I have to solve a system of non-linear equations for a wide range of parameter space. I'm using FindRoot, which is sensitive to initial start point so I have to do it by hand and by trial and error and plotting, rather than putting the equations in a loop or in a table.
So what I want to do is create a database or a Matrix with a fixed number of columns but variable number of rows so I can keep appending it with new results as and when I solve for them.
Right now I've used something like:
{{{xx, yy}} = {x, y} /. FindRoot[{f1(x,y) == 0,f2(x,y)==0}, {x,a},{y,b}],
g(xx,yy)} >>> "Attempt1.txt"
Where I am solving for two variables and then storing the variables and also a function g(xx,yy) of the variables.
This seems to work for me but the result is not a Matrix any more but the data is stored as some text type thing.
Is there anyway I can get this to stay a matrix or a database where I keep adding rows to it each time I solve for FindRoot by hand? Again, I need to do FindRoot by hand because it is sensitive to the start points and I don't know the good start points without first plotting it.
Thanks a lot
Unless I'm not understanding what you're trying to do, this should work
results = {};
results = Append[Flatten[{{xx, yy} = {x, y} /. FindRoot[{f1(x,y) == 0,f2(x,y)==0}, {x,a},{y,b}],g(xx,yy)}],results];
Then every time you're tying to add a line to the matrix results by hand, you would just type
results = Append[Flatten[{{xx, yy} = {x, y} /. FindRoot[{f1(x,y) == 0,f2(x,y)==0}, {x,a},{y,b}],g(xx,yy)}],results];
By the way, to get around the problem of sensitivity to the initial a and b values, you could explore the parameter space in a loop, varying the parameters slowly and using the solution of x and y from the previous loop iteration for your new a and b values each time.
What you want to do can be achieved by using Read instead of Get. While Get reads the complete file in one run, Read can be adjusted to extract a single Expression, Byte, Number and many more. So what you should do is open your file and read expression after expression and pack it inside a list.
PutAppend[{{1, 2}, {3, 4}}, "tmp.mx"]
PutAppend[{{5, 6}, {7, 8}}, "tmp.mx"]
PutAppend[{{9, 23}, {11, 12}}, "tmp.mx"]
PutAppend[{{13, 14}, {15, 16}}, "tmp.mx"]
stream = OpenRead["tmp.mx"];
mat = ArrayPad[
NestWhileList[Read[stream, Expression] &,
stream = OpenRead["tmp.mx"], # =!= EndOfFile &], -1];
Close[stream];
And now you have in mat a list containing all lines. The ArrayPad, which cuts off one element at each end is necessary because the first element contains the output of the OpenRead and the last element contains EndOfFile. If you are not familiar with functional constructs like NestWhileList then you can put it in a loop as you like, since it is really just the iterated calls to Read
stream = OpenRead["tmp.mx"];
mat = {};
AppendTo[mat, Read[stream, Expression]];
AppendTo[mat, Read[stream, Expression]];
AppendTo[mat, Read[stream, Expression]];
AppendTo[mat, Read[stream, Expression]];
Close[stream];
Is there a way to make MatrixForm display a row vector horizontally on the line and not vertically as it does for column vectors? As this confuses me sometimes. Do you think it will be hard to write wrapper around matrix form to adjust this behavior?
For example, here is a 2 by 2 matrix. The rows display the same as the columns. Both are shown vertical.
Question: Is it possible to make MatrixForm display row vectors laid out horizontally and not vertically?
Sorry if this was asked before, a quick search shows nothing.
thanks
update (1)
fyi, this is in Matlab, it prints rows horizontally and column vertically automatically, I was hoping for something like this. But I'll use suggestion by Heike below for now as it solves this at the cost of little extra typing.
update (2)
Using Hilderic solution is nice also, I always had hard time printing 3D matrix in a way I can read it. Here it is now using the {} trick
For both arrayname[[All,1]] and arrayname[[1,All]], Part delivers a vector, and MatrixForm has no way of determining which "orientation" it has. Accordingly, it always prints vectors as columns.
About the only thing you can do is provide your own output routine for row vectors, e.g., by wrapping it in an enclosing list, converting it back to a (single-row) matrix:
rowVector[a_List] := MatrixForm[{a}]
columnVector = MatrixForm (*for symmetry*)
It's still up to you to remember whether a vector came from a row or a column, though.
Or you could just cook up your own RowForm function, e.g.:
RowForm[(m_)?VectorQ] := Row[{"(",Row[m," "],
")"}, "\[MediumSpace]"];
Then
RowForm[twoRowsMatrix[[All,1]]]
looks kind of o.k.
Alternatively, if you really just care about displaying vectors, you could do:
twoRowsMatrix = {{a11, a12}, {a21, a22}};
TakeColumn[m_?MatrixQ, i_] := (Print[MatrixForm[#]]; #) &#m[[All, i]];
TakeRow[m_?MatrixQ, i_] := (Print[MatrixForm[{#}]]; #) &#m[[i]];
TakeColumn[twoRowsMatrix, 1]
TakeRow[twoRowsMatrix, 1]
If you don't care about the () part, then you can append with ,{}, wrap in curly brackets, and use TableForm or Grid instead:
vec = {x, y, z};
TableForm[{vec, {}}]
Grid[{vec, {}}]
When I get worried about this, I use {{a,b,c}} to specify a row of a,b,c (they can be any kind of list) and Transpose[{{a,b,c}}] to specify a column of a,b,c.
MatrixForm[a = RandomInteger[{0, 6}, {2, 2}]]
MatrixForm[b = RandomInteger[{0, 6}, {2, 2}]]
MatrixForm[c = RandomInteger[{0, 6}, {2, 2}]]
w = {a, b, c};
MatrixForm[w]
w = {{a, b, c}};
MatrixForm[w]
w = Transpose[{{a, b, c}}];
MatrixForm[w]
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!
I use the following function to convert a number to a string for display purposes (don't use scientific notation, don't use a trailing dot, round as specified):
(* Show Number. Convert to string w/ no trailing dot. Round to the nearest r. *)
Unprotect[Round]; Round[x_,0] := x; Protect[Round];
shn[x_, r_:0] := StringReplace[
ToString#NumberForm[Round[N#x,r], ExponentFunction->(Null&)], re#"\\.$"->""]
(Note that re is an alias for RegularExpression.)
That's been serving me well for years.
But sometimes I don't want to specify the number of digits to round to, rather I want to specify a number of significant figures.
For example, 123.456 should display as 123.5 but 0.00123456 should display as 0.001235.
To get really fancy, I might want to specify significant digits both before and after the decimal point.
For example, I might want .789 to display as 0.8 but 789.0 to display as 789 rather than 800.
Do you have a handy utility function for this sort of thing, or suggestions for generalizing my function above?
Related: Suppressing a trailing "." in numerical output from Mathematica
UPDATE: I tried asking a general version of this question here:
https://stackoverflow.com/questions/5627185/displaying-numbers-to-non-technical-users
dreeves, I think I finally understand what you want, and you already had it, pretty much. If not, please try again to explain what I am missing.
shn2[x_, r_: 0] :=
StringReplace[
ToString#NumberForm[x, r, ExponentFunction -> (Null &)],
RegularExpression#"\\.0*$" -> ""]
Testing:
shn2[#, 4] & /# {123.456, 0.00123456}
shn2[#, {3, 1}] & /# {789.0, 0.789}
shn2[#, {10, 2}] & /# {0.1234, 1234.}
shn2[#, {4, 1}] & /# {12.34, 1234.56}
Out[1]= {"123.5", "0.001235"}
Out[2]= {"789", "0.8"}
Out[3]= {"0.12", "1234"}
Out[4]= {"12.3", "1235"}
This may not be the complete answer (you need to convert from/to string), but this function takes arguments a number x and significant figures sig wanted. The number of digits it keeps is the maximum of sig or the number of digits to the left of the decimal.
A[x_,sig_]:=NumberForm[x, Max[Last[RealDigits[x]], sig]]
RealDigits
Here's a possible generalization of my original function.
(I've determined that it's not equivalent to Mr Wizard's solution but I'm not sure yet which I think is better.)
re = RegularExpression;
(* Show Number. Convert to string w/ no trailing dot. Use at most d significant
figures after the decimal point. Target t significant figures total (clipped
to be at least i and at most i+d, where i is the number of digits in integer
part of x). *)
shn[x_, d_:5, t_:16] := ToString[x]
shn[x_?NumericQ, d_:5, t_:16] := With[{i= IntegerLength#IntegerPart#x},
StringReplace[ToString#NumberForm[N#x, Clip[t, {i,i+d}],
ExponentFunction->(Null&)],
re#"\\.$"->""]]
Testing:
Here we specify 4 significant digits, but never dropping any to the left of the decimal point and never using more than 2 significant digits to the right of the decimal point.
(# -> shn[#, 2, 4])& /#
{123456, 1234.4567, 123.456, 12.345, 1.234, 1.0001, 0.123, .0001234}
{ 123456 -> "123456",
1234.456 -> "1234",
123.456 -> "123.5"
12.345 -> "12.35",
1.234 -> "1.23",
1.0001 -> "1",
0.123 -> "0.12",
0.0001234 -> "0.00012" }