For loop to change the values of a four dimensional table - wolfram-mathematica

I would like your help on something,
I have a Table:
InitialMatrix[x_, y_, age_, disease_] :=
ReplacePart[Table[Floor[Divide[dogpopulation/cellsno,9]], {x}, {y}, {age}, {disease}],
{{_, _, 1, _} -> 0, {_, _, 3, _} -> 6}];
I was trying to set up a condition to change all the values inside the table to sumthing else, according to a value, I tried:
listInitial={};
For[a = 1, a < 4, a++,
For[b = 1, b < 4, b++,
For[x = 1, x < 4, x = x + 1,
For[z = 1, z < 4, z = z + 1,
listInitial =
If[Random[] > psurvival,
ReplacePart[ InitialMatrix[3, 3, 3, 3], {a, b, x, z} ->
InitialMatrix[3, 3, 3, 3][[a]][[b]][[x]][[z]] - 1],
InitialMatrix[3, 3, 3, 3], {a, b, x, z} ->
InitialMatrix[3, 3, 3, 3][[a]][[b]][[x]][[z]]]]]]]
but it only changes the last part of my table, finally I decided to use the following code instead of the for loop,
SetAttributes[myFunction, Listable]
myFunction[x_] :=
If[Random[] > psurvival, If [x - 1 < 0 , x , x - 1], x]
myFunction[InitialMatrix[3, 3, 3, 3]] // TableForm
but now I want to change specific parts inside the table, for example I want all the part
{__,__,3,_} to change I tried to choose the range with MapAt but again I think I need to do a loop, and I cannot, can any one please help me?
For[x = 1, x < 4, x++,
listab[MapAt[f, InitialMatrix[3, 3, 3, 3], {x, 3, 3}]//TableForm]]

If you check out the documentation for MapAt, you will see that you can address multiple elements at various depths of your tensor, using various settings of the third argument. Note also the use of Flatten's second argument. I think this is what you are looking for.
MapAt[g, InitialMatrix[3, 3, 3, 3],
Flatten[Table[{i, j, 3, k}, {i, 3}, {j, 3}, {k, 3}], 2]]
http://reference.wolfram.com/mathematica/ref/MapAt.html
http://reference.wolfram.com/mathematica/ref/Flatten.html
Since this seems to be your second attempt to ask a question involving a really complicated For loop, may I just emphasise that you almost never need a For or Do loop in Mathematica in the circumstances where you would use one in, say, Fortran or C. Certainly not for most construction of lists. Table works. So do things like Listable functions (which I know you know) and commands like NestList, FoldList and Array.
You will probably also find this tutorial useful.
http://reference.wolfram.com/mathematica/tutorial/SelectingPartsOfExpressionsWithFunctions.html

I used the following code as an answer, I am not sure whether is the best solution or not, but it works!!
InitialTable[x_, y_, z_, w_] :=
MapAt[g,ReplacePart[
InitialMatrix[3, 3, 3, 3] +
ReplacePart[
Table[If[RandomReal[] > psurvival, -1,
0], {3}, {3}, {3}, {3}], {{_, _, 1, _} -> 0, {_, _, 2, _} ->
0}], {{_, _, 1, 2} -> 0, {_, _, 1, 3} -> 0}],
Flatten[Table[{i, j, 3, l}, {i, x}, {j, y}, {l, w}], 2]];
g[x_] := If[x < 0, 0, x];

Related

Mathematica Nested Manipulates Lag

Ok! I'm working towards building a nested manipulate command that will solve n number of damped oscillating masses in series (with fixed endpoints). I have everything pretty much working but I have one problem - when I increase the number of oscillators, my initial conditions lag behind a bit. For example, if I set n to 4, Mathematica says I still only have 2 initial conditions (the starting number - position and velocity for one oscillator). When I then move to 3, I now have 8 (from my 4 oscillators) - which is too many for the state space equations, and it all fails. What is going on?
(Yes, I know that my initial conditions aren't going to be put in correctly yet, I'm just trying to get them to match up first).
coupledSMD[n_, m_, k_, b_, f_, x0_, v0_, tmax_] :=
Module[{aM, bM, cM, dM},
aM = Join[Table[Boole[i == j - n], {i, n}, {j, 2 n}],
Join[
If[n != 1,DiagonalMatrix[-2 k/m Table[1, {n}]] +
k/m DiagonalMatrix[Table[1, {n - 1}], 1] +
k/m DiagonalMatrix[Table[1, {n - 1}], -1],
{{-2 k/m}}],
If[n != 1,DiagonalMatrix[-2 b/m Table[1, {n}]] +
b/m DiagonalMatrix[Table[1, {n - 1}], 1] +
b/m DiagonalMatrix[Table[1, {n - 1}], -1],
{{-2 b/m}}], 2]];
bM = Join[Table[0, {n}, {1}], Table[1/m, {n}, {1}]];
cM = Table[Boole[i == j], {i, n}, {j, 2 n}];
dM = Table[0, {n}, {1}];
OutputResponse[
{StateSpaceModel[{aM, bM, cM, dM}], Flatten[Join[x0, v0]]},
f, {t, 0, tmax}]
]
Manipulate[
With[{
x0s = Table[Subscript[x, i, 0], {i, 1, n}],
v0s = Table[Subscript[v, i, 0], {i, 1, n}],
initialx = Sequence ## Table[{{Subscript[x, i, 0], 0}, -10, 10}, {i, 1, n}],
initialv = Sequence ## Table[{{Subscript[v, i, 0], 0}, -10, 10}, {i, 1, n}]},
Manipulate[
myplot = coupledSMD[n, m, k, b, f, x0s, v0s, tmax];
Plot[myplot, {t, 0, tmax}, PlotRange -> yheight {-1, 1},
PlotLegends -> Table[Subscript[x, i, 0], {i, 1, n}]],
Style["Initial Positions", Bold],
initialx,
Delimiter,
Style["Initial Velocities", Bold],
initialv,
Delimiter,
Style["System conditions", Bold],
{{m, 1, "Mass(kg)"}, 0.1, 10, Appearance -> "Labeled"},
{{k, 1, "Spring Constant(N/m)"}, 0.1, 10, Appearance -> "Labeled"},
{{b, 0, "Damping Coefficient(N.s/m)"}, 0, 1, Appearance -> "Labeled"},
{{f, 0, "Applied Force"}, 0, 10, Appearance -> "Labeled"},
Delimiter,
Style["Plot Ranges", Bold],
{tmax, 10, 100},
{{yheight, 10}, 1, 100},
Delimiter,
ControlPlacement -> Flatten[{Table[Right, {2 n + 2}], Table[Left, {8}]}]
]],
{n, 1, 4, 1}
]
Edit: Updated the code. It works now, but I'm still getting the errors. I'm guessing that it has something to do with a time lag in the updating process? - That some parts are getting updated before others. Again, it seems to be working perfectly, except it throws these errors (the errors seem superfluous to me, as if they are remnants in the code, but not actually causing a problem)
But I don't really know what I'm talking about :)

form a matrix by picking row with maximum value at first place from a matrix whose each element has various rows with 3 columns each

I have a matrix Inter[i,j] whose each element has a set of rows with 3 columns each.
for eg. Inter[15,40]= {{0,0,0},{1,1,3}}, Inter[32,64]={{0,0,0},{1,1,3},{2,2,3}} and so on
Now i want a matrix MaxGG[i,j] by picking a row from Inter[i,j] that has maximum value at the first place.
for eg MaxGG[15,40] should give me {1,1,3}. similarly MaxGG[32,64]={2,2,3}.
When i am giving this code,it does not give me the right answer:
MaxGG[i_, j_] := Cases[Inter[i, j], {Max#Inter[i, j][[All, 1]], _, _}];
Please suggest me what is wrong with this code and please provide me with the correct code for this.
Thanks in advance :)
The main problem is that Cases will return a list of all matching cases :-
Inter[15, 40] = {{0, 0, 0}, {1, 1, 3}};
Inter[32, 64] = {{0, 0, 0}, {1, 1, 3}, {2, 2, 3}};
MaxGG[i_, j_] := Cases[Inter[i, j], {Max#Inter[i, j][[All, 1]], _, _}]
MaxGG[32, 64]
{{2, 2, 3}}
You can lose the outer brackets by selecting the first match :-
MaxGG[i_, j_] := First#Cases[Inter[i, j], {Max#Inter[i, j][[All, 1]], _, _}]
MaxGG[32, 64]
{2, 2, 3}
The function can also be simplified :-
MaxGG[i_, j_] := First#Reverse#Sort#Inter[i, j]
MaxGG[32, 64]
{2, 2, 3}

Deriving Mean of Values within a Tensor

I have a 20000 x 185 x 5 tensor, which looks like
{{{a1_1,a2_1,a3_1,a4_1,a5_1},{b1_1,b2_1,b3_1,b4_1,b5_1}...
(continue for 185 times)}
{{a1_2,a2_2,a3_2,a4_2,a5_2},{b1_2,b2_2,b3_2,b4_2,b5_2}...
...
...
...
{{a1_20000,a2_20000,a3_20000,a4_20000,a5_20000},
{b1_20000,b2_20000,b3_20000,b4_20000,b5_20000}... }}
The 20000 represents iteration number, the 185 represents individuals, and each individual has 5 attributes. I need to construct a 185 x 5 matrix that stores the mean value for each individual's 5 attributes, averaged across the 20000 iterations.
Not sure what the best way to do this is. I know Mean[ ] works on matrices, but with a Tensor, the derived values might not be what I need. Also, Mathematica ran out of memory if I tried to do Mean[tensor]. Please provide some help or advice. Thank you.
When in doubt, drop the size of the dimensions. (You can still keep them distinct to easily see where things end up.)
(* In[1]:= *) data = Array[a, {4, 3, 2}]
(* Out[1]= *) {{{a[1, 1, 1], a[1, 1, 2]}, {a[1, 2, 1],
a[1, 2, 2]}, {a[1, 3, 1], a[1, 3, 2]}}, {{a[2, 1, 1],
a[2, 1, 2]}, {a[2, 2, 1], a[2, 2, 2]}, {a[2, 3, 1],
a[2, 3, 2]}}, {{a[3, 1, 1], a[3, 1, 2]}, {a[3, 2, 1],
a[3, 2, 2]}, {a[3, 3, 1], a[3, 3, 2]}}, {{a[4, 1, 1],
a[4, 1, 2]}, {a[4, 2, 1], a[4, 2, 2]}, {a[4, 3, 1], a[4, 3, 2]}}}
(* In[2]:= *) Dimensions[data]
(* Out[2]= *) {4, 3, 2}
(* In[3]:= *) means = Mean[data]
(* Out[3]= *) {
{1/4 (a[1, 1, 1] + a[2, 1, 1] + a[3, 1, 1] + a[4, 1, 1]),
1/4 (a[1, 1, 2] + a[2, 1, 2] + a[3, 1, 2] + a[4, 1, 2])},
{1/4 (a[1, 2, 1] + a[2, 2, 1] + a[3, 2, 1] + a[4, 2, 1]),
1/4 (a[1, 2, 2] + a[2, 2, 2] + a[3, 2, 2] + a[4, 2, 2])},
{1/4 (a[1, 3, 1] + a[2, 3, 1] + a[3, 3, 1] + a[4, 3, 1]),
1/4 (a[1, 3, 2] + a[2, 3, 2] + a[3, 3, 2] + a[4, 3, 2])}
}
(* In[4]:= *) Dimensions[means]
(* Out[4]= *) {3, 2}
Mathematica ran out of memory if I tried to do Mean[tensor]
This is probably because intermediate results are larger than the final result. This is likely if the elements are not type Real or Integer. Example:
a = Tuples[{x, Sqrt[y], z^x, q/2, Mod[r, 1], Sin[s]}, {2, 4}];
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
{109125576, 124244808}
{269465456, 376960648}
If they are, and are in packed array form, perhaps the elements are such that the array in unpacked during processing.
Here is an example where the tensor is a packed array of small numbers, and unpacking does not occur.
a = RandomReal[99, {20000, 185, 5}];
PackedArrayQ[a]
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
True
{163012808, 163016952}
{163018944, 163026688}
Here is the same size of tensor with very large numbers.
a = RandomReal[$MaxMachineNumber, {20000, 185, 5}];
Developer`PackedArrayQ[a]
{MemoryInUse[], MaxMemoryUsed[]}
b = Mean[a];
{MemoryInUse[], MaxMemoryUsed[]}
True
{163010680, 458982088}
{163122608, 786958080}
To elaborate a little on the other answers, there is no reason to expect Mathematica functions to operate materially differently on tensors than matrices because Mathemetica considers them both to be nested Lists, that are just of different nesting depth. How functions behave with lists depends on whether they're Listable, which you can check using Attributes[f], where fis the function you are interested in.
Your data list's dimensionality isn't actually that big in the scheme of things. Without seeing your actual data it is hard to be sure, but I suspect the reason you are running out of memory is that some of your data is non-numerical.
I don't know what you're doing incorrectly (your code will help). But Mean[] already works as you want it to.
a = RandomReal[1, {20000, 185, 5}];
b = Mean#a;
Dimensions#b
Out[1]= {185, 5}
You can even check that this is correct:
{Max#b, Min#b}
Out[2]={0.506445, 0.494061}
which is the expected value of the mean given that RandomReal uses a uniform distribution by default.
Assume you have the following data :
a = Table[RandomInteger[100], {i, 20000}, {j, 185}, {k, 5}];
In a straightforward manner You can find a table which stores the means of a[[1,j,k]],a[[2,j,k]],...a[[20000,j,k]]:
c = Table[Sum[a[[i, j, k]], {i, Length[a]}], {j, 185}, {k, 5}]/
Length[a] // N; // Timing
{37.487, Null}
or simply :
d = Total[a]/Length[a] // N; // Timing
{0.702, Null}
The second way is about 50 times faster.
c == d
True
To extend on Brett's answer a bit, when you call Mean on a n-dimensional tensor then it averages over the first index and returns an n-1 dimensional tensor:
a = RandomReal[1, {a1, a2, a3, ... an}];
Dimensions[a] (* This would have n entries in it *)
b = Mean[a];
Dimensions[b] (* Has n-1 entries, where averaging was done over the first index *)
In the more general case where you may wish to average over the i-th argument, you would have to transpose the data around first. For example, say you want to average the 3nd of 5 dimensions. You would need the 3rd element first, followed by the 1st, 2nd, 4th, 5th.
a = RandomReal[1, {5, 10, 2, 40, 10}];
b = Transpose[a, {2, 3, 4, 1, 5}];
c = Mean[b]; (* Now of dimensions {5, 10, 40, 10} *)
In other words, you would make a call to Transpose where you placed the i-th index as the first tensor index and moved everything before it ahead one. Anything that comes after the i-th index stays the same.
This tends to come in handy when your data comes in odd formats where the first index may not always represent different realizations of a data sample. I've had this come up, for example, when I had to do time averaging of large wind data sets where the time series came third (!) in terms of the tensor representation that was available.
You could imagine the generalizedTenorMean would look something like this then:
Clear[generalizedTensorMean];
generalizedTensorMean[A_, i_] :=
Module[{n = Length#Dimensions#A, ordering},
ordering =
Join[Table[x, {x, 2, i}], {1}, Table[x, {x, i + 1, n}]];
Mean#Transpose[A, ordering]]
This reduces to the plain-old-mean when i == 1. Try it out:
A = RandomReal[1, {2, 4, 6, 8, 10, 12, 14}];
Dimensions#A (* {2, 4, 6, 8, 10, 12, 14} *)
Dimensions#generalizedTensorMean[A, 1] (* {4, 6, 8, 10, 12, 14} *)
Dimensions#generalizedTensorMean[A, 7] (* {2, 4, 6, 8, 10, 12} *)
On a side note, I'm surprised that Mathematica doesn't support this by default. You don't always want to average over the first level of a list.

Optimising the game of life

I'm writing a game of life program in mathematica however there is a caveat in that I need to be able to apply the reproduction rules to some percentage of the cells, I want to try a new method using MapAt but liveNeighbors doesn't work elementwise, and I can't think of a way of fixing it without doing exactly what I did before (lots of messy indexing), does anyone have any suggestions? (I am assuming this will be more efficient then the old method, which is listed below, if not please let me know, I am just a beginner!).
What I am trying to do:
Map[ArrayPlot,FixedPointList[MapAt[update[#,liveNeighbors[#]]&,#,coords]&,Board, 1]]
What I have done already:
LifeGame[ n_Integer?Positive, steps_] := Module [{Board, liveNeighbors, update},
Board = Table [Random [Integer], {n}, {n}];
liveNeighbors[ mat_] :=
Apply[Plus,Map[RotateRight[mat,#]&,{{-1,-1},{-1, 0},{-1,1}, {0, -1}, {0, 1}, {1, -1}, {1, 0}, {1, 1}}]];
update[1, 2] := 1;
update[_, 3] := 1;
update[ _, _] := 0;
SetAttributes[update, Listable];
Seed = RandomVariate[ProbabilityDistribution[0.7 UnitStep[x] + 0.3 UnitStep[x - 1], {x, 0, 1, 1}], {n, n}];
FixedPointList[Table[If[Seed[[i, j]] == 1,update[#[[i, j]], liveNeighbors[#][[i, j]]],#[[i, j]]], {i, n}, {j, n}]&, Board, steps]]]
Thanks!
In[156]:=
LifeGame2[n_Integer?Positive, steps_] :=
Module[{Board, liveNeighbors, update},
Board = RandomInteger[1, {n, n}];
liveNeighbors[mat_] :=
ListConvolve[{{1, 1, 1}, {1, 0, 1}, {1, 1, 1}},
ArrayPad[mat, 1, "Periodic"]];
SetAttributes[update, Listable];
Seed = RandomVariate[BernoulliDistribution[0.3], {n, n}];
update[0, el_, nei_] := el;
update[1, 1, 2] := 1;
update[1, _, 3] := 1;
update[1, _, _] := 0;
FixedPointList[MapThread[update, {Seed, #, liveNeighbors[#]}, 2] &,
Board, steps]
]
This implementation does the same as yours, except is quite a lot faster:
In[162]:= AbsoluteTiming[
res1 = BlockRandom[SeedRandom[11]; LifeGame[20, 100]];]
Out[162]= {6.3476347, Null}
In[163]:= Timing[BlockRandom[Seed[11]; LifeGame2[20, 100]] == res1]
Out[163]= {0.047, True}
Assuming you don't have to roll your own code for a homework problem, have you considered just using the built-in CellularAutomaton function?
Straight from the documentation, the 2D CA rule:
GameOfLife = {224, {2, {{2, 2, 2}, {2, 1, 2}, {2, 2, 2}}}, {1, 1}};
And iterate over a 100x100 grid for 100 steps:
ArrayPlot[CellularAutomaton[GameOfLife, RandomInteger[1, {100, 100}], {{{100}}}]]
It would at least give you a baseline for a speed comparison.
Instead of MapAt, you could use Part with the Span syntax to replace a whole subarray at once:
a = ConstantArray[0, {5, 5}];
a[[2 ;; 4, 2 ;; 4]] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}
HTH!
Here you have my golfed version.

enumerating all partitions in Mathematica

Like Select[Tuples[Range[0, n], d], Total[#] == n &], but faster?
Update
Here are the 3 solutions and plot of their times, IntegerPartitions followed by Permutations seems to be fastest. Timing at 1, 7, 0.03 for recursive, FrobeniusSolve and IntegerPartition solutions respectively
partition[n_, 1] := {{n}};
partition[n_, d_] :=
Flatten[Table[
Map[Join[{k}, #] &, partition[n - k, d - 1]], {k, 0, n}], 1];
f[n_, d_, 1] := partition[n, d];
f[n_, d_, 2] := FrobeniusSolve[Array[1 &, d], n];
f[n_, d_, 3] :=
Flatten[Permutations /# IntegerPartitions[n, {d}, Range[0, n]], 1];
times = Table[First[Log[Timing[f[n, 8, i]]]], {i, 1, 3}, {n, 3, 8}];
Needs["PlotLegends`"];
ListLinePlot[times, PlotRange -> All,
PlotLegend -> {"Recursive", "Frobenius", "IntegerPartitions"}]
Exp /# times[[All, 6]]
Your function:
In[21]:= g[n_, d_] := Select[Tuples[Range[0, n], d], Total[#] == n &]
In[22]:= Timing[g[15, 4];]
Out[22]= {0.219, Null}
Try FrobeniusSolve:
In[23]:= f[n_, d_] := FrobeniusSolve[ConstantArray[1, d], n]
In[24]:= Timing[f[15, 4];]
Out[24]= {0.031, Null}
The results are the same:
In[25]:= f[15, 4] == g[15, 4]
Out[25]= True
You can make it faster with IntegerPartitions, though you don't get the results in the same order:
In[43]:= h[n_, d_] :=
Flatten[Permutations /# IntegerPartitions[n, {d}, Range[0, n]], 1]
The sorted results are the same:
In[46]:= Sort[h[15, 4]] == Sort[f[15, 4]]
Out[46]= True
It is much faster:
In[59]:= {Timing[h[15, 4];], Timing[h[23, 5];]}
Out[59]= {{0., Null}, {0., Null}}
Thanks to phadej's fast answer for making me look again.
Note you only need the call to Permutations (and Flatten) if you actually want all the differently ordered permutations, i.e. if you want
In[60]:= h[3, 2]
Out[60]= {{3, 0}, {0, 3}, {2, 1}, {1, 2}}
instead of
In[60]:= etc[3, 2]
Out[60]= {{3, 0}, {2, 1}}
partition[n_, 1] := {{n}}
partition[n_, d_] := Flatten[ Table[ Map[Join[{k}, #] &, partition[n - k, d - 1]], {k, 0, n}], 1]
This is even quicker than FrobeniusSolve :)
Edit: If written in Haskell, it's probably clearer what is happening - functional as well:
partition n 1 = [[n]]
partition n d = concat $ map outer [0..n]
where outer k = map (k:) $ partition (n-k) (d-1)

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