你可以给我建议我的代码来改善时间计算。因为这段代码需要花费很多时间。有一段时间我不会得到结果。我的情况是如何组合RandomMatriceA [1 12]和RandomMatriceB [0 2]。
示例：
ProcessTime=[2 2 2 3 3] %number of consecutive value RandomMatriceA
RandomMatriceA=[1 2 3 4 5;
3 4 2 1 5;
1 4 2 3 5;
5 1 2 3 4]
RandomMatriceB=[0 2 0 0 1;%humber of consecutive zeros
0 0 0 1 1;
1 2 1 0 0;
0 0 1 1 1]
Result =[1 1 0 0 2 2 3 3 4 4 4 0 5 0 0 0;
3 3 4 4 4 2 2 0 1 1 0 5 5 5 0 0;
0 1 1 0 0 4 4 4 0 2 2 3 3 5 5 5;
5 5 5 1 1 0 2 2 0 3 3 0 4 4 4 0]
NewMatric{1}=[1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0;
0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0;
NewMatric{2}= just only identify value 2;
NewMatric{3}= for value 3
NewMatric{4}= value 4;
实际上，我使用以下代码来强制NewMatric中的每一列必须具有每列值1的总和等于2.
如果你们其中一个想给我建议改善计算时间，那么非常感谢你。
clc
clear all
A=2;
B=12;
C=4;
ProcessTime= [ 11 11 11 11 4 4 4 4 4 4 4 2 ]; %Converting Matrice
RandomMatriceA=zeros(A,B,C);
RandomMatriceB=zeros(A,B,C);
SumRandomMatriceB=zeros(1,A,C);
ConvertMatriceA=zeros(A,B,C);
ProcessRandomMatriceA=zeros(A,B,C);
StartProcess=zeros(A,B,C);
ZerosCoordinate=zeros(A,B,C);
total=zeros(1,1,C);
%make first random matrice
for ii=1:C;
for a=1:A
RandomMatriceA(a,:,ii)=randperm(B); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
end
%make second random matrice
for ii=1:C;
RandomMatriceB(:,:,ii)=randint(A,B,[0 2]);
end
%Make sure sum all element per row in RandomMatriceB <=11
for ii=1:C;
SumRandomMatriceB(1,:,ii)=sum(RandomMatriceB(:,:,ii),2);
for jj=1:A,
while SumRandomMatriceB(1,jj,ii) > 11
RandomMatriceB(jj,:,ii)=randint(1,B,[0 2]);
SumRandomMatriceB(1,:,ii)=sum(RandomMatriceB(:,:,ii),2);
end
end
end
%After making RandomMatriceA and RandomMatriceB, then Make New Matrice
%which has size (8,85,10)
%To know varible of new matrice which is result of combining
%RandomMatriceA,RandomMatriceB and ProcessTime
for ii=1:C;
for a=1:A,
for b=1:B,
ConvertMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii));%FirstVarible:Consecutive The Number of value in all element of RandomMatriceA
ProcessRandomMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii))+RandomMatriceB(a,b,ii);
EndProcess=cumsum(ProcessRandomMatriceA,2);%secondVaribale:to know in which column The Consecutive value RandomMatrice will be end
StartProcess(a,b,ii)=EndProcess(a,b,ii)-(ProcessTime(RandomMatriceA(a,b,ii))-1);%ThirdVariable:to know in which column The Consecutive Value will be start
end
end
end
%To know in which column and row in new matrice will have zeros
for ii=1:C,
for a=1:A;
for b=1:B
if RandomMatriceB (a,b,ii)~=0
ZerosCoordinate(a,b,ii)=StartProcess(a,b,ii)-RandomMatriceB(a,b,ii);%To know in which column in new matrice will have value zeros
else
ZerosCoordinate(a,b,ii)=RandomMatriceB(a,b,ii);
end
end
end
end
%to force/make column 43th will be filled by zeros in every row.
for ii=1:C;
while size(find(ZerosCoordinate(:,:,ii)==43),1)<A, %looping untuk mendapatkan hari libur diminggu ke 43
for a=1:A;
if numel(find(ZerosCoordinate(a,:,ii)==43))==1, %mengharuskan adanya libur di minggu ke 43
continue;
else
%make sure sum of element new random row in RandomMatriceB
%<=11
newRow = randi([0 2],1,B);
sumnewRow=sum(newRow,2);
while sumnewRow >11,
newRow = randi([0 2],1,B);
sumnewRow=sum(newRow,2);
end
%replace newRow to row in RandomMatriceB
RandomMatriceB(a,:,ii)=newRow; %proses pergantian baris yang tidak sesuai dengan konstrain
%random row in RandomMatriceA
RandomMatriceA(a,:,ii)=randperm(B); %randoming baris kromosom yang tidak sesuai dengan konstrain
%Try to know variabel to make NewMatrice
for b=1:B;
ConvertMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii));
ProcessRandomMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii))+RandomMatriceB(a,b,ii);
EndProcess=cumsum(ProcessRandomMatriceA,2);
StartProcess(a,b,ii)=EndProcess(a,b,ii)-(ProcessTime(RandomMatriceA(a,b,ii))-1);
end
for b=1:B
%To know column and row which has value zeros
if RandomMatriceB(a,b,ii)~=0
ZerosCoordinate(a,b,ii)=StartProcess(a,b,ii)-RandomMatriceB(a,b,ii);
else
ZerosCoordinate(a,b,ii)=RandomMatriceB(a,b,ii);
end
end
end
end
end
end
%Make New Matrice (8,85,10) by considering variable above
max_row_length = max(StartProcess(:,B) + ConvertMatriceA(:, B) - 1); %to know row length
result = zeros(size(StartProcess, 1),max_row_length, size(StartProcess,3)); %preallocation size
for row = 1 : size(StartProcess, 1)
for column = 1 : size(StartProcess, 2)
for depth = 1 : size(StartProcess,3)
s = StartProcess(row, column,depth);
c = ConvertMatriceA(row, column,depth);
v = RandomMatriceA(row,column,depth);
result(row, s : s + c - 1,depth) = v;%this matrix is new matrix (8,85,10) which is result of combining RandomMatriceA,RandomMatriceB and ProcessTime
end
end
end
%Make A new Matrice per Value (B).
newMatrice = arrayfun(@(b) (result == b).*cumsum((result == b),2),nonzeros(unique(result)), 'UniformOutput', false); %indikator pemakaian bagian
%to know how many number of value 1 per column in matrice
%bagian1,bagian2,bagian3 and bagian4
SumOnePerColumn=cell(1,4);
SumOnePerColumn{1}=sum(newMatrice{1}==1);
SumOnePerColumn{2}=sum(newMatrice{2}==1);
SumOnePerColumn{3}=sum(newMatrice{3}==1);
SumOnePerColumn{4}=sum(newMatrice{4}==1);
%to count how many column which has sum of value one per column ~=2&~=0,
gab=cell(1,4);
gabb{1}=sum(SumOnePerColumn{1}~=0 &SumOnePerColumn{1}~=2);
gabb{2}=sum(SumOnePerColumn{2}~=0 &SumOnePerColumn{2}~=2);
gabb{3}=sum(SumOnePerColumn{3}~=0 &SumOnePerColumn{3}~=2);
gabb{4}=sum(SumOnePerColumn{4}~=0 &SumOnePerColumn{4}~=2);
for yyy=1:C;
total(1,1,yyy)= gabb{1}(1,1,yyy)+gabb{2}(1,1,yyy)+gabb{3}(1,1,yyy)+gabb{4}(1,1,yyy);%Total
end
for ii=1:C;
if total(:,:,ii)==0;
continue;
else
while total(:,:,ii)~=0;%to make total=0 by randoming RandomMatriceA and RandomMatriceB
for a=1:A
RandomMatriceA(a,:,ii)=randperm(B); %batasan tidak boleh satu kelompok melakukan lebih dari satu aktivitas dalam satu waktu
end
%make second random matrice
RandomMatriceB(:,:,ii)=randint(A,B,[0 2]);
%Make sure sum all element per row in RandomMatriceB <=11
SumRandomMatriceB(:,:,ii)=sum(RandomMatriceB(:,:,ii),2);
for jj=1:A,
while SumRandomMatriceB(1,jj,ii) > 11
RandomMatriceB(jj,:,ii)=randint(1,B,[0 2]);
SumRandomMatriceB(1,:,ii)=sum(RandomMatriceB(:,:,ii),2);
end
end
for a=1:A,
for b=1:B,
ConvertMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii));%FirstVarible:Consecutive The Number of value in all element of RandomMatriceA
ProcessRandomMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii))+RandomMatriceB(a,b,ii);
EndProcess=cumsum(ProcessRandomMatriceA,2);%secondVaribale:to know in which column The Consecutive value RandomMatrice will be end
StartProcess(a,b,ii)=EndProcess(a,b,ii)-(ProcessTime(RandomMatriceA(a,b,ii))-1);%ThirdVariable:to know in which column The Consecutive Value will be start
end
for bb=1:B
if RandomMatriceB (a,bb,ii)~=0
ZerosCoordinate(a,bb,ii)=StartProcess(a,bb,ii)-RandomMatriceB(a,bb,ii);%To know in which column in new matrice will have value zeros
else
ZerosCoordinate(a,bb,ii)=RandomMatriceB(a,bb,ii);
end
end
end
while size(find(ZerosCoordinate(:,:,ii)==43),1)<A, %looping untuk mendapatkan hari libur diminggu ke 43
for a=1:A;
if numel(find(ZerosCoordinate(a,:,ii)==43))==1, %mengharuskan adanya libur di minggu ke 43
continue;
else
%make sure sum of element new random row in RandomMatriceB
%<=11
newRow = randi([0 2],1,B);
sumnewRow=sum(newRow,2);
while sumnewRow >11,
newRow = randi([0 2],1,B);
sumnewRow=sum(newRow,2);
end
%replace newRow to row in RandomMatriceB
RandomMatriceB(a,:,ii)=newRow; %proses pergantian baris yang tidak sesuai dengan konstrain
%random row in RandomMatriceA
RandomMatriceA(a,:,ii)=randperm(B); %randoming baris kromosom yang tidak sesuai dengan konstrain
%Try to know variabel to make NewMatrice
for b=1:B;
ConvertMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii));
ProcessRandomMatriceA(a,b,ii)=ProcessTime(RandomMatriceA(a,b,ii))+RandomMatriceB(a,b,ii);
EndProcess=cumsum(ProcessRandomMatriceA,2);
StartProcess(a,b,ii)=EndProcess(a,b,ii)-(ProcessTime(RandomMatriceA(a,b,ii))-1);
end
for b=1:B
%To know column and row which has value zeros
if RandomMatriceB(a,b,ii)~=0
ZerosCoordinate(a,b,ii)=StartProcess(a,b,ii)-RandomMatriceB(a,b,ii);
else
ZerosCoordinate(a,b,ii)=RandomMatriceB(a,b,ii);
end
end
end
end
end
%To make one combining matrice to replace combining matrice which
%doesnt have total=0
max_row= 85;
DummyResult = zeros(A, max_row);
for bbb= 1 : B
for aaa= 1 : A
su = StartProcess(aaa,bbb,ii);
cu = ConvertMatriceA(aaa,bbb,ii);
vu = RandomMatriceA(aaa,bbb,ii);
DummyResult(aaa, su : su + cu - 1) = vu;
end
end
indhasil= arrayfun(@(b) (DummyResult == b).*cumsum((DummyResult == b),2),nonzeros(unique(DummyResult)), 'UniformOutput', false); %indikator pemakaian bagian
SumOnePerColumnIndhasil=cell(1,4);
SumOnePerColumnIndhasil{1}=sum(indhasil{1}==1);
SumOnePerColumnIndhasil{2}=sum(indhasil{2}==1);
SumOnePerColumnIndhasil{3}=sum(indhasil{3}==1);
SumOnePerColumnIndhasil{4}=sum(indhasil{4}==1);
%to know how many column that has not yet had
%SumOnePerColumnIndhasil= 2 and 0.
gabbh=cell(1,4);
gabbh{1}=sum(SumOnePerColumnIndhasil{1}~=0 &SumOnePerColumnIndhasil{1}~=2);
gabbh{2}=sum(SumOnePerColumnIndhasil{2}~=0 &SumOnePerColumnIndhasil{2}~=2);
gabbh{3}=sum(SumOnePerColumnIndhasil{3}~=0 &SumOnePerColumnIndhasil{3}~=2);
gabbh{4}=sum(SumOnePerColumnIndhasil{4}~=0 &SumOnePerColumnIndhasil{4}~=2);
total(1,1,ii)=gabbh{1}+gabbh{2}+gabbh{3}+gabbh{4};
end
end
end
答案 0 :(得分：6)
在我的计算机上，代码非常快，因为立即因错误而中止; - ）
一些问题：
第25行：没有randint()函数（我正在使用Matlab R2011b）。
第30行：此SumRandomMatriceB(:,:,ii)=sum(RandomMatriceB(:,:,C),2);没有
工作是因为
size(SumRandomMatriceB(:,:,ii)) == 1x12但是
size(sum(RandomMatriceB(:,:,1),2)) == 8 x 1，因此维度不匹配。
第58行：没有名为libur的函数或矩阵。
第89行：从未创建ProcessMatriceA矩阵。
第90行：变量bb未被删除。
我只使用了一半代码并且已经找到五个严重错误。老实说，我没心情为你解决这些错误。但是，当您有代码的运行版本时，我建议您输入
profile viewer
在命令行中。它将调出剖析器。只需告诉它您的代码的文件名 它将带来非常好的时序图和统计数据。在进行性能分析之后，您可以非常具体地询问代码的哪一部分很慢，如果可能的话，我很乐意提供解决方法。