a), i would like to preserve/save the results from the optimization for further use. Is there an elegant way to do this, or do I have to write some kind of loop to extract the values myself?
Now that I've seen the full text of your comment on my original comment, I suspect that you do understand the differences between Set and SetDelayed well enough. I think what you may be looking for is memoisation, sometimes implemented a bit like this;
f[x_,y_] := f[x,y] = Max[0,Min[a-y,1-x-y]]
When you evaluate, for example f[3,4] for the first time it will evaluate to the entire expression to the right of the :=. The rhs is the assignment f[3,4] = Max[0,Min[a-y,1-x-y]]. Next time you evaluate f[3,4] Mathematica already has a value for it so doesn't need to recompute it, it just recalls it. In this example the stored value would be Max[0,Min[a-4,-6]] of course.
I remain a little uncertain of what you are trying to do so this answer may not be any use to you at all.
Simple approach
results = Table[{x, y, a} /. m[a][[2]], {a, 0.01, 1, .01}]
ListPlot[{#[[3]], #[[1]]} & /# results, Joined -> True]
(The Set = is ok here so long as 'a' is not previosly defined )
If you want to utilise Plot[]s automatic evaluation take a look at Reap[]/Sow[]
{p, data} = Reap[Plot[x /. Sow[m[a]][[2]], {a, 0.01, 1}]];
Show[p]
(this takes a few minutes as the function output is a mess..).
hmm try this again: assuming you want x,y,a and the minimum value:
{p, data} = Reap[Plot[x /. Sow[{a, m[a]}][[2, 2]], {a, 0.01, .1}]];
Show[p]
results = {#[[1]], x /. #[[2, 2]], y /. #[[2, 2]], #[[2, 1]]} & /# data[[1]]
BTW Your function appears to be independent of x over some ranges which is why the plot is a mess..
I often have a list of pairs, as
data = {{0,0.0},{1,12.4},{2,14.6},{3,25.1}}
and I want to do something, for instance Rescale, to all of the second elements without touching the first elements. The neatest way I know is:
Transpose[MapAt[Rescale, Transpose[data], 2]]
There must be a way to do this without so much Transposeing. My wish is for something like this to work:
MapAt[Rescale, data, {All, 2}]
But my understanding is that MapAt takes Position-style specifications instead of Part-style specifications. What's the proper solution?
To clarify,
I'm seeking a solution where I don't have to repeat myself, so lacking double Transpose or double [[All,2]], because I consider repetition a signal I'm not doing something the easiest way. However, if eliminating the repetition requires the introduction of intermediate variables or a named function or other additional complexity, maybe the transpose/untranspose solution is already correct.
Use Part:
data = {{0, 0.0}, {1, 12.4}, {2, 14.6}, {3, 25.1}}
data[[All, 2]] = Rescale # data[[All, 2]];
data
Create a copy first if you need to. (data2 = data then data2[[All, 2]] etc.)
Amending my answer to keep up with ruebenko's, this can be made into a function also:
partReplace[dat_, func_, spec__] :=
Module[{a = dat},
a[[spec]] = func # a[[spec]];
a
]
partReplace[data, Rescale, All, 2]
This is quite general is design.
I am coming late to the party, and what I will describe will differ very little with what #Mr. Wizard has, so it is best to consider this answer as a complementary to his solution. My partial excuses are that first, the function below packages things a bit differently and closer to the syntax of MapAt itself, second, it is a bit more general and has an option to use with Listable function, and third, I am reproducing my solution from the past Mathgroup thread for exactly this question, which is more than 2 years old, so I am not plagiarizing :)
So, here is the function:
ClearAll[mapAt,MappedListable];
Protect[MappedListable];
Options[mapAt] = {MappedListable -> False};
mapAt[f_, expr_, {pseq : (All | _Integer) ..}, OptionsPattern[]] :=
Module[{copy = expr},
copy[[pseq]] =
If[TrueQ[OptionValue[MappedListable]] && Head[expr] === List,
f[copy[[pseq]]],
f /# copy[[pseq]]
];
copy];
mapAt[f_, expr_, poslist_List] := MapAt[f, expr, poslist];
This is the same idea as what #Mr. Wizard used, with these differences: 1. In case when the spec is not of the prescribed form, regular MapAt will be used automatically 2. Not all functions are Listable. The solution of #Mr.Wizard assumes that either a function is Listable or we want to apply it to the entire list. In the above code, you can specify this by the MappedListable option.
I will also borrow a few examples from my answer in the above-mentioned thread:
In[18]:= mat=ConstantArray[1,{5,3}];
In[19]:= mapAt[#/10&,mat,{All,3}]
Out[19]= {{1,1,1/10},{1,1,1/10},{1,1,1/10},{1,1,1/10},{1,1,1/10}}
In[20]:= mapAt[#/10&,mat,{3,All}]
Out[20]= {{1,1,1},{1,1,1},{1/10,1/10,1/10},{1,1,1},{1,1,1}}
Testing on large lists shows that using Listability improves the performance, although not so dramatically here:
In[28]:= largemat=ConstantArray[1,{150000,15}];
In[29]:= mapAt[#/10&,largemat,{All,3}];//Timing
Out[29]= {0.203,Null}
In[30]:= mapAt[#/10&,largemat,{All,3},MappedListable->True];//Timing
Out[30]= {0.094,Null}
This is likely because for the above function (#/10&), Map (which is used internally in mapAt for the MappedListable->False (default) setting, was able to auto-compile. In the example below, the difference is more substantial:
ClearAll[f];
f[x_] := 2 x - 1;
In[54]:= mapAt[f,largemat,{All,3}];//Timing
Out[54]= {0.219,Null}
In[55]:= mapAt[f,largemat,{All,3},MappedListable->True];//Timing
Out[55]= {0.031,Null}
The point is that, while f was not declared Listable, we know that its body is built out of Listable functions, and thus it can be applied to the entire list - but OTOH it can not be auto-compiled by Map. Note that adding Listable attribute to f would have been completely wrong here and would destroy the purpose, leading to mapAt being slow in both cases.
How about
Transpose[{#[[All, 1]], Rescale[#[[All, 2]]]} &#data]
which returns what you want (ie, it does not alter data)
If no Transpose is allowed,
Thread[Join[{#[[All, 1]], Rescale[#[[All, 2]]]} &#data]]
works.
EDIT: As "shortest" is now the goal, best from me so far is:
data\[LeftDoubleBracket]All, 2\[RightDoubleBracket] = Rescale[data[[All, 2]]]
at 80 characters, which is identical to Mr.Wizard's... So vote for his answer.
Here is another approach:
op[data_List, fun_] :=
Join[data[[All, {1}]], fun[data[[All, {2}]]], 2]
op[data, Rescale]
Edit 1:
An extension from Mr.Wizard, that does not copy it's data.
SetAttributes[partReplace, HoldFirst]
partReplace[dat_, func_, spec__] := dat[[spec]] = func[dat[[spec]]];
used like this
partReplace[data, Rescale, All, 2]
Edit 2:
Or like this
ReplacePart[data, {All, 2} -> Rescale[data[[All, 2]]]]
This worked for me and a friend
In[128]:= m = {{x, sss, x}, {y, sss, y}}
Out[128]= {{2, sss, 2}, {y, sss, y}}
In[129]:= function[ins1_] := ToUpperCase[ins1];
fatmap[ins2_] := MapAt[function, ins2, 2];
In[131]:= Map[fatmap, m]
Out[131]= {{2, ToUpperCase[sss], 2}, {y, ToUpperCase[sss], y}}
Some types of objects have special input/output formatting in Mathematica. This includes Graphics, raster images, and, as of Mathematica 8, graphs (Graph[]). Unfortunately large graphs may take a very long time to visualize, much longer than most other operations I'm doing on them during interactive work.
How can I prevent auto-layout of Graph[] objects in StandardForm and TraditionalForm, and have them displayed as e.g. -Graph-, preferably preserving the interpretability of the output (perhaps using Interpretation?). I think this will involve changing Format and/or MakeBoxes in some way, but I was unsuccessful in getting this to work.
I would like to do this in a reversible way, and preferably define a function that will return the original interactive graph display when applied to a Graph object (not the same as GraphPlot, which is not interactive).
On a related note, is there a way to retrieve Format/MakeBoxes definitions associated with certain symbols? FormatValues is one relevant function, but it is empty for Graph.
Sample session:
In[1]:= Graph[{1->2, 2->3, 3->1}]
Out[1]= -Graph-
In[2]:= interactiveGraphPlot[%] (* note that % works *)
Out[2]= (the usual interactive graph plot should be shown here)
Though I do not have Mathematica 8 to try this in, one possibility is to use this construct:
Unprotect[Graph]
MakeBoxes[g_Graph, StandardForm] /; TrueQ[$short] ^:=
ToBoxes#Interpretation[Skeleton["Graph"], g]
$short = True;
Afterward, a Graph object should display in Skeleton form, and setting $short = False should restore default behavior.
Hopefully this works to automate the switching:
interactiveGraphPlot[g_Graph] := Block[{$short}, Print[g]]
Mark's concern about modifying Graph caused me to consider the option of using $PrePrint. I think this should also prevent the slow layout step from taking place. It may be more desirable, assuming you are not already using $PrePrint for something else.
$PrePrint =
If[TrueQ[$short], # /. _Graph -> Skeleton["Graph"], #] &;
$short = True
Also conveniently, at least with Graphics (again I cannot test with Graph in v7) you can get the graphic with simply Print. Here, shown with Graphics:
g = Plot[Sin[x], {x, 0, 2 Pi}]
(* Out = <<"Graphics">> *)
Then
Print[g]
I left the $short test in place for easy switching via a global symbol, but one could leave it out and use:
$PrePrint = # /. _Graph -> Skeleton["Graph"] &;
And then use $PrePrint = . to reset the default functionality.
You can use GraphLayout option of Graph as well as graph-constructors to suppress the rendering. A graph can still be visualized with GraphPlot. Try the following
{gr1, gr2, gr3} = {RandomGraph[{100, 120}, GraphLayout -> None],
PetersenGraph[10, 3, GraphLayout -> None],
Graph[{1 -> 2, 2 -> 3, 3 -> 1}, GraphLayout -> None]}
In order to make working easier, you can use SetOptions to set GraphLayout option to None for all graph constructors you are interested in.
Have you tried simply suppressing the output? I don't think that V8's Graph command does any layout, if you do so. To explore this, we can generate a large list of edges and compare the timings of graph[edges];, Graph[edges];, and GraphPlot[edges];
In[23]:= SeedRandom[1];
edges = Union[Rule ### (Sort /#
RandomInteger[{1, 5000}, {50000, 2}])];
In[25]:= t = AbsoluteTime[];
graph[edges];
In[27]:= AbsoluteTime[] - t
Out[27]= 0.029354
In[28]:= t = AbsoluteTime[];
Graph[edges];
In[30]:= AbsoluteTime[] - t
Out[30]= 0.080434
In[31]:= t = AbsoluteTime[];
GraphPlot[edges];
In[33]:= AbsoluteTime[] - t
Out[33]= 4.934918
The inert graph command is, of course, the fastest. The Graph command takes much longer, but no where near as long as the GraphPlot command. Thus, it seems to me that Graph is not, in fact, computing the layout, as GraphPlot does.
The logical question is, what is Graph spending it's time on. Let's examine the InputForm of Graph output in a simple case:
Graph[{1 -> 2, 2 -> 3, 3 -> 1, 1 -> 4}] // InputForm
Out[123]//InputForm=
Graph[{1, 2, 3, 4},
{DirectedEdge[1, 2],
DirectedEdge[2, 3],
DirectedEdge[3, 1],
DirectedEdge[1, 4]}]
Note that the vertices of the graph have been determined and I think this is what Graph is doing. In fact, the amount of time it took to compute Graph[edges] in the first example, comparable to the fastest way that I can think to do this:
Union[Sequence ### edges]; // Timing
This took 0.087045 seconds.
What is the simplest way to map an arbitrarily funky nested list expr to a function unflatten so that expr==unflatten##Flatten#expr?
Motivation:
Compile can only handle full arrays (something I just learned -- but not from the error message), so the idea is to use unflatten together with a compiled version of the flattened expression:
fPrivate=Compile[{x,y},Evaluate#Flatten#expr];
f[x_?NumericQ,y_?NumericQ]:=unflatten##fPrivate[x,y]
Example of a solution to a less general problem:
What I actually need to do is to calculate all the derivatives for a given multivariate function up to some order. For this case, I hack my way along like so:
expr=Table[D[x^2 y+y^3,{{x,y},k}],{k,0,2}];
unflatten=Module[{f,x,y,a,b,sslot,tt},
tt=Table[D[f[x,y],{{x,y},k}],{k,0,2}] /.
{Derivative[a_,b_][_][__]-> x[a,b], f[__]-> x[0,0]};
(Evaluate[tt/.MapIndexed[#1->sslot[#2[[1]]]&,
Flatten[tt]]/. sslot-> Slot]&) ]
Out[1]= {x^2 y + y^3, {2 x y, x^2 + 3 y^2}, {{2 y, 2 x}, {2 x, 6 y}}}
Out[2]= {#1, {#2, #3}, {{#4, #5}, {#5, #7}}} &
This works, but it is neither elegant nor general.
Edit: Here is the "job security" version of the solution provided by aaz:
makeUnflatten[expr_List]:=Module[{i=1},
Function#Evaluate#ReplaceAll[
If[ListQ[#1],Map[#0,#1],i++]&#expr,
i_Integer-> Slot[i]]]
It works a charm:
In[2]= makeUnflatten[expr]
Out[2]= {#1,{#2,#3},{{#4,#5},{#6,#7}}}&
You obviously need to save some information about list structure, because Flatten[{a,{b,c}}]==Flatten[{{a,b},c}].
If ArrayQ[expr], then the list structure is given by Dimensions[expr] and you can reconstruct it with Partition. E.g.
expr = {{a, b, c}, {d, e, f}};
dimensions = Dimensions[expr]
{2,3}
unflatten = Fold[Partition, #1, Reverse[Drop[dimensions, 1]]]&;
expr == unflatten # Flatten[expr]
(The Partition man page actually has a similar example called unflatten.)
If expr is not an array, you can try this:
expr = {a, {b, c}};
indexes = Module[{i=0}, If[ListQ[#1], Map[#0, #1], ++i]& #expr]
{1, {2, 3}}
slots = indexes /. {i_Integer -> Slot[i]}
{#1, {#2, #3}}
unflatten = Function[Release[slots]]
{#1, {#2, #3}} &
expr == unflatten ## Flatten[expr]
I am not sure what you are trying to do with Compile. It is used when you want to evaluate procedural or functional expressions very quickly on numerical values, so I don't think it is going to help here. If repeated calculations of D[f,...] are impeding your performance, you can precompute and store them with something like
Table[d[k]=D[f,{{x,y},k}],{k,0,kk}];
Then just call d[k] to get the kth derivative.
I just wanted to update the excellent solutions by aaz and Janus. It seems that, at least in Mathematica 9.0.1.0 on Mac OSX, the assignment (see aaz's solution)
{i_Integer -> Slot[i]}
fails. If, however, we use
{i_Integer :> Slot[i]}
instead, we succeed. The same holds, of course, for the ReplaceAll call in Janus's "job security" version.
For good measure, I include my own function.
unflatten[ex_List, exOriginal_List] :=
Module[
{indexes, slots, unflat},
indexes =
Module[
{i = 0},
If[ListQ[#1], Map[#0, #1], ++i] &#exOriginal
];
slots = indexes /. {i_Integer :> Slot[i]};
unflat = Function[Release[slots]];
unflat ## ex
];
(* example *)
expr = {a, {b, c}};
expr // Flatten // unflatten[#, expr] &
It might seem a little like a cheat to use the original expression in the function, but as aaz points out, we need some information from the original expression. While you don't need it all, in order to have a single function that can unflatten, all is necessary.
My application is similar to Janus's: I am parallelizing calls to Simplify for a tensor. Using ParallelTable I can significantly improve performance, but I wreck the tensor structure in the process. This gives me a quick way to reconstruct my original tensor, simplified.
I have a basic problem in Mathematica which has puzzled me for a while. I want to take the m'th derivative of x*Exp[t*x], then evaluate this at x=0. But the following does not work correct. Please share your thoughts.
D[x*Exp[t*x], {x, m}] /. x -> 0
Also what does the error mean
General::ivar: 0 is not a valid variable.
Edit: my previous example (D[Exp[t*x], {x, m}] /. x -> 0) was trivial. So I made it harder. :)
My question is: how to force it to do the derivative evaluation first, then do substitution.
As pointed out by others, (in general) Mathematica does not know how to take the derivative an arbitrary number of times, even if you specify that number is a positive integer.
This means that the D[expr,{x,m}] command remains unevaluated and then when you set x->0, it's now trying to take the derivative with respect to a constant, which yields the error message.
In general, what you want is the m'th derivative of the function evaluated at zero.
This can be written as
Derivative[m][Function[x,x Exp[t x]]][0]
or
Derivative[m][# Exp[t #]&][0]
You then get the table of coefficients
In[2]:= Table[%, {m, 1, 10}]
Out[2]= {1, 2 t, 3 t^2, 4 t^3, 5 t^4, 6 t^5, 7 t^6, 8 t^7, 9 t^8, 10 t^9}
But a little more thought shows that you really just want the m'th term in the series, so SeriesCoefficient does what you want:
In[3]:= SeriesCoefficient[x*Exp[t*x], {x, 0, m}]
Out[3]= Piecewise[{{t^(-1 + m)/(-1 + m)!, m >= 1}}, 0]
The final output is the general form of the m'th derivative. The PieceWise is not really necessary, since the expression actually holds for all non-negative integers.
Thanks to your update, it's clear what's happening here. Mathematica doesn't actually calculate the derivative; you then replace x with 0, and it ends up looking at this:
D[Exp[t*0],{0,m}]
which obviously is going to run into problems, since 0 isn't a variable.
I'll assume that you want the mth partial derivative of that function w.r.t. x. The t variable suggests that it might be a second independent variable.
It's easy enough to do without Mathematica: D[Exp[t*x], {x, m}] = t^m Exp[t*x]
And if you evaluate the limit as x approaches zero, you get t^m, since lim(Exp[t*x]) = 1. Right?
Update: Let's try it for x*exp(t*x)
the mth partial derivative w.r.t. x is easily had from Wolfram Alpha:
t^(m-1)*exp(t*x)(t*x + m)
So if x = 0 you get m*t^(m-1).
Q.E.D.
Let's see what is happening with a little more detail:
When you write:
D[Sin[x], {x, 1}]
you get an expression in with x in it
Cos[x]
That is because the x in the {x,1} part matches the x in the Sin[x] part, and so Mma understands that you want to make the derivative for that symbol.
But this x, does NOT act as a Block variable for that statement, isolating its meaning from any other x you have in your program, so it enables the chain rule. For example:
In[85]:= z=x^2;
D[Sin[z],{x,1}]
Out[86]= 2 x Cos[x^2]
See? That's perfect! But there is a price.
The price is that the symbols inside the derivative get evaluated as the derivative is taken, and that is spoiling your code.
Of course there are a lot of tricks to get around this. Some have already been mentioned. From my point of view, one clear way to undertand what is happening is:
f[x_] := x*Exp[t*x];
g[y_, m_] := D[f[x], {x, m}] /. x -> y;
{g[p, 2], g[0, 1]}
Out:
{2 E^(p t) t + E^(p t) p t^2, 1}
HTH!