필터 지우기
필터 지우기

Accessing data from cells inside of cells

조회 수: 3 (최근 30일)
Thomas Veith
Thomas Veith 2019년 6월 3일
댓글: Star Strider 2019년 6월 3일
So I have the code below, which solves an ODE 1000 times with random variables at 10 different intervals (tspans). The result I end up with a 1000x1 cell array of 10x1 cell arrays. Each entry of the 10x1 cell arrays represents a 41x2 numeric array, which is the solution for x1 and x2 in each of the ten tspans. What I would like to end up with is a 410x1000 numeric array, where each column represents ten column 2 entries (x2 solutions) from the 41x2 numeric arrays per each 10x1 cell. Is this possible? Thanks in advance.
n = 1000;
result = cell(n,1);
for k=1:n
tspan1=[0 1];A01=rand;P01=rand;g1=rand;p1=rand;B1=rand;
[t1,x1] = ode45(@(t1,x1) [-g1*x1(1) + p1*x1(1); -x1(1)*x1(2)+ B1*x1(2)], tspan1, [A01 P01]);
tspan2=[1 2];A02=rand;P02=rand;g2=rand;p2=rand;B2=rand;
[t2,x2] = ode45(@(t2,x2) [-g2*x2(1) + p2*x2(1); -x2(1)*x2(2)+ B2*x2(2)], tspan2, [A02 P02]);
tspan3=[2 3];A03=rand;P03=rand;g3=rand;p3=rand;B3=rand;
[t3,x3] = ode45(@(t3,x3) [-g3*x3(1) + p3*x3(1); -x3(1)*x3(2)+ B3*x3(2)], tspan3, [A03 P03]);
tspan4=[3 4];A04=rand;P04=rand;g4=rand;p4=rand;B4=rand;
[t4,x4] = ode45(@(t4,x4) [-g4*x4(1) + p4*x4(1); -x4(1)*x4(2)+ B4*x4(2)], tspan4, [A04 P04]);
tspan5=[4 5];A05=rand;P05=rand;g5=rand;p5=rand;B5=rand;
[t5,x5] = ode45(@(t5,x5) [-g5*x5(1) + p5*x5(1); -x5(1)*x5(2)+ B5*x5(2)], tspan5, [A05 P05]);
tspan6=[5 6];A06=rand;P06=rand;g6=rand;p6=rand;B6=rand;
[t6,x6] = ode45(@(t6,x6) [-g6*x6(1) + p6*x6(1); -x6(1)*x6(2)+ B6*x6(2)], tspan6, [A06 P06]);
tspan7=[6 7];A07=rand;P07=rand;g7=rand;p7=rand;B7=rand;
[t7,x7] = ode45(@(t7,x7) [-g7*x7(1) + p7*x7(1); -x7(1)*x7(2)+ B7*x7(2)], tspan7, [A07 P07]);
tspan8=[7 8];A08=rand;P08=rand;g8=rand;p8=rand;B8=rand;
[t8,x8] = ode45(@(t8,x8) [-g8*x8(1) + p8*x8(1); -x8(1)*x8(2)+ B8*x8(2)], tspan8, [A08 P08]);
tspan9=[8 9];A09=rand;P09=rand;g9=rand;p9=rand;B9=rand;
[t9,x9] = ode45(@(t9,x9) [-g9*x9(1) + p9*x9(1); -x9(1)*x9(2)+ B9*x9(2)], tspan9, [A09 P09]);
tspan10=[9 10];A10=rand;P10=rand;g10=rand;p10=rand;B10=rand;
[t10,x10] = ode45(@(t10,x10) [-g10*x10(1) + p10*x10(1); -x10(1)*x10(2)+ B10*x10(2)], tspan10, [A10 P10]);
x={x1;x2;x3;x4;x5;x6;x7;x8;x9;x10};
result{k} = x;
end

채택된 답변

Star Strider
Star Strider 2019년 6월 3일
There is no guarantee that the row lengths of each ‘x’ are going to be the same using a two-element ‘tspan’ vector, so what you outlined may not be possible.
However if you fix the number of elements in ‘tspan’, what you want to do is straightforward:
n = 1000;
tsv = linspace(0, 1, 50);
xtm = zeros(numel(tsv)*10,2,n);
for k1 = 1:n
xim = zeros(9*numel(tsv), 2);
for k2 = 1:10
tspan=tsv+(k2-1);
A=rand; P=rand; g=rand; p=rand; B=rand;
[t,x] = ode45(@(t,x) [-g*x(1) + p*x(1); -x(1)*x(2)+ B*x(2)], tspan, [A P]);
xim((1:numel(tsv))+numel(tsv)*(k2-1),:) = x;
end
xtm(:,:,k1) = xim;
end
The ‘xtm’ matrix will be (500x2x1000), representing the concatenated rows, the two columns, and the 1000 trials. It is also easy to index into.
  댓글 수: 2
Thomas Veith
Thomas Veith 2019년 6월 3일
Thank you so much! Very helpful!
Star Strider
Star Strider 2019년 6월 3일
As always, my pleasure!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Shifting and Sorting Matrices에 대해 자세히 알아보기

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by