I'm trying to create three separate graphs, this code will give you the idea:
f[t_] := Sin[10 t] + Cos[15 t];
Slider[Dynamic[dx], {0.01, 1}]
var = Dynamic[Fourier[Table[f[t], {t, 0, 100, dx}]]];
ListLinePlot[Abs[var]]
ListLinePlot[Re[var]]
ListLinePlot[Im[var]]
This won't work because var hasn't been evaluated an so ListLinePlot/Abs/Re/Im does not recognize it as a list of numbers. Dynamic has to wrap ListLinePlot.
Wrapping ListLinePlot and everything else with Dynamic works. But then I would have to calculate Fourier[Table[... once for each graph. Per principle, I don't want to have this duplication of code.
This is a way that avoids duplication of code but is not as semantic as my proposed not working example above, plus it puts all series in one graph and not in three separate:
Dynamic[
ListLinePlot[
(#[Fourier[
Table[f[t], {t, 0, 100, dx}]
]]) & /# {Abs,Re,Min}, DataRange -> {0, 100}
]
]
Hopefully you can see now what I am trying to achieve. Something like my first piece of code except it should work. How can I do that?
In most cases you only need to wrap Dynamic around the expression that needs to be recomputed. As you noticed, if you wrap Dynamic around the contents of var, it will not work because ListPlot will see a Dynamic head, not the list, when you pass var to it. What needs to be recomputed in this case is the complete ListPlot.
The correct solution is to use a delayed definition for var (i.e. := instead of =) and wrap Dynamic around ListPlot:
f[t_] := Sin[10 t] + Cos[15 t];
Slider[Dynamic[dx], {0.01, 1}]
var := Fourier[Table[f[t], {t, 0, 100, dx}]];
Dynamic#ListLinePlot[Abs[var]]
Dynamic#ListLinePlot[Re[var]]
Dynamic#ListLinePlot[Im[var]]
People often get confused with Dynamic because it sometimes shows up deep within in expression, e.g. in your Slider example. But there Dynamic has a different function: setting a value.
Generally, unless used to set a value, Dynamic always needs to be the outermost head in an expression. (There are some exceptions, notably when we're handling expressions that directly correspond to what is shown on screen, and are handled by the front end, such as graphics primitives: Slider[Dynamic[x], {0, 5}], Graphics[{Disk[], Dynamic#Disk[{x, 0}]}] will work.)
Dynamic affects only the way expressions are displayed in the front end, not how the kernel sees them. Here's an example:
x=1
arr = {Dynamic[x], 2, 3}
The Front End will display arr as {1, 2, 3}, but the kernel still sees it as {Dynamic[x], 2, 3}. So if we calculate Total[arr], the front end will display it as 1 + 5 but the kernel will see if as Dynamic[x] + 5. I hope this clarifies the situation a bit.
Note: I did not want to use Manipulate in this solution because the OP didn't use it either. Manipulate is just a high level convenience function and everything it does can be achieved with Dynamic and some controls such as Slider.
You probably want something like this:
f[t_] := Sin[10 t] + Cos[15 t]
DynamicModule[{var},
Manipulate[
var = Fourier[Table[f[t], {t, 0, 100, dx}]];
{ListLinePlot[Abs[var]],
ListLinePlot[Re[var]],
ListLinePlot[Im[var]]},
{dx, 0.01, 1}
]]
Untested:
f[t_] := Sin[10 t] + Cos[15 t];
Slider[Dynamic[dx], {0.01, 1}]
Dynamic[var = Fourier[Table[f[t], {t, 0, 100, dx}]]];
Dynamic[ListLinePlot[Abs[var]]]
Dynamic[ListLinePlot[Re[var]]]
Dynamic[ListLinePlot[Im[var]]]
I think this should calculate Fourier just once. From my understanding, the ListLinePlots should be triggered by the change of var after evaluating Fourier (note that the assignment of var is inside the Dynamic).
My question sounds more general, but I have a specific example. I have a list of data in form:
plotDataAll={{DateList1, integerValue1}, {DateList2, integerValue2}...}
The dates are sorted chronologically, that is plotDataAll[[2,1]] is a more recent time then plotDataAll[[1,1]].
I want to create plots of specific periods, 24h ago, 1 week ago, etc. For that I need just a portion of data. Here's how I got what I wanted:
mostRecentDate=Max[Map[AbsoluteTime, plotDataAll[[All,1]]]];
plotDataLast24h=Select[plotDataAll,AbsoluteTime[#[[1]]]>(mostRecentDate-86400.)&];
plotDataLastWeek=Select[plotDataAll,AbsoluteTime[#[[1]]]>(mostRecentDate-604800.)&];
plotDataLastMonth=Select[plotDataAll,AbsoluteTime[#[[1]]]>(mostRecentDate-2.592*^6)&];
plotDataLast6M=Select[plotDataAll,AbsoluteTime[#[[1]]]>(mostRecentDate-1.5552*^7)&];
Then I used DateListPlot to plot the data. This becomes slow if you need to do this for many sets of data.
What comes to my mind, if I could find the index of first element in list that satisfies the date condition, because it's chronologically sorted, the rest of them should satisfy the condition as well. So I would have:
plotDataLast24h=plotDataAll[[beginningIndexThatSatisfiesLast24h;;Length[plotDataAll]]
But how do I get the index of the first element that satisfies the condition?
If you have a faster way to do this, please share your answer. Also, if you have a simple, faster, but sub-optimal solution, that's fine too.
EDIT:
Time data is not in regular intervals.
If your data is at regular intervals you should be able to know how many elements constitute a day, week, etc. and use Part.
plotDataAll2[[knownIndex;;-1]]
or more specifically if the data was hourly:
plotDataAll2[[-25;;-1]]
would give you the last 24 hours. If the spacing is irregular then use Select or Pick. Date and time functions in Mma are horrendously slow unfortunately. If you are going to do a lot of date and time calculation better to do a conversion to AbsoluteTime just once and then work with that. You will also notice that your DateListPlots render much faster if you use AbsoluteTime.
plotDataAll2=plotDataAll;
plotDataAll2[[All,1]]=AbsoluteTime/#plotDataAll2[[All,1]];
mostRecentDate=plotDataAll2[[-1,1]]
On my computer Pick is about 3 times faster but there may be other improvements you can make to the code below:
selectInterval[data_, interval_] := (tmp = data[[-1, 1]] - interval;
Select[data, #[[1]] > tmp &])
pickInterval[data_, interval_] := (tmp = data[[-1, 1]] - interval;
Pick[data, Sign[data[[All, 1]] - tmp], 1])
So to find data within the last week:
Timing[selectInterval[plotDataAll2, 604800]]
Timing[pickInterval[plotDataAll2, 604800]]
The thing that you want to avoid is checking all the values in the data table. Since the data is sequential you can just start checking from the back and stop when you have found the correct index.
Schematically:
tab = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
i = j = Length#tab;
While[tab[[i]] > 5, --i];
tab[[i ;; j]]
-> {5, 6, 7, 8, 9}
sustitute > 5 for whatever you want to check for. I didn't have time to test this right now but in your case, e.g.,
maxDate=AbsoluteTime#plotDataAll[[-1,1]]; (* no need to find Max if data is sequential*)
i24h = iWeek = iMonth = iMax = Length#plotDataAll;
While[AbsoluteTime#plotDataAll[[i24h,1]] > maxDate-86400.,--i24h];
While[AbsoluteTime#plotDataAll[[iWeek,1]] > maxDate-604800.,--iWeek];
While[AbsoluteTime#plotDataAll[[iMonth,1]] > maxDate-2.592*^6.,--iMonth];
While[AbsoluteTime#plotDataAll[[i6Month,1]] > maxDate-1.5552*^7.,--i6Month];
Then, e.g.,
DateListPlot#plotDataAll[[i24h;;iMax]]
If you want to start somewhere in the middle of plotDataAll just use a While to first find the starting point and set iMax and maxDate apropriately.
For large data sets this may be one of the few instances where a loop construct is better than MMA's inbuilt functions. That, however, may be my own ignorance and if anyone here knows of a MMA inbuilt function that does this sort of "stop when match found" comparison better than While.
EDIT: Timing comparisons
I played around a bit with Mike's and my solution and compared it to the OP's method. Here is the toy code I used for each solution
tab = Range#1000000;
(* My solution *)
i = j = tab[[-1]];
While[tab[[i]] > j - 24, --i];
tab[[i ;; j]]
(* Mike's solution *)
tmp = tab[[-1]] - 24;
Pick[tab, Sign[tab[[All]] - tmp], 1]
(* Enedene's solution *)
j = tab[[-1]];
Select[tab, # > (j - 24) &]
Here are the results (OS X, MMA 8.0.4, Core2Duo 2.0GHz)
As you can see, Mike's solution has a definite advantage over enedene's solution but, as I surmised originally, the downside of using inbuilt functions like Pick is that they still perform a comparative check on all the element in a list which is highly superfluous in this instance. My solution has constant time due to the fact that no unneccessary checks are made.
Given a set of numbers and a set of binary operations,
what is the fastest way to create random expression trees or exhaustively check every possible combination in Mathematica?
What I am trying to solve is given:
numbers={25,50,75,100,3,6} (* each can ONLY be used ONCE *)
operators={Plus,Subtract,Times,Divide} (* each can be used repeatedly *)
target=99
find expression trees that would evaluate to target.
I have two solutions whose performances I give for the case where expression trees contain exactly 4 of the numbers and 3 operators:
random sample & choice: ~25K trees / second
exhaustive scan: 806400 trees in ~2.15 seconds
(timed on a laptop with: Intel(R) Core(TM)2 Duo CPU T9300 # 2.50GHz, 3GB ram, no parallelization used yet but would be most welcome in answers)
My notebooks are a bit messy at the moment. So I would first love to pose the question and hope for original ideas and answers while I clean up my code for sharing.
Largest possible case is where every expression tree uses up all the (6) numbers and 'Length[numbers]-1' (5) operators.
Performance of methods in the largest case is:
random sample & choice: ~21K trees / second
exhaustive scan: 23224320 trees in ~100 seconds
Also I am using Mathematica 8.0.1 so I am more than all ears if there are any ways to do it in OpenCL or using compiled functions wiht CompilationTarget->"C", etc.
OK, this is not elegant or fast, and it's buggy, but it works (sometimes). It uses a monte carlo method, implementing the metropolis algorithm for a weight function that I (arbitrarily) selected just to see if this would work. This was some time ago for a similar problem; I suppose my mathematica skills have improved as it looks ugly now, but I have no time to fix it at the moment.
Execute this (it looks more reasonable when you paste it into a notebook):
ClearAll[swap];
swap[lst_, {p1_, p2_}] :=
ReplacePart[
lst, {p1 \[Rule] lst\[LeftDoubleBracket]p2\[RightDoubleBracket],
p2 \[Rule] lst\[LeftDoubleBracket]p1\[RightDoubleBracket]}]
ClearAll[evalops];
(*first element of opslst is Identity*)
evalops[opslst_, ord_, nums_] :=
Module[{curval}, curval = First#nums;
Do[curval =
opslst\[LeftDoubleBracket]p\[RightDoubleBracket][curval,
nums\[LeftDoubleBracket]ord\[LeftDoubleBracket]p\
\[RightDoubleBracket]\[RightDoubleBracket]], {p, 2, Length#nums}];
curval]
ClearAll[randomizeOrder];
randomizeOrder[ordlst_] :=
swap[ordlst, RandomInteger[{1, Length#ordlst}, 2]]
ClearAll[randomizeOps];
(*never touch the first element*)
randomizeOps[oplst_, allowedOps_] :=
ReplacePart[
oplst, {RandomInteger[{2, Length#oplst}] \[Rule] RandomChoice[ops]}]
ClearAll[takeMCstep];
takeMCstep[goal_, opslst_, ord_, nums_, allowedops_] :=
Module[{curres, newres, newops, neword, p},
curres = evalops[opslst, ord, nums];
newops = randomizeOps[opslst, allowedops];
neword = randomizeOrder[ord];
newres = evalops[newops, neword, nums];
Switch[Abs[newres - goal],
0, {newops,
neword}, _, (p = Abs[curres - goal]/Abs[newres - goal];
If[RandomReal[] < p, {newops, neword}, {opslst, ord}])]]
then to solve your actual problem, do
ops = {Times, Plus, Subtract, Divide}
nums = {25, 50, 75, 100, 3, 6}
ord = Range[Length#nums]
(*the first element is identity to simplify the logic later*)
oplist = {Identity}~Join~RandomChoice[ops, Length#nums - 1]
out = NestList[
takeMCstep[
99, #\[LeftDoubleBracket]1\[RightDoubleBracket], #\
\[LeftDoubleBracket]2\[RightDoubleBracket], nums, ops] &, {oplist,
ord}, 10000]
and then to see that it worked,
ev = Map[evalops[#\[LeftDoubleBracket]1\[RightDoubleBracket], #\
\[LeftDoubleBracket]2\[RightDoubleBracket], nums] &, out];
ev // Last // N
ev // ListPlot[#, PlotMarkers \[Rule] None] &
giving
thus, it obtained the correct order of operators and numbers after around 2000 tries.
As I said, it's ugly, inefficient, and badly programmed as it was a quick-and-dirty adaptation of a quick-and-dirty hack. If you're interested I can clean up and explain the code.
This was a fun question. Here's my full solution:
ExprEval[nums_, ops_] := Fold[
#2[[1]][#1, #2[[2]]] &,
First#nums,
Transpose[{ops, Rest#nums}]]
SymbolicEval[nums_, ops_] := ExprEval[nums, ToString /# ops]
GetExpression[nums_, ops_, target_] := Select[
Tuples[ops, Length#nums - 1],
(target == ExprEval[nums, #]) &]
Usage example:
nums = {-1, 1, 2, 3};
ops = {Plus, Subtract, Times, Divide};
solutions = GetExpression[nums, ops, 3]
ExprEval[nums, #] & /# solutions
SymbolicEval[nums, #] & /# solutions
Outputs:
{{Plus, Times, Plus}, {Plus, Divide, Plus}, {Subtract, Plus,
Plus}, {Times, Plus, Times}, {Divide, Plus, Times}}
{3, 3, 3, 3, 3}
{"Plus"["Times"["Plus"[-1, 1], 2], 3],
"Plus"["Divide"["Plus"[-1, 1], 2], 3],
"Plus"["Plus"["Subtract"[-1, 1], 2], 3],
"Times"["Plus"["Times"[-1, 1], 2], 3],
"Times"["Plus"["Divide"[-1, 1], 2], 3]}
How it works
The ExprEval function takes in the numbers and operations, and applies them using (I think) RPN:
ExprEval[{1, 2, 3}, {Plus, Times}] == (1 + 2) * 3
It does this by continually folding pairs of numbers using the appropriate operation.
Now that I have a way to evaluate an expression tree, I just needed to generate them. Using Tuples, I'm able to generate all the different operators that I would intersperse between the numbers.
Once you get all possible operations, I used Select to pick out the the ones that evaluate to the target number.
Drawbacks
The solution above is really slow. Generating all the possible tuples is exponential in time. If there are k operations and n numbers, it's on the order of O(k^n).
For n = 10, it took 6 seconds to complete on Win 7 x64, Core i7 860, 12 GB RAM. The timings of the runs match the theoretical time complexity almost exactly:
Red line is the theoretical, blue is experimental. The x-axis is size of the nums input and the y-axis is the time in seconds to enumerate all solutions.
The above solution also solves the problem using a functional programming style. It looks pretty, but the thing also sucks up a butt ton of memory since it's storing the full results at nearly every step.
It doesn't even make use of parallelization, and I'm not entirely certain how you would even parallelize the solution I produced.
Some limitations
Mr. Wizard brought to my attention that this code only solves for only particular set of solutions. Given some input such as {a, b, c, d, e, ... } it only permutes the operators in between the numbers. It doesn't permute the ordering of the numbers. If it were to permute the numbers as well, the time complexity would rise up to O(k^n * n!) where k is the number of operators and n is the length of the input number array.
The following will produce the set of solutions for any permutation of the input numbers and operators:
(* generates a lists of the form
{
{number permutation, {{op order 1}, {op order 2}, ... }
}, ...
}*)
GetAllExpressions[nums_, ops_, target_] :=
ParallelMap[{#, GetExpression[#, ops, target]} &,
Tuples[nums, Length#nums]]
I have the following problem.
I am developing a stochastic simulator which samples configurations of the system at random and stores the statistics of how many times each configuration has been visited at certain time instances. Roughly the code works like this
f[_Integer][{_Integer..}] :=0
...
someplace later in the code, e.g.,
index = get index;
c = get random configuration (i.e. a tuple of integers, say a pair {n1, n2});
f[index][c] = f[index][c] + 1;
which tags that configuration c has occurred once more in the simulation at time instance index.
Once the code has finished there is a list of definitions for f that looks something like this (I typed it by hand just to emphasize the most important parts)
?f
f[1][{1, 2}] = 112
f[1][{3, 4}] = 114
f[2][{1, 6}] = 216
f[2][{2, 7}] = 227
...
f[index][someconfiguration] = some value
...
f[_Integer][{_Integer..}] :=0
Please note that pattern free definitions that come first can be rather sparse. Also one cannot know which values and configurations will be picked.
The problem is to efficiently extract down values for a desired index, for example issue something like
result = ExtractConfigurationsAndOccurences[f, 2]
which should give a list with the structure
result = {list1, list2}
where
list1 = {{1, 6}, {2, 7}} (* the list of configurations that occurred during the simulation*)
list2 = {216, 227} (* how many times each of them occurred *)
The problem is that ExtractConfigurationsAndOccurences should be very fast. The only solution I could come up with was to use SubValues[f] (which gives the full list) and filter it with Cases statement. I realize that this procedure should be avoided at any cost since there will be exponentially many configurations (definitions) to test, which slows down the code considerably.
Is there a natural way in Mathematica to do this in a fast way?
I was hoping that Mathematica would see f[2] as a single head with many down values but using DownValues[f[2]] gives nothing. Also using SubValues[f[2]] results in an error.
This is a complete rewrite of my previous answer. It turns out that in my previous attempts, I overlooked a much simpler method based on a combination of packed arrays and sparse arrays, that is much faster and more memory - efficient than all previous methods (at least in the range of sample sizes where I tested it), while only minimally changing the original SubValues - based approach. Since the question was asked about the most efficient method, I will remove the other ones from the answer (given that they are quite a bit more complex and take a lot of space. Those who would like to see them can inspect past revisions of this answer).
The original SubValues - based approach
We start by introducing a function to generate the test samples of configurations for us. Here it is:
Clear[generateConfigurations];
generateConfigurations[maxIndex_Integer, maxConfX_Integer, maxConfY_Integer,
nconfs_Integer] :=
Transpose[{
RandomInteger[{1, maxIndex}, nconfs],
Transpose[{
RandomInteger[{1, maxConfX}, nconfs],
RandomInteger[{1, maxConfY}, nconfs]
}]}];
We can generate a small sample to illustrate:
In[3]:= sample = generateConfigurations[2,2,2,10]
Out[3]= {{2,{2,1}},{2,{1,1}},{1,{2,1}},{1,{1,2}},{1,{1,2}},
{1,{2,1}},{2,{1,2}},{2,{2,2}},{1,{2,2}},{1,{2,1}}}
We have here only 2 indices, and configurations where both "x" and "y" numbers vary from 1 to 2 only - 10 such configurations.
The following function will help us imitate the accumulation of frequencies for configurations, as we increment SubValues-based counters for repeatedly occurring ones:
Clear[testAccumulate];
testAccumulate[ff_Symbol, data_] :=
Module[{},
ClearAll[ff];
ff[_][_] = 0;
Do[
doSomeStuff;
ff[#1][#2]++ & ## elem;
doSomeMoreStaff;
, {elem, data}]];
The doSomeStuff and doSomeMoreStaff symbols are here to represent some code that might preclude or follow the counting code. The data parameter is supposed to be a list of the form produced by generateConfigurations. For example:
In[6]:=
testAccumulate[ff,sample];
SubValues[ff]
Out[7]= {HoldPattern[ff[1][{1,2}]]:>2,HoldPattern[ff[1][{2,1}]]:>3,
HoldPattern[ff[1][{2,2}]]:>1,HoldPattern[ff[2][{1,1}]]:>1,
HoldPattern[ff[2][{1,2}]]:>1,HoldPattern[ff[2][{2,1}]]:>1,
HoldPattern[ff[2][{2,2}]]:>1,HoldPattern[ff[_][_]]:>0}
The following function will extract the resulting data (indices, configurations and their frequencies) from the list of SubValues:
Clear[getResultingData];
getResultingData[f_Symbol] :=
Transpose[{#[[All, 1, 1, 0, 1]], #[[All, 1, 1, 1]], #[[All, 2]]}] &#
Most#SubValues[f, Sort -> False];
For example:
In[10]:= result = getResultingData[ff]
Out[10]= {{2,{2,1},1},{2,{1,1},1},{1,{2,1},3},{1,{1,2},2},{2,{1,2},1},
{2,{2,2},1},{1,{2,2},1}}
To finish with the data-processing cycle, here is a straightforward function to extract data for a fixed index, based on Select:
Clear[getResultsForFixedIndex];
getResultsForFixedIndex[data_, index_] :=
If[# === {}, {}, Transpose[#]] &[
Select[data, First## == index &][[All, {2, 3}]]];
For our test example,
In[13]:= getResultsForFixedIndex[result,1]
Out[13]= {{{2,1},{1,2},{2,2}},{3,2,1}}
This is presumably close to what #zorank tried, in code.
A faster solution based on packed arrays and sparse arrays
As #zorank noted, this becomes slow for larger sample with more indices and configurations. We will now generate a large sample to illustrate that (note! This requires about 4-5 Gb of RAM, so you may want to reduce the number of configurations if this exceeds the available RAM):
In[14]:=
largeSample = generateConfigurations[20,500,500,5000000];
testAccumulate[ff,largeSample];//Timing
Out[15]= {31.89,Null}
We will now extract the full data from the SubValues of ff:
In[16]:= (largeres = getResultingData[ff]); // Timing
Out[16]= {10.844, Null}
This takes some time, but one has to do this only once. But when we start extracting data for a fixed index, we see that it is quite slow:
In[24]:= getResultsForFixedIndex[largeres,10]//Short//Timing
Out[24]= {2.687,{{{196,26},{53,36},{360,43},{104,144},<<157674>>,{31,305},{240,291},
{256,38},{352,469}},{<<1>>}}}
The main idea we will use here to speed it up is to pack individual lists inside the largeres, those for indices, combinations and frequencies. While the full list can not be packed, those parts individually can:
In[18]:= Timing[
subIndicesPacked = Developer`ToPackedArray[largeres[[All,1]]];
subCombsPacked = Developer`ToPackedArray[largeres[[All,2]]];
subFreqsPacked = Developer`ToPackedArray[largeres[[All,3]]];
]
Out[18]= {1.672,Null}
This also takes some time, but it is a one-time operation again.
The following functions will then be used to extract the results for a fixed index much more efficiently:
Clear[extractPositionFromSparseArray];
extractPositionFromSparseArray[HoldPattern[SparseArray[u___]]] := {u}[[4, 2, 2]]
Clear[getCombinationsAndFrequenciesForIndex];
getCombinationsAndFrequenciesForIndex[packedIndices_, packedCombs_,
packedFreqs_, index_Integer] :=
With[{positions =
extractPositionFromSparseArray[
SparseArray[1 - Unitize[packedIndices - index]]]},
{Extract[packedCombs, positions],Extract[packedFreqs, positions]}];
Now, we have:
In[25]:=
getCombinationsAndFrequenciesForIndex[subIndicesPacked,subCombsPacked,subFreqsPacked,10]
//Short//Timing
Out[25]= {0.094,{{{196,26},{53,36},{360,43},{104,144},<<157674>>,{31,305},{240,291},
{256,38},{352,469}},{<<1>>}}}
We get a 30 times speed-up w.r.t. the naive Select approach.
Some notes on complexity
Note that the second solution is faster because it uses optimized data structures, but its complexity is the same as that of Select- based one, which is, linear in the length of total list of unique combinations for all indices. Therefore, in theory, the previously - discussed solutions based on nested hash-table etc may be asymptotically better. The problem is, that in practice we will probably hit the memory limitations long before that. For the 10 million configurations sample, the above code was still 2-3 times faster than the fastest solution I posted before.
EDIT
The following modification:
Clear[getCombinationsAndFrequenciesForIndex];
getCombinationsAndFrequenciesForIndex[packedIndices_, packedCombs_,
packedFreqs_, index_Integer] :=
With[{positions =
extractPositionFromSparseArray[
SparseArray[Unitize[packedIndices - index], Automatic, 1]]},
{Extract[packedCombs, positions], Extract[packedFreqs, positions]}];
makes the code twice faster still. Moreover, for more sparse indices (say, calling the sample-generation function with parameters like generateConfigurations[2000, 500, 500, 5000000] ), the speed-up with respect to the Select- based function is about 100 times.
I'd probably use SparseArrays here (see update below), but if you insist on using functions and *Values to store and retrieve values an approach would be to have the first part (f[2] etc.) replaced by a symbol you create on the fly like:
Table[Symbol["f" <> IntegerString[i, 10, 3]], {i, 11}]
(* ==> {f001, f002, f003, f004, f005, f006, f007, f008, f009, f010, f011} *)
Symbol["f" <> IntegerString[56, 10, 3]]
(* ==> f056 *)
Symbol["f" <> IntegerString[56, 10, 3]][{3, 4}] = 12;
Symbol["f" <> IntegerString[56, 10, 3]][{23, 18}] = 12;
Symbol["f" <> IntegerString[56, 10, 3]] // Evaluate // DownValues
(* ==> {HoldPattern[f056[{3, 4}]] :> 12, HoldPattern[f056[{23, 18}]] :> 12} *)
f056 // DownValues
(* ==> {HoldPattern[f056[{3, 4}]] :> 12, HoldPattern[f056[{23, 18}]] :> 12} *)
Personally I prefer Leonid's solution, as it's much more elegant but YMMV.
Update
On OP's request, about using SparseArrays:
Large SparseArrays take up a fraction of the size of standard nested lists. We can make f to be a large (100,000 entires) sparse array of sparse arrays:
f = SparseArray[{_} -> 0, 100000];
f // ByteCount
(* ==> 672 *)
(* initialize f with sparse arrays, takes a few seconds with f this large *)
Do[ f[[i]] = SparseArray[{_} -> 0, {100, 110}], {i,100000}] // Timing//First
(* ==> 18.923 *)
(* this takes about 2.5% of the memory that a normal array would take: *)
f // ByteCount
(* ==> 108000040 *)
ConstantArray[0, {100000, 100, 100}] // ByteCount
(* ==> 4000000176 *)
(* counting phase *)
f[[1]][[1, 2]]++;
f[[1]][[1, 2]]++;
f[[1]][[42, 64]]++;
f[[2]][[100, 11]]++;
(* reporting phase *)
f[[1]] // ArrayRules
f[[2]] // ArrayRules
f // ArrayRules
(*
==>{{1, 2} -> 2, {42, 64} -> 1, {_, _} -> 0}
==>{{100, 11} -> 1, {_, _} -> 0}
==>{{1, 1, 2} -> 2, {1, 42, 64} -> 1, {2, 100, 11} -> 1, {_, _, _} -> 0}
*)
As you can see, ArrayRules makes a nice list with contributions and counts. This can be done for each f[i] separately or the whole bunch together (last line).
In some scenarios (depending upon the performance needed to generate the values), the following easy solution using an auxiliary list (f[i,0]) may be useful:
f[_Integer][{_Integer ..}] := 0;
f[_Integer, 0] := Sequence ## {};
Table[
r = RandomInteger[1000, 2];
f[h = RandomInteger[100000]][r] = RandomInteger[10];
f[h, 0] = Union[f[h, 0], {r}];
, {i, 10^6}];
ExtractConfigurationsAndOccurences[f_, i_] := {f[i, 0], f[i][#] & /# f[i, 0]};
Timing#ExtractConfigurationsAndOccurences[f, 10]
Out[252]= {4.05231*10^-15, {{{172, 244}, {206, 115}, {277, 861}, {299,
862}, {316, 194}, {361, 164}, {362, 830}, {451, 306}, {614,
769}, {882, 159}}, {5, 2, 1, 5, 4, 10, 4, 4, 1, 8}}}
Many thanks for everyone on the help provided. I've been thinking a lot about everybody's input and I believe that in the simulation setup the following is the optimal solution:
SetAttributes[linkedList, HoldAllComplete];
temporarySymbols = linkedList[];
SetAttributes[bookmarkSymbol, Listable];
bookmarkSymbol[symbol_]:=
With[{old = temporarySymbols}, temporarySymbols= linkedList[old,symbol]];
registerConfiguration[index_]:=registerConfiguration[index]=
Module[
{
cs = linkedList[],
bookmarkConfiguration,
accumulator
},
(* remember the symbols we generate so we can remove them later *)
bookmarkSymbol[{cs,bookmarkConfiguration,accumulator}];
getCs[index] := List ## Flatten[cs, Infinity, linkedList];
getCsAndFreqs[index] := {getCs[index],accumulator /# getCs[index]};
accumulator[_]=0;
bookmarkConfiguration[c_]:=bookmarkConfiguration[c]=
With[{oldCs=cs}, cs = linkedList[oldCs, c]];
Function[c,
bookmarkConfiguration[c];
accumulator[c]++;
]
]
pattern = Verbatim[RuleDelayed][Verbatim[HoldPattern][HoldPattern[registerConfiguration [_Integer]]],_];
clearSimulationData :=
Block[{symbols},
DownValues[registerConfiguration]=DeleteCases[DownValues[registerConfiguration],pattern];
symbols = List ## Flatten[temporarySymbols, Infinity, linkedList];
(*Print["symbols to purge: ", symbols];*)
ClearAll /# symbols;
temporarySymbols = linkedList[];
]
It is based on Leonid's solution from one of previous posts, appended with belsairus' suggestion to include extra indexing for configurations that have been processed. Previous approaches are adapted so that configurations can be naturally registered and extracted using the same code more or less. This is hitting two flies at once since bookkeeping and retrieval and strongly interrelated.
This approach will work better in the situation when one wants to add simulation data incrementally (all curves are normally noisy so one has to add runs incrementally to obtain good plots). The sparse array approach will work better when data are generated in one go and then analyzed, but I do not remember being personally in such a situation where I had to do that.
Also, I was rather naive thinking that the data extraction and generation could be treated separately. In this particular case it seems one should have both perspectives in mind. I profoundly apologise for bluntly dismissing any previous suggestions in this direction (there were few implicit ones).
There are some open/minor problems that I do not know how to handle, e.g. when clearing the symbols I cannot clear headers like accumulator$164, I can only clean subvalues associated with it. Have not clue why. Also, if With[{oldCs=cs}, cs = linkedList[oldCs, c]]; is changed into something like cs = linkedList[cs, c]]; configurations are not stored. Have no clue either why the second option does not work. But these minor problems are well defined satellite issues that one can address in the future. By and large the problem seems solved by the generous help from all involved.
Many thanks again for all the help.
Regards
Zoran
p.s. There are some timings, but to understand what is going on I will append the code that is used for benchmarking. In brief, idea is to generate lists of configurations and just Map through them by invoking registerConfiguration. This essentially simulates data generation process. Here is the code used for testing:
fillSimulationData[sampleArg_] :=MapIndexed[registerConfiguration[#2[[1]]][#1]&, sampleArg,{2}];
sampleForIndex[index_]:=
Block[{nsamples,min,max},
min = Max[1,Floor[(9/10)maxSamplesPerIndex]];
max = maxSamplesPerIndex;
nsamples = RandomInteger[{min, max}];
RandomInteger[{1,10},{nsamples,ntypes}]
];
generateSample :=
Table[sampleForIndex[index],{index, 1, nindexes}];
measureGetCsTime :=((First # Timing[getCs[#]])& /# Range[1, nindexes]) // Max
measureGetCsAndFreqsTime:=((First # Timing[getCsAndFreqs[#]])& /# Range[1, nindexes]) // Max
reportSampleLength[sampleArg_] := StringForm["Total number of confs = ``, smallest accumulator length ``, largest accumulator length = ``", Sequence## {Total[#],Min[#],Max[#]}& [Length /# sampleArg]]
The first example is relatively modest:
clearSimulationData;
nindexes=100;maxSamplesPerIndex = 1000; ntypes = 2;
largeSample1 = generateSample;
reportSampleLength[largeSample1];
Total number of confs = 94891, smallest accumulator length 900, largest accumulator length = 1000;
First # Timing # fillSimulationData[largeSample1]
gives 1.375 secs which is fast I think.
With[{times = Table[measureGetCsTime, {50}]},
ListPlot[times, Joined -> True, PlotRange -> {0, Max[times]}]]
gives times around 0.016 secs, and
With[{times = Table[measureGetCsAndFreqsTime, {50}]},
ListPlot[times, Joined -> True, PlotRange -> {0, Max[times]}]]
gives same times. Now the real killer
nindexes = 10; maxSamplesPerIndex = 100000; ntypes = 10;
largeSample3 = generateSample;
largeSample3 // Short
{{{2,2,1,5,1,3,7,9,8,2},92061,{3,8,6,4,9,9,7,8,7,2}},8,{{4,10,1,5,9,8,8,10,8,6},95498,{3,8,8}}}
reported as
Total number of confs = 933590, smallest accumulator length 90760, largest accumulator length = 96876
gives generation times of ca 1.969 - 2.016 secs which is unbeliavably fast. I mean this is like going through the gigantic list of ca one million elements and applying a function to each element.
The extraction times for configs and {configs, freqs} are roughly 0.015 and 0.03 secs respectivelly.
To me this is a mind blowing speed I would never expect from Mathematica!