CSV import, only one column (Matlab/Octave) - matlab

I have since several days problems with reading my measurement csv files and make some simple calculations. I hope someone can help me.
My Aim
Read CSV data file, as followed:
Open with Excel:
date: 20140202 time: 083736 Cycles total: 74127 T_zer: 56 T_op1: 90.000
Actu state: stoppes ! T1: -23 T2: -12 T3: -32 T4: -65
*-*
324203 0 34724 0 0 0 2
431040 0 0 0 0 0 1
230706 0 0 0 0 0 1
340810 0 0 0 0 0 1
..............
....
.
-->Here 1st question: If I open with editor, I can only see one delimiter, its ";". But there must be two? One for row , one for columns? How can Excel separate it correctly into row and col, if there is only ";" ?
However... now I tried to csvread this file with octave. There I get it into octave, but everything only in one column:/. For me it would be very comfortable Octave could read it into a 7x X Matrix. In this case I can handle the data easy.
Here my Code:
clc
clear all
[fname,pname] =uigetfile();
fname;
extra="/";
pname;
b=strcat(pname,extra,fname);
m = csvread(b);
Result:
m as double with 4003x1. 4003 is corretct, but everything in one colum:/
m =
0
0
0
454203
561040
340706
I tried now to handle this problem up to several days, but no result.

Not a Octave expert, but looks like you can use the dlmread function to read a CSV files, it has many parameters which can help you read the file correctly.
start reading the data from row X (and not from the start)
only have Y columns
defined the separator between fields

Related

Having trouble in using nlinfit function in MATLAB

Kindly please help me with the problem as I need to use nlinfit function for fitting unknown parameters but it is showing some error. Although yesterday I was getting some values for parameters to be fitted but now if I am running it is having some issue for the function output to be used in fitted with NaN answer for last iteration only. X data is a concatenated matrix of three columns as independent variable and yk is dependent variable, taua is a matrix of initial guesses of number of parameters to be fitted.
function [yk]=activity_coefficientE(taua,x)
T=523;
alpha12=0.3; alpha13=0.3; alpha21=0.3; alpha23=0.3; alpha31=0.3; alpha32=0.3;
alpha18=0.2; alpha81=0.2; alpha28=0.2; alpha82=0.2; alpha38=0.2; alpha83=0.3;
alpha19=0.2; alpha91=0.2; alpha29=0.2; alpha92=0.2; alpha39=0.2; alpha93=0.2;
alpha110=0.2;alpha101=0.2;alpha210=0.2;alpha102=0.2;alpha310=0.2;alpha103=0.2;
alpha113=0.2;alpha131=0.2;alpha213=0.2;alpha132=0.2;alpha313=0.2;alpha133=0.2;
alpha114=0.2;alpha141=0.2;alpha214=0.2;alpha142=0.2;alpha314=0.2;alpha143=0.2;
alpha115=0.2;alpha151=0.2;alpha215=0.2;alpha152=0.2;alpha315=0.2;alpha153=0.2;
alpha117=0.2;alpha171=0.2;alpha217=0.2;alpha172=0.2;alpha317=0.2;alpha173=0.2;
alpha118=0.2;alpha181=0.2;alpha218=0.2;alpha182=0.2;alpha318=0.2;alpha183=0.2;
alpha810=0.2;alpha915=0.2;alpha1314=0.2;alpha108=0.2;alpha159=0.2;alpha1413=0.2;
alpha1718=0.2;alpha1817=0.2;
tau12=0; tau13=0; tau21=0; tau23=0; tau31=0; tau32=0;
%taua=randi([-5,5],1,112)
tau18=taua(1)+taua(57)/T;
tau81=taua(2)+taua(58)/T;
tau28=taua(3)+taua(59)/T;
tau82=taua(4)+taua(60)/T;
tau38=taua(5)+taua(61)/T;
tau83=taua(6)+taua(62)/T;
tau19=taua(7)+taua(63)/T;
tau91=taua(8)+taua(64)/T;
tau29=taua(9)+taua(65)/T;
tau92=taua(10)+taua(66)/T;
tau39=taua(11)+taua(67)/T;
tau93=taua(12)+taua(68)/T;
tau110=taua(13)+taua(69)/T;
tau101=taua(14)+taua(70)/T;
tau210=taua(15)+taua(71)/T;
tau102=taua(16)+taua(72)/T;
tau310=taua(17)+taua(73)/T;
tau103=taua(18)+taua(74)/T;
tau113=taua(19)+taua(75)/T;
tau131=taua(20)+taua(76)/T;
tau213=taua(21)+taua(77)/T;
tau132=taua(22)+taua(78)/T;
tau313=taua(23)+taua(79)/T;
tau133=taua(24)+taua(80)/T;
tau114=taua(25)+taua(81)/T;
tau141=taua(26)+taua(82)/T;
tau214=taua(27)+taua(83)/T;
tau142=taua(28)+taua(84)/T;
tau314=taua(29)+taua(85)/T;
tau143=taua(30)+taua(86)/T;
tau115=taua(31)+taua(87)/T;
tau151=taua(32)+taua(88)/T;
tau215=taua(33)+taua(89)/T;
tau152=taua(34)+taua(90)/T;
tau315=taua(35)+taua(91)/T;
tau153=taua(36)+taua(92)/T;
tau117=taua(37)+taua(93)/T;
tau171=taua(38)+taua(94)/T;
tau217=taua(39)+taua(95)/T;
tau172=taua(40)+taua(96)/T;
tau317=taua(41)+taua(97)/T;
tau173=taua(42)+taua(98)/T;
tau118=taua(43)+taua(99)/T;
tau181=taua(44)+taua(100)/T;
tau218=taua(45)+taua(101)/T;
tau182=taua(46)+taua(102)/T;
tau318=taua(47)+taua(103)/T;
tau183=taua(48)+taua(104)/T;
tau810=taua(49)+taua(105)/T;
tau108=taua(50)+taua(106)/T;
tau915=taua(51)+taua(107)/T;
tau159=taua(52)+taua(108)/T;
tau1314=taua(53)+taua(109)/T;
tau1413=taua(54)+taua(110)/T;
tau1718=taua(55)+taua(111)/T;
tau1817=taua(56)+taua(112)/T;
G12=exp(-(tau12*alpha12));
G21=exp(-(tau21*alpha21));
G13=exp(-(tau13*alpha13));
G31=exp(-(tau31*alpha31));
G23=exp(-(tau23*alpha23));
G32=exp(-(tau32*alpha32));
G18=exp(-(tau18*alpha18));
G81=exp(-(tau81*alpha81));
G28=exp(-(tau28*alpha28));
G82=exp(-(tau82*alpha82));
G38=exp(-(tau38*alpha83));
G83=exp(-(tau83*alpha83));
G19=exp(-(tau19*alpha19));
G91=exp(-(tau91*alpha91));
G29=exp(-(tau29*alpha29));
G92=exp(-(tau92*alpha92));
G39=exp(-(tau39*alpha39));
G93=exp(-(tau93*alpha93));
G110=exp(-(tau110*alpha110));
G101=exp(-(tau101*alpha101));
G210=exp(-(tau210*alpha210));
G102=exp(-(tau102*alpha102));
G310=exp(-(tau310*alpha310));
G103=exp(-(tau103*alpha103));
G113=exp(-(tau113*alpha113));
G131=exp(-(tau131*alpha131));
G213=exp(-(tau213*alpha213));
G132=exp(-(tau132*alpha132));
G313=exp(-(tau313*alpha313));
G133=exp(-(tau133*alpha133));
G114=exp(-(tau114*alpha114));
G141=exp(-(tau141*alpha141));
G214=exp(-(tau214*alpha214));
G142=exp(-(tau142*alpha142));
G314=exp(-(tau314*alpha314));
G143=exp(-(tau143*alpha143));
G115=exp(-(tau115*alpha115));
G151=exp(-(tau151*alpha151));
G215=exp(-(tau215*alpha215));
G152=exp(-(tau152*alpha152));
G315=exp(-(tau315*alpha315));
G153=exp(-(tau153*alpha153));
G117=exp(-(tau117*alpha117));
G171=exp(-(tau171*alpha171));
G217=exp(-(tau217*alpha217));
G172=exp(-(tau172*alpha172));
G317=exp(-(tau317*alpha317));
G173=exp(-(tau173*alpha173));
G118=exp(-(tau118*alpha118));
G181=exp(-(tau181*alpha181));
G218=exp(-(tau218*alpha218));
G182=exp(-(tau182*alpha182));
G318=exp(-(tau318*alpha318));
G183=exp(-(tau183*alpha183));
G810=exp(-(tau810*alpha810));
G108=exp(-(tau108*alpha108));
G915=exp(-(tau915*alpha915));
G159=exp(-(tau159*alpha159));
G1314=exp(-(tau1314*alpha1314));
G1413=exp(-(tau1413*alpha1413));
G1718=exp(-(tau1718*alpha1718));
G1817=exp(-(tau1817*alpha1817));
%calculating mole fractions of ionic species
x1=x(:,1);
x2=x(:,2);
x3=x(:,3);
%x1=[0.1577 0.1492 0.1462 0.1366 0.1299 0.1180 0.0863 0.0761 0.0550 ];
%x2=[0.8278 0.7945 0.7678 0.7450 0.6979 0.6309 0.4611 0.4114 0.2952 ];
%x3=[0.0145 0.0563 0.0860 0.1184 0.1722 0.2511 0.4526 0.5125 0.6498 ];
A=[0.0674243 0.0773881 0.0843400 0.0865343 0.0899223 0.0882858 0.0715087 0.0643867 0.0483658];
B=[0.0141081 0.0479814 0.0643151 0.0737477 0.0820756 0.0838701 0.0701576 0.0634457 0.0479639];
C=[0.0565665 0.0450072 0.0387724 0.0313828 0.02506094 0.0186280 0.0092734 0.0073438 0.0041595 ];
D=[0.0336447 0.0267694 0.0230611 0.0186659 0.0149058 0.0110795 0.0055157 0.0043679 0.0024739 ];
E=[0.0008148 0.0008756 0.00087131 0.0008794 0.0008711 0.0008441 0.0007384 0.0006997 0.0005980 ];
N=length(A);
x1n=zeros(N,1);x2n=zeros(N,1);x3n=zeros(N,1);
X1=zeros(N,1);X2=zeros(N,1);X3=zeros(N,1);X4=zeros(N,1);X5=zeros(N,1);X6=zeros(N,1);X7=zeros(N,1);
X12=zeros(N,1);X16=zeros(N,1);
for i=1:N
x1n(i)=(x1(i)-A(i)-D(i)-2*E(i)-C(i)+3*B(i))
x2n(i)=(x2(i)-A(i)-C(i)-D(i))
x3n(i)=(x3(i)-B(i))
X1(i)=(x1n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X2(i)=(x2n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X3(i)=(x3n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X4(i)=(A(i)+D(i)+E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X5(i)=(C(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X6(i)=(A(i)-B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X7(i)=(B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X12(i)=(E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
X16(i)=(C(i)+D(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)))
end
yc4=X4./(X4+X5);
yc5=X5./(X4+X5);
yc6=X6./(X6+X7+X12+X16);
yc7=X7./(X6+X7+X12+X16);
yc12=X12./(X6+X7+X12+X16);
yc16=X16./(X6+X7+X12+X16);
alpha14=yc6.*alpha18+yc7.*alpha19+yc12.*alpha113+yc16.*alpha117;
%alpha41=alpha14;
alpha24=yc6.*alpha28+yc7.*alpha29+yc12.*alpha213+yc16.*alpha217;
%alpha42=alpha24;
alpha34=yc6.*alpha38+yc7.*alpha39+yc12.*alpha313+yc16.*alpha317;
%alpha43=alpha34;
alpha15=yc6.*alpha110+yc7.*alpha115+yc12.*alpha114+yc16.*alpha118;
%alpha51=alpha15;
alpha25=yc6.*alpha210+yc7.*alpha215+yc12.*alpha214+yc16.*alpha218;
%alpha52=alpha25;
alpha35=yc6.*alpha310+yc7.*alpha315+yc12.*alpha314+yc16.*alpha318;
%alpha53=alpha35;
alpha16=yc4.*alpha81+yc5.*alpha101;
%alpha61=alpha16;
alpha26=yc4.*alpha82+yc5.*alpha102;
%alpha62=alpha26;
alpha36=yc4.*alpha83+yc5.*alpha103;
%alpha63=alpha36;
alpha17=yc4.*alpha91+yc5.*alpha151;
%alpha71=alpha17;
alpha27=yc4.*alpha92+yc5.*alpha152;
%alpha72=alpha27;
alpha37=yc4.*alpha93+yc5.*alpha153;
%alpha73=alpha37;
alpha112=yc4.*alpha131+yc5.*alpha141;
%alpha121=alpha112;
alpha212=yc4.*alpha132+yc5.*alpha142;
%alpha122=alpha212;
alpha312=yc4.*alpha133+yc5.*alpha143;
%alpha123=alpha312;
alpha116=yc4.*alpha171+yc5.*alpha181;
%alpha161=alpha116;
alpha216=yc4.*alpha172+yc5.*alpha182;
%alpha162=alpha216;
alpha316=yc4.*alpha173+yc5.*alpha183;
%alpha163=alpha316;
alpha46=yc5.*alpha810;
%alpha64=alpha46;
alpha47=yc5.*alpha915;
%alpha74=alpha47;
alpha412=yc5.*alpha1314;
%alpha124=alpha412;
alpha416=yc5.*alpha1718;
%alpha164=alpha416;
alpha56=yc4.*alpha108;
%alpha65=alpha56;
alpha57=yc4.*alpha159;
%alpha75=alpha57;
alpha512=yc4.*alpha1413;
%alpha125=alpha512;
alpha516=yc4.*alpha1817;
%alpha165=alpha516;
G14=yc6.*G18+yc7.*G19+yc12.*G113+yc16.*G117;
%G41=G14;
G24=yc6.*G28+yc7.*G29+yc12.*G213+yc16.*G217;
%G42=G24;
G34=yc6.*G38+yc7.*G39+yc12.*G313+yc16.*G317;
%G43=G34;
G15=yc6.*G110+yc7.*G115+yc12.*G114+yc16.*G118;
%G51=G15;
G25=yc6.*G210+yc7.*G215+yc12.*G214+yc16.*G218;
%G52=G25;
G35=yc6.*G310+yc7.*G315+yc12.*G314+yc16.*G318;
%G53=G35;
G16=yc4.*G81+yc5.*G101;
%G61=G16;
G26=yc4.*G82+yc5.*G102;
%G62=G26;
G36=yc4.*G83+yc5.*G103;
%G63=G36;
G17=yc4.*G91+yc5.*G151;
%G71=G17;
G27=yc4.*G92+yc5.*G152;
%G72=G27;
G37=yc4.*G93+yc5.*G153;
%G73=G37;
G112=yc4.*G131+yc5.*G141;
%G121=G112;
G212=yc4.*G132+yc5.*G142;
%G122=G212;
G312=yc4.*G133+yc5.*G143;
%G123=G312;
G116=yc4.*G171+yc5.*G181;
%G161=G116;
G216=yc4.*G172+yc5.*G182;
%G162=G216;
G316=yc4.*G173+yc5.*G183;
%G163=G316;
G46=yc5.*G810;
%G64=G46;
G47=yc5.*G915;
%G74=G47;
G412=yc5.*G1314;
%G124=G412;
G416=yc5.*G1718;
%G164=G416;
G56=yc4.*G108;
%G65=G56;
G57=yc4.*G159;
%G75=G57;
G512=yc4.*G1413;
%G125=G512;
G516=yc4.*G1817;
%G165=G516;
tau14=-log(G14)./alpha14;
%tau41=tau14;
tau24=-log(G24)./alpha24;
%tau42=tau24;
tau34=-log(G34)./alpha34;
%tau43=tau34;
tau15=-log(G15)./alpha15;
%tau51=tau15;
tau25=-log(G25)./alpha25;
%tau52=tau25;
tau35=-log(G35)./alpha35;
%tau53=tau35;
tau16=-log(G16)./alpha16;
%tau61=tau16;
tau26=-log(G26)./alpha26;
%tau62=tau26;
tau36=-log(G36)./alpha36;
%tau63=tau36;
tau17=-log(G17)./alpha17;
%tau71=tau17;
tau27=-log(G27)./alpha27;
%tau72=tau27;
tau37=-log(G37)./alpha37;
%tau73=tau37;
tau112=-log(G112)./alpha112;
%tau121=tau112;
tau212=-log(G212)./alpha212;
%tau122=tau212;
tau312=-log(G312)./alpha312;
%tau123=tau312;
tau116=-log(G116)./alpha116;
%tau161=tau116;
tau216=-log(G216)./alpha216;
%tau162=tau216;
tau316=-log(G316)./alpha316;
%tau163=tau316;
tau46=-log(G46)./alpha46;
%tau64=tau46;
tau47=-log(G47)./alpha47;
%tau74=tau47;
tau412=-log(G412)./alpha412;
%tau124=tau412;
tau416=-log(G416)./alpha416;
%tau164=tau416;
tau56=-log(G56)./alpha56;
%tau65=tau56;
tau57=-log(G57)./alpha57;
%tau75=tau57;
tau512=-log(G512)./alpha512;
%tau125=tau512;
tau516=-log(G516)./alpha516;
%tau165=tau516;
ln_y1_1=G12.*X2.*tau12+ G31.*X3.*tau13+ G14.*X4.*tau14+G15.*X5.*tau15+G16.*X6.*tau16+G17.*X7.*tau17+G112.*X12.*tau112+G116.*X16.*tau116;
ln_y1_2=G12.*X2+ G13.*X3+ G14.*X4+G15.*X5+G16.*X6+G17.*X7+G112.*X12+G116.*X16;
ln_y2_1=G21.*X1.*tau12+ G32.*X3.*tau32+ G24.*X4.*tau24+G25.*X5.*tau25+G26.*X6.*tau26+G27.*X7.*tau27+G212.*X12.*tau212+G216.*X16.*tau216;
ln_y2_2=G12.*X1+ G23.*X3+G24.*X4+G25.*X5+G26.*X6+G27.*X7+G212.*X12+G216.*X16;
ln_y3_1=G13.*X1.*tau13+ G23.*X3.*tau23+ G34.*X4.*tau34+G35.*X5.*tau35+G36.*X6.*tau36+G37.*X7.*tau37+G312.*X12.*tau312+G316.*X16.*tau316;
ln_y3_2=G13.*X1+ G23.*X3+ G34.*X4+G35.*X5+G36.*X6+G37.*X7+G312.*X12+G316.*X16;
ln_y4_1=G14.*X1.*tau14+G24.*X2.*tau24+G34.*X3.*tau34+G46.*X6.*tau46+G47.*X7.*tau47+G412.*X12.*tau412+G416.*X16.*tau416;
ln_y4_2=G14.*X1+G24.*X2+G34.*X3+G46.*X6+G47.*X7+G412.*X12+G416.*X16;
ln_y5_1=G15.*X1.*tau15+G25.*X2.*tau25+G35.*X3.*tau35+G56.*X6.*tau56+G57.*X7.*tau57+G512.*X12.*tau512+G516.*X16.*tau516;
ln_y5_2=G15.*X1+G25.*X2+G35.*X3+G56.*X6+G57.*X7+G512.*X12+G516.*X16;
ln_y6_1=G16.*X1.*tau16+G26.*X2.*tau26+G36.*X3.*tau36+G46.*X4.*tau46+G56.*X5.*tau56;
ln_y6_2=G16.*X1+G26.*X2+G36.*X3+G46.*X4+G56.*X5;
ln_y7_1=G17.*X1.*tau17+G27.*X2.*tau27+G37.*X3.*tau37+G47.*X4.*tau47+G57.*X5.*tau57;
ln_y7_2=G17.*X1+G27.*X2+G37.*X3+G47.*X4+G57.*X5;
ln_y12_1=G112.*X1.*tau112+G212.*X2.*tau212+G312.*X3.*tau312+G412.*X4.*tau412+G512.*X5.*tau512;
ln_y12_2=G112.*X1+G212.*X2+G312.*X3+G412.*X4+G512.*X5;
ln_y16_1=G116.*X1.*tau116+G216.*X2.*tau216+G316.*X3.*tau316+G416.*X4.*tau416+G516.*X5.*tau516;
ln_y16_2=G116.*X1+G216.*X2+G316.*X3+G416.*X4+G516.*X5;
ln_y1_3=(((X2.*G12)./ln_y2_2).*(tau12-(ln_y2_1)./(ln_y2_2)))+(((X3.*G13)./ln_y3_2).*(tau13-(ln_y3_1)./(ln_y3_2)));
ln_y1_4=(((X6.*G16)./ln_y6_2).*(tau16- (ln_y6_1./ln_y6_2))) + (((X7.*G17)./ln_y7_2).*(tau17- (ln_y7_1./ln_y7_2)))+(((X12.*G12)./ln_y12_2).*(tau112- (ln_y12_1./ln_y12_2)))+(((X16.*G16)./ln_y16_2).*(tau116- (ln_y16_1./ln_y16_2)));
ln_y1_5=(((X4.*G14)./ln_y4_2).*(tau14- (ln_y4_1./ln_y4_2))) + (((X5.*G15)./ln_y5_2).*(tau15- (ln_y5_1./ln_y5_2)));
yk=exp((ln_y1_1./ln_y1_2) + ln_y1_3 + ln_y1_4+ ln_y1_5) % activity coefficient for H2O
end
........................................
Another function where above function to be called.....
% calling the function act_coeff to estimate the binary interaction parameters
for i=1:112
filename = 'EagelsDATA.xlsx'; %reading VLE data from excel file
Data = xlsread(filename);
x(:,1) = Data([10:15 17:19],16);
x(:,2) = Data([10:15 17:19],1);
x(:,3)= Data([10:15 17:19],2);
taua=(randi([-5,5],1,112));
yk=[0.0606 (values calculated from above function and will be used for fitting)
0.4327
0.6545
0.9417
1.2570
1.6881
1.9108
1.7777
1.3821]
% taua =[ -2 3 4 -3 -4 1 4 -2 4 -4 -1 4 5 -3 3 2 -5 3 -4
% 1 4 1 5 -1 -1 -3 2 -3 4 3 4 2 5 4 -2 4 3 -1
% 1 0 -5 -5 -5 -3 4 2 1 4 0 2 -3 -4 5 0 -3 2 5
% 1 0 5 1 -3 5 4 1 5 2 3 2 0 -5 -4 -2 1 -2 5
%-5 5 -2 -2 4 1 -1 3 -1 1 5 -1 0 -1 4 5 5 1 4
% 1 0 4 -4 4 0 -1 -2 -5 -3 -4 -5
% -5 0 -2 0 -5] (random values for which yk was calculted from the command
taua= randi([-5,5],1,112))
try % try-catch used to continue the loop without stopping on encountering an error
[taua1]= nlinfit(x,yk,#activity_coefficientE,taua)
catch exception
continue
end
end
I am not able to attach excel sheet here so data from excel sheet is as:
x =[0.1577 0.1492 0.1462 0.1366 0.1299 0.1180 0.0863 0.0761 0.0550; column 1
0.8278 0.7945 0.7678 0.7450 0.6979 0.6309 0.4611 0.4114 0.2952 ; column 2
0.0145 0.0563 0.0860 0.1184 0.1722 0.2511 0.4526 0.5125 0.6498 ]; column 3
I found 3 major problems with what you did.
Problem #1 - errors
The reason you get the error is because your function "activity_coefficientE" can sometimes return NaN or inf values. My suggestion is to look for these values and set the value of "yk" to a large value so that the optimizer in "nlinfit" will stay away from coefficients that produce infinite or NaN values. This is the code at the bottom of the function so that you avoid crashes:
if any(~isfinite(yk))
yk = 10 * ones(size(yk));
end
Problem #2 - random initial guesses
The trouble with using random numbers for your initial conditions is that every time you run it you get a different answer, so sometimes it works and sometimes it doesn't. If you set the random number generator seed, you can get the same random numbers each time you run the script. If you change you seed, you can get a different set of random numbers. I shortened your main script to this, where I try 100 different random seeds (and store the results of each attempt) to see what answers result:
for i=1:100
rng(i)
taua = randi([-5,5],1,112);
taua1(i, :) = nlinfit(x,yk,#activity_coefficientE,taua);
end
Each row of "taua1" is a set of 111 coefficients.
Problem #3 - Trying to fit 9 points with 112 coefficients
Every time nlinfit is called, you get this warning:
Warning: Rank deficient
because you have more coefficients (112) that you are asking nlinfit to find than data points you are fitting (9). It's like trying to find the 2nd order equation that best fits 2 points, there are an infinite number of solutions. When curve fitting you should have more data points than coefficients to make sure you're not fitting noise. You need more data points in "yk" and "x" and/or fewer coefficients to fit. I've done a lot of curve fitting and I've never seen an equation with 112 coefficients, so I am thinking that you are not solving the problem correctly. Perhaps the 112 coefficients aren't really independent or there are 112 data points and 9 coefficients that you want to find.
For completeness, here is my edited version of the activity_coefficientE.m function that I created to work on this solution. In general, I never see Matlab code with this many variables with similar names. Much of this code could be greatly simplified by using vector operations. Most of my changes involve formatting, adding semicolons, and the checks for non-finite values at the end.
function yk=activity_coefficientE(taua,x)
T=523;
alpha12=0.3; alpha13=0.3; alpha21=0.3; alpha23=0.3; alpha31=0.3; alpha32=0.3;
alpha18=0.2; alpha81=0.2; alpha28=0.2; alpha82=0.2; alpha38=0.2; alpha83=0.3;
alpha19=0.2; alpha91=0.2; alpha29=0.2; alpha92=0.2; alpha39=0.2; alpha93=0.2;
alpha110=0.2;alpha101=0.2;alpha210=0.2;alpha102=0.2;alpha310=0.2;alpha103=0.2;
alpha113=0.2;alpha131=0.2;alpha213=0.2;alpha132=0.2;alpha313=0.2;alpha133=0.2;
alpha114=0.2;alpha141=0.2;alpha214=0.2;alpha142=0.2;alpha314=0.2;alpha143=0.2;
alpha115=0.2;alpha151=0.2;alpha215=0.2;alpha152=0.2;alpha315=0.2;alpha153=0.2;
alpha117=0.2;alpha171=0.2;alpha217=0.2;alpha172=0.2;alpha317=0.2;alpha173=0.2;
alpha118=0.2;alpha181=0.2;alpha218=0.2;alpha182=0.2;alpha318=0.2;alpha183=0.2;
alpha810=0.2;alpha915=0.2;alpha1314=0.2;alpha108=0.2;alpha159=0.2;alpha1413=0.2;
alpha1718=0.2;alpha1817=0.2;
tau12=0; tau13=0; tau21=0; tau23=0; tau31=0; tau32=0;
tau18=taua(1)+taua(57)/T;
tau81=taua(2)+taua(58)/T;
tau28=taua(3)+taua(59)/T;
tau82=taua(4)+taua(60)/T;
tau38=taua(5)+taua(61)/T;
tau83=taua(6)+taua(62)/T;
tau19=taua(7)+taua(63)/T;
tau91=taua(8)+taua(64)/T;
tau29=taua(9)+taua(65)/T;
tau92=taua(10)+taua(66)/T;
tau39=taua(11)+taua(67)/T;
tau93=taua(12)+taua(68)/T;
tau110=taua(13)+taua(69)/T;
tau101=taua(14)+taua(70)/T;
tau210=taua(15)+taua(71)/T;
tau102=taua(16)+taua(72)/T;
tau310=taua(17)+taua(73)/T;
tau103=taua(18)+taua(74)/T;
tau113=taua(19)+taua(75)/T;
tau131=taua(20)+taua(76)/T;
tau213=taua(21)+taua(77)/T;
tau132=taua(22)+taua(78)/T;
tau313=taua(23)+taua(79)/T;
tau133=taua(24)+taua(80)/T;
tau114=taua(25)+taua(81)/T;
tau141=taua(26)+taua(82)/T;
tau214=taua(27)+taua(83)/T;
tau142=taua(28)+taua(84)/T;
tau314=taua(29)+taua(85)/T;
tau143=taua(30)+taua(86)/T;
tau115=taua(31)+taua(87)/T;
tau151=taua(32)+taua(88)/T;
tau215=taua(33)+taua(89)/T;
tau152=taua(34)+taua(90)/T;
tau315=taua(35)+taua(91)/T;
tau153=taua(36)+taua(92)/T;
tau117=taua(37)+taua(93)/T;
tau171=taua(38)+taua(94)/T;
tau217=taua(39)+taua(95)/T;
tau172=taua(40)+taua(96)/T;
tau317=taua(41)+taua(97)/T;
tau173=taua(42)+taua(98)/T;
tau118=taua(43)+taua(99)/T;
tau181=taua(44)+taua(100)/T;
tau218=taua(45)+taua(101)/T;
tau182=taua(46)+taua(102)/T;
tau318=taua(47)+taua(103)/T;
tau183=taua(48)+taua(104)/T;
tau810=taua(49)+taua(105)/T;
tau108=taua(50)+taua(106)/T;
tau915=taua(51)+taua(107)/T;
tau159=taua(52)+taua(108)/T;
tau1314=taua(53)+taua(109)/T;
tau1413=taua(54)+taua(110)/T;
tau1718=taua(55)+taua(111)/T;
tau1817=taua(56)+taua(112)/T;
G12=exp(-(tau12*alpha12));
G21=exp(-(tau21*alpha21));
G13=exp(-(tau13*alpha13));
G31=exp(-(tau31*alpha31));
G23=exp(-(tau23*alpha23));
G32=exp(-(tau32*alpha32));
G18=exp(-(tau18*alpha18));
G81=exp(-(tau81*alpha81));
G28=exp(-(tau28*alpha28));
G82=exp(-(tau82*alpha82));
G38=exp(-(tau38*alpha83));
G83=exp(-(tau83*alpha83));
G19=exp(-(tau19*alpha19));
G91=exp(-(tau91*alpha91));
G29=exp(-(tau29*alpha29));
G92=exp(-(tau92*alpha92));
G39=exp(-(tau39*alpha39));
G93=exp(-(tau93*alpha93));
G110=exp(-(tau110*alpha110));
G101=exp(-(tau101*alpha101));
G210=exp(-(tau210*alpha210));
G102=exp(-(tau102*alpha102));
G310=exp(-(tau310*alpha310));
G103=exp(-(tau103*alpha103));
G113=exp(-(tau113*alpha113));
G131=exp(-(tau131*alpha131));
G213=exp(-(tau213*alpha213));
G132=exp(-(tau132*alpha132));
G313=exp(-(tau313*alpha313));
G133=exp(-(tau133*alpha133));
G114=exp(-(tau114*alpha114));
G141=exp(-(tau141*alpha141));
G214=exp(-(tau214*alpha214));
G142=exp(-(tau142*alpha142));
G314=exp(-(tau314*alpha314));
G143=exp(-(tau143*alpha143));
G115=exp(-(tau115*alpha115));
G151=exp(-(tau151*alpha151));
G215=exp(-(tau215*alpha215));
G152=exp(-(tau152*alpha152));
G315=exp(-(tau315*alpha315));
G153=exp(-(tau153*alpha153));
G117=exp(-(tau117*alpha117));
G171=exp(-(tau171*alpha171));
G217=exp(-(tau217*alpha217));
G172=exp(-(tau172*alpha172));
G317=exp(-(tau317*alpha317));
G173=exp(-(tau173*alpha173));
G118=exp(-(tau118*alpha118));
G181=exp(-(tau181*alpha181));
G218=exp(-(tau218*alpha218));
G182=exp(-(tau182*alpha182));
G318=exp(-(tau318*alpha318));
G183=exp(-(tau183*alpha183));
G810=exp(-(tau810*alpha810));
G108=exp(-(tau108*alpha108));
G915=exp(-(tau915*alpha915));
G159=exp(-(tau159*alpha159));
G1314=exp(-(tau1314*alpha1314));
G1413=exp(-(tau1413*alpha1413));
G1718=exp(-(tau1718*alpha1718));
G1817=exp(-(tau1817*alpha1817));
%calculating mole fractions of ionic species
x1=x(:,1);
x2=x(:,2);
x3=x(:,3);
A=[0.0674243 0.0773881 0.0843400 0.0865343 0.0899223 0.0882858 0.0715087 0.0643867 0.0483658];
B=[0.0141081 0.0479814 0.0643151 0.0737477 0.0820756 0.0838701 0.0701576 0.0634457 0.0479639];
C=[0.0565665 0.0450072 0.0387724 0.0313828 0.02506094 0.0186280 0.0092734 0.0073438 0.0041595 ];
D=[0.0336447 0.0267694 0.0230611 0.0186659 0.0149058 0.0110795 0.0055157 0.0043679 0.0024739 ];
E=[0.0008148 0.0008756 0.00087131 0.0008794 0.0008711 0.0008441 0.0007384 0.0006997 0.0005980 ];
N=length(A);
x1n=zeros(N,1);x2n=zeros(N,1);x3n=zeros(N,1);
X1=zeros(N,1);X2=zeros(N,1);X3=zeros(N,1);X4=zeros(N,1);X5=zeros(N,1);X6=zeros(N,1);X7=zeros(N,1);
X12=zeros(N,1);X16=zeros(N,1);
for i=1:N
x1n(i)=(x1(i)-A(i)-D(i)-2*E(i)-C(i)+3*B(i));
x2n(i)=(x2(i)-A(i)-C(i)-D(i));
x3n(i)=(x3(i)-B(i));
X1(i)=(x1n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X2(i)=(x2n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X3(i)=(x3n(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X4(i)=(A(i)+D(i)+E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X5(i)=(C(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X6(i)=(A(i)-B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X7(i)=(B(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X12(i)=(E(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
X16(i)=(C(i)+D(i)/(x1n(i)+x2n(i)+x3n(i)+2*A(i)+4*B(i)+2*C(i)+2*D(i)+2*E(i)));
end
yc4=X4./(X4+X5);
yc5=X5./(X4+X5);
yc6=X6./(X6+X7+X12+X16);
yc7=X7./(X6+X7+X12+X16);
yc12=X12./(X6+X7+X12+X16);
yc16=X16./(X6+X7+X12+X16);
alpha14=yc6.*alpha18+yc7.*alpha19+yc12.*alpha113+yc16.*alpha117;
alpha24=yc6.*alpha28+yc7.*alpha29+yc12.*alpha213+yc16.*alpha217;
alpha34=yc6.*alpha38+yc7.*alpha39+yc12.*alpha313+yc16.*alpha317;
alpha15=yc6.*alpha110+yc7.*alpha115+yc12.*alpha114+yc16.*alpha118;
alpha25=yc6.*alpha210+yc7.*alpha215+yc12.*alpha214+yc16.*alpha218;
alpha35=yc6.*alpha310+yc7.*alpha315+yc12.*alpha314+yc16.*alpha318;
alpha16=yc4.*alpha81+yc5.*alpha101;
alpha26=yc4.*alpha82+yc5.*alpha102;
alpha36=yc4.*alpha83+yc5.*alpha103;
alpha17=yc4.*alpha91+yc5.*alpha151;
alpha27=yc4.*alpha92+yc5.*alpha152;
alpha37=yc4.*alpha93+yc5.*alpha153;
alpha112=yc4.*alpha131+yc5.*alpha141;
alpha212=yc4.*alpha132+yc5.*alpha142;
alpha312=yc4.*alpha133+yc5.*alpha143;
alpha116=yc4.*alpha171+yc5.*alpha181;
alpha216=yc4.*alpha172+yc5.*alpha182;
alpha316=yc4.*alpha173+yc5.*alpha183;
alpha46=yc5.*alpha810;
alpha47=yc5.*alpha915;
alpha412=yc5.*alpha1314;
alpha416=yc5.*alpha1718;
alpha56=yc4.*alpha108;
alpha57=yc4.*alpha159;
alpha512=yc4.*alpha1413;
alpha516=yc4.*alpha1817;
G14=yc6.*G18+yc7.*G19+yc12.*G113+yc16.*G117;
G24=yc6.*G28+yc7.*G29+yc12.*G213+yc16.*G217;
G34=yc6.*G38+yc7.*G39+yc12.*G313+yc16.*G317;
G15=yc6.*G110+yc7.*G115+yc12.*G114+yc16.*G118;
G25=yc6.*G210+yc7.*G215+yc12.*G214+yc16.*G218;
G35=yc6.*G310+yc7.*G315+yc12.*G314+yc16.*G318;
G16=yc4.*G81+yc5.*G101;
G26=yc4.*G82+yc5.*G102;
G36=yc4.*G83+yc5.*G103;
G17=yc4.*G91+yc5.*G151;
G27=yc4.*G92+yc5.*G152;
G37=yc4.*G93+yc5.*G153;
G112=yc4.*G131+yc5.*G141;
G212=yc4.*G132+yc5.*G142;
G312=yc4.*G133+yc5.*G143;
G116=yc4.*G171+yc5.*G181;
G216=yc4.*G172+yc5.*G182;
G316=yc4.*G173+yc5.*G183;
G46=yc5.*G810;
G47=yc5.*G915;
G412=yc5.*G1314;
G416=yc5.*G1718;
G56=yc4.*G108;
G57=yc4.*G159;
G512=yc4.*G1413;
G516=yc4.*G1817;
tau14=-log(G14)./alpha14;
tau24=-log(G24)./alpha24;
tau34=-log(G34)./alpha34;
tau15=-log(G15)./alpha15;
tau25=-log(G25)./alpha25;
tau35=-log(G35)./alpha35;
tau16=-log(G16)./alpha16;
tau26=-log(G26)./alpha26;
tau36=-log(G36)./alpha36;
tau17=-log(G17)./alpha17;
tau27=-log(G27)./alpha27;
tau37=-log(G37)./alpha37;
tau112=-log(G112)./alpha112;
tau212=-log(G212)./alpha212;
tau312=-log(G312)./alpha312;
tau116=-log(G116)./alpha116;
tau216=-log(G216)./alpha216;
tau316=-log(G316)./alpha316;
tau46=-log(G46)./alpha46;
tau47=-log(G47)./alpha47;
tau412=-log(G412)./alpha412;
tau416=-log(G416)./alpha416;
tau56=-log(G56)./alpha56;
tau57=-log(G57)./alpha57;
tau512=-log(G512)./alpha512;
tau516=-log(G516)./alpha516;
ln_y1_1=G12.*X2.*tau12+ G31.*X3.*tau13+ G14.*X4.*tau14+G15.*X5.*tau15+G16.*X6.*tau16+G17.*X7.*tau17+G112.*X12.*tau112+G116.*X16.*tau116;
ln_y1_2=G12.*X2+ G13.*X3+ G14.*X4+G15.*X5+G16.*X6+G17.*X7+G112.*X12+G116.*X16;
ln_y2_1=G21.*X1.*tau12+ G32.*X3.*tau32+ G24.*X4.*tau24+G25.*X5.*tau25+G26.*X6.*tau26+G27.*X7.*tau27+G212.*X12.*tau212+G216.*X16.*tau216;
ln_y2_2=G12.*X1+ G23.*X3+G24.*X4+G25.*X5+G26.*X6+G27.*X7+G212.*X12+G216.*X16;
ln_y3_1=G13.*X1.*tau13+ G23.*X3.*tau23+ G34.*X4.*tau34+G35.*X5.*tau35+G36.*X6.*tau36+G37.*X7.*tau37+G312.*X12.*tau312+G316.*X16.*tau316;
ln_y3_2=G13.*X1+ G23.*X3+ G34.*X4+G35.*X5+G36.*X6+G37.*X7+G312.*X12+G316.*X16;
ln_y4_1=G14.*X1.*tau14+G24.*X2.*tau24+G34.*X3.*tau34+G46.*X6.*tau46+G47.*X7.*tau47+G412.*X12.*tau412+G416.*X16.*tau416;
ln_y4_2=G14.*X1+G24.*X2+G34.*X3+G46.*X6+G47.*X7+G412.*X12+G416.*X16;
ln_y5_1=G15.*X1.*tau15+G25.*X2.*tau25+G35.*X3.*tau35+G56.*X6.*tau56+G57.*X7.*tau57+G512.*X12.*tau512+G516.*X16.*tau516;
ln_y5_2=G15.*X1+G25.*X2+G35.*X3+G56.*X6+G57.*X7+G512.*X12+G516.*X16;
ln_y6_1=G16.*X1.*tau16+G26.*X2.*tau26+G36.*X3.*tau36+G46.*X4.*tau46+G56.*X5.*tau56;
ln_y6_2=G16.*X1+G26.*X2+G36.*X3+G46.*X4+G56.*X5;
ln_y7_1=G17.*X1.*tau17+G27.*X2.*tau27+G37.*X3.*tau37+G47.*X4.*tau47+G57.*X5.*tau57;
ln_y7_2=G17.*X1+G27.*X2+G37.*X3+G47.*X4+G57.*X5;
ln_y12_1=G112.*X1.*tau112+G212.*X2.*tau212+G312.*X3.*tau312+G412.*X4.*tau412+G512.*X5.*tau512;
ln_y12_2=G112.*X1+G212.*X2+G312.*X3+G412.*X4+G512.*X5;
ln_y16_1=G116.*X1.*tau116+G216.*X2.*tau216+G316.*X3.*tau316+G416.*X4.*tau416+G516.*X5.*tau516;
ln_y16_2=G116.*X1+G216.*X2+G316.*X3+G416.*X4+G516.*X5;
ln_y1_3=(((X2.*G12)./ln_y2_2).*(tau12-(ln_y2_1)./(ln_y2_2)))+(((X3.*G13)./ln_y3_2).*(tau13-(ln_y3_1)./(ln_y3_2)));
ln_y1_4=(((X6.*G16)./ln_y6_2).*(tau16- (ln_y6_1./ln_y6_2))) + (((X7.*G17)./ln_y7_2).*(tau17- (ln_y7_1./ln_y7_2)))+(((X12.*G12)./ln_y12_2).*(tau112- (ln_y12_1./ln_y12_2)))+(((X16.*G16)./ln_y16_2).*(tau116- (ln_y16_1./ln_y16_2)));
ln_y1_5=(((X4.*G14)./ln_y4_2).*(tau14- (ln_y4_1./ln_y4_2))) + (((X5.*G15)./ln_y5_2).*(tau15- (ln_y5_1./ln_y5_2)));
yk=exp((ln_y1_1./ln_y1_2) + ln_y1_3 + ln_y1_4+ ln_y1_5)'; % activity coefficient for H2O
if any(~isfinite(yk))
yk = 10 * ones(size(yk));
end

"Heap exhausted, game over" message in wxMaxima - Does ccl will work for me?

everyone,
I'm trying to do some calculations and plot the results, but it seems that these are too heavy for Maxima. When I try to calculate N1 and N2 the program crashes when parameter j is too high or when I try to plot them, the program displays the following error message: "Heap exhausted, game over." What should I do? I've seen some people saying to try to compile Maxima with ccl, but I don't know how to do it or if it will work.
I usually receive error messages like:
Message from maxima's stderr stream: Heap exhausted during garbage collection: 0 bytes available, 16 requested.
Gen Boxed Unboxed LgBox LgUnbox Pin Alloc Waste Trig WP GCs Mem-age
0 0 0 0 0 0 0 0 20971520 0 0 0,0000
1 0 0 0 0 0 0 0 20971520 0 0 0,0000
2 0 0 0 0 0 0 0 20971520 0 0 0,0000
3 16417 2 0 0 43 1075328496 707088 293986768 16419 1 0,8032
4 13432 21 0 1141 70 955593760 838624 2000000 14594 0 0,2673
5 0 0 0 0 0 0 0 2000000 0 0 0,0000
6 741 184 34 28 0 63259792 1424240 2000000 987 0 0,0000
7 0 0 0 0 0 0 0 2000000 0 0 0,0000
Total bytes allocated = 2094182048
Dynamic-space-size bytes = 2097152000
GC control variables:
*GC-INHIBIT* = true
*GC-PENDING* = true
*STOP-FOR-GC-PENDING* = false
fatal error encountered in SBCL pid 13884(tid 0000000001236360):
Heap exhausted, game over.
Here goes the code:
enter code here
a: 80$;
b: 6*a$;
h1: 80$;
t: 2$;
j: 5$;
carga: 250$;
sig: -carga/2$;
n: 2*q*%pi/b$;
m: i*%pi/a$;
i: 2*p-1$;
i1: 2*p1-1$;
/*i1: p1$;*/
Φ: a/b$;
τ: cosh(x) - (x/sinh(x))$;
σ: sinh(x) - (x/cosh(x))$;
Ψ: sinh(x)/τ$;
Χ: cosh(x)/σ$;
Λ0: 1/(((i/2)^2+Φ^2*q^2)^2)$;
Λ1: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ0, p, 1, j)$;
Λ2: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ1, q1, 1, j)$;
Λ3: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ2, p, 1, j)$;
Λ4: sum(((q1^3*subst([x=(q1*%pi*Φ)],Χ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ3, q1, 1, j)$;
Λ5: sum((((i/2)^3*subst([x=(i*%pi/(2*Φ))],Ψ))/(((i/2)^2+Φ^2*q1^2)^2))*Λ4, p, 1, j)$;
Ζ0: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ0, q, 1, j)$;
Ζ2: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ2, q, 1, j)$;
Ζ4: sum(((q^3*subst([x=(q*%pi*Φ)],Χ))/(((i1/2)^2+Φ^2*q^2)^2))*Λ4, q, 1, j)$;
E: 200000$;
ν: 0.3$;
λ: (ν*E)/((1+ν)*(1-2*ν))$;
μ: E/(2*(1+ν))$;
a0: float(1/(b/2)*integrate(0, y, -(b/2), -h1/2)+1/b*integrate(sig, y, -h1/2, h1/2)+1/(b/2)*integrate(0, y, h1/2, (b/2)))$;
aq: float(1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q*y*%pi/(b/2)), y, h1/2, (b/2)))$;
aq1: float(1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, -(b/2), - h1/2)+1/(b/2)*integrate(sig*cos(q1*y*%pi/(b/2)), y, -h1/2, h1/2)+1/(b/2)*integrate(0*cos(q1*y*%pi/(b/2)), y, h1/2, (b/2)))$;
Bq: aq/((λ+μ)*subst([x=q*%pi*Φ],σ))+((16*Φ^4*q^2*(-1)^q)/((λ+μ)*%pi^2*subst([x=q*%pi*Φ],σ)))*sum(q1*aq1*(-1) ^q1*subst([x=q1*%pi*Φ],Χ)*(Λ1+(16*Φ^4/(%pi^2))*Λ3+((16*Φ^4/(%pi^2))^2)*Λ5), q1, 1, j)+(8*λ*Φ^3*q^2*(-1)^q*a0)/((λ+μ)*(λ+2*μ)*(%pi^3)*subst([x=q*%pi*Φ],σ))*sum(subst([x=i*%pi/(2*Φ)],Ψ)/(i/ 2)*(Λ0+(16*Φ^4/(%pi^2))*Λ2+((16*Φ^4/(%pi^2))^2)*Λ4), p, 1, j)$;
βp: -(2*λ*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*(i/2)^2*%pi^2*subst([x=i*%pi/(2*Φ)],τ))-((32*λ*Φ^4*(i/2)^2*a0*(-1)^((i-1)/2))/((λ+μ)*(λ+2*μ)*%pi^2*subst([x=i*%pi/(2*Φ)],τ)))*sum(((subst([x=i1*%pi/(2*Φ)],Ψ))/(i1/2))*(Ζ0+Ζ2*((16*Φ^4)/%pi^2)+Ζ4*(((16*Φ^4)/%pi^2)^2)),p1,1,j)-((4*Φ*(i/2)^2*(-1)^((i-1)/2))/((λ+μ)*%pi*subst([x=i*%pi/(2*Φ)],τ)))*sum(q*aq*(-1)^q*subst([x=q*%pi*Φ],Χ)*(Λ0+Λ2*(16*Φ^4/%pi^2)+Λ4*(16*Φ^4/%pi^2)^2),q,1,j)$;
N1: (2*a0/a)*x+(λ+μ)*sum(Bq*((1+((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)-n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1-((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)+m*y*sinh(m*y))*sin(m*x),p,1,j)$;
N2: ((2*λ*a0)/(a*(λ+2*μ)))*x+(λ+μ)*sum(Bq*((1-((n*a*sinh(n*a/2))/(2*cosh(n*a/2))))*sinh(n*x)+n*x*cosh(n*x))*cos(n*y),q,1,j)+(λ+μ)*sum(βp*((1+((m*b*cosh(m*b/2))/(2*sinh(m*b/2))))*cosh(m*y)-m*y*sinh(m*y))*sin(m*x),p,1,j);
wxplot3d(N1, [x,-a/2,a/2], [y,-b/2,b/2])$;
wxplot3d(N2, [x,-a/2,a/2], [y,-b/2,b/2])$;
This is not a complete answer, since I don't know how this should work with wxMaxima: I would suggest that you ask the developers. However it's too long for a comment and I think might be useful to people, and it does answer the question of how you solve the heap-size limit for Maxima itself when using SBCL, at least when run on Linux or some other platform with a command-line.
As a note, I suspect that the underlying problem is not the heap size, but that the calculation is blowing up in some horrible way: the best fix is probably to understand what's blowing up and fix that. See Robert Dodier's answer, which is probably going to be a lot more helpful. However, if heap size is the problem, this is how you deal with it for Maxima.
The trick is that you can tell SBCL what the heap limit should be by passing it the --dynamic-space-size <MB> argument, and you can pas arguments through the maxima wrapper to do this.
Here is a transcript of Maxima, being run on Linux, with SBCL as a back end (this is a version built from source: the packaged version will I assume be the same):
$ maxima
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
1073741824
So, on this system the defaule heap limit is 1GB (this is SBCL's default limit on the platform).
Now we can pass the -X <lisp options> aka --lisp-options=<lisp options> option to the maxima wrapper to pass the appropriate option through to sbcl:
$ maxima -X '--dynamic-space-size 2000'
Lisp options: (--dynamic-space-size 2000)
Maxima 5.43.2 http://maxima.sourceforge.net
using Lisp SBCL 2.0.0
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) :lisp (sb-ext:dynamic-space-size)
2097152000
As you can see this has doubled the heap size.
If someone knows the answer for wxMaxima then please do add an edit to this answer: I can't experiment it because all my Linux VMs are headless.
Also not a complete answer here, but some more notes and pointers which I hope will help.
To make the problem easier for Maxima to digest, use only exact numbers (integers and ratios), and avoid float and numer. (Plotting functions will apply float and numer automatically.) I changed 0.3 to 3/10 and cut out the calls to float.
Also, try setting j to a smaller number (I tried j equal to 1) to try to work all the way through the problem before increasing it to 5 again.
Also, replace all sum and integrate with 'sum and 'integrate (i.e. noun expressions instead of verb expressions). Take a look at the summands and integrands to see if they look right. You can evaluate the sums and/or integrals or both via ev(expr, sum) or ev(expr, integrate) or ev(expr, nouns) to evaluate 'sum, 'integrate, or all noun expressions, respectively.
With j equal to 1, I get the following expression for N1:
(2500000*((-(13*cosh(%pi/6)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(120000*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
+(13*sinh(3*%pi)
*((2754990144*cosh(%pi/6)^3*sinh(3*%pi)^2)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^3
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(17006112*cosh(%pi/6)^2*sinh(3*%pi))
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+(104976*cosh(%pi/6))
/(625*(sinh(%pi/6)-%pi/(6*cosh(%pi/6))))))
/(22680000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+13/(35000*%pi^2*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))))
*sin((%pi*(2*p-1)*x)/80)
*((%pi*(2*p-1)*y*sinh((%pi*(2*p-1)*y)/80))/80
+(1-(3*%pi*(2*p-1)*cosh(3*%pi*(2*p-1)))
/sinh(3*%pi*(2*p-1)))
*cosh((%pi*(2*p-1)*y)/80)))
/13
+(2500000*((-(13*cosh(%pi/6)
*((344373768*cosh(%pi/6)^2*sinh(3*%pi)^3)
/(244140625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))
^3)
+(2125764*cosh(%pi/6)*sinh(3*%pi)^2)
/(390625*%pi^2
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(13122*sinh(3*%pi))
/(625*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))))
/(1620000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2))
+(13*sinh(3*%pi)
*((8503056*cosh(%pi/6)^2*sinh(3*%pi)^2)
/(9765625*%pi^4
*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))^2
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi))^2)
+(52488*cosh(%pi/6)*sinh(3*%pi))
/(15625*%pi^2*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
+324/25))
/(3780000*%pi^3*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))
*(cosh(3*%pi)-(3*%pi)/sinh(3*%pi)))
-13/(20000*%pi*(sinh(%pi/6)-%pi/(6*cosh(%pi/6)))))
*(((%pi*sinh(%pi/6))/(6*cosh(%pi/6))+1)
*sinh((%pi*x)/240)
-(%pi*x*cosh((%pi*x)/240))/240)*cos((%pi*y)/240))
/13-(25*x)/48$
Now in order to plot that, it should be a function of x and y only. However listofvars reports that it contains x, y, and p. Hmm. I see that βp has a summation over p1 but it contains Ζ0, which contains Λ0, which contains p. Is the summation over p1 supposed to be over p? Is the summand supposed to contain p1 instead of p?
Likewise it appears that N2, after evaluating the sums and integrals with j equal to 1, contains p.
Maybe you need to rework the formulas somewhat? I don't know what the correct form might be.

about torch.nn.CrossEntropyLoss parameter shape

i'm learning pytorch, and taking the anpr project,which is based tensorflow
(https://github.com/matthewearl/deep-anpr,
http://matthewearl.github.io/2016/05/06/cnn-anpr/)
as a exercise, transplant it to pytorch platform.
there is a problem,i'm using nn.CrossEntropyLoss() as loss function:
criterion=nn.CrossEntropyLoss()
the output.data of model is:
- 1.00000e-02 *
- 2.5552 2.7582 2.5368 ... 5.6184 1.2288 -0.0076
- 0.7033 1.3167 -1.0966 ... 4.7249 1.3217 1.8367
- 0.7592 1.4777 1.8095 ... 0.8733 1.2417 1.1521
- 0.1040 -0.7054 -3.4862 ... 4.7703 2.9595 1.4263
- [torch.FloatTensor of size 4x253]
and targets.data is:
- 1 0 0 ... 0 0 0
- 1 0 0 ... 0 0 0
- 1 0 0 ... 0 0 0
- 1 0 0 ... 0 0 0
- [torch.DoubleTensor of size 4x253]
when i call:
loss=criterion(output,targets)
error occured,information is:
TypeError: FloatClassNLLCriterion_updateOutput received an invalid combination of arguments - got (int, torch.FloatTensor, **torch.DoubleTensor**, torch.FloatTensor, bool, NoneType, torch.FloatTensor), but expected (int state, torch.FloatTensor input, **torch.LongTensor** target, torch.FloatTensor output, bool sizeAverage, [torch.FloatTensor weights or None], torch.FloatTensor total_weight)
'expected torch.LongTensor'......'got torch.DoubleTensor',but if i convert the targets into LongTensor:
torch.LongTensor(numpy.array(targets.data.numpy(),numpy.long))
call loss=criterion(output,targets), the error is:
RuntimeError: multi-target not supported at /data/users/soumith/miniconda2/conda-bld/pytorch-0.1.10_1488752595704/work/torch/lib/THNN/generic/ClassNLLCriterion.c:20
my last exercise is mnist, a example from pytorch,i made a bit modification,batch_size is 4,the loss function:
loss = F.nll_loss(outputs, labels)
outputs.data:
- -2.3220 -2.1229 -2.3395 -2.3391 -2.5270 -2.3269 -2.1055 -2.2321 -2.4943 -2.2996
-2.3653 -2.2034 -2.4437 -2.2708 -2.5114 -2.3286 -2.1921 -2.1771 -2.3343 -2.2533
-2.2809 -2.2119 -2.3872 -2.2190 -2.4610 -2.2946 -2.2053 -2.3192 -2.3674 -2.3100
-2.3715 -2.1455 -2.4199 -2.4177 -2.4565 -2.2812 -2.2467 -2.1144 -2.3321 -2.3009
[torch.FloatTensor of size 4x10]
labels.data:
- 8
- 6
- 0
- 1
- [torch.LongTensor of size 4]
the labels, for a input image,must be a single element, in upper example, there is 253 numbers, and in 'mnist',there is only one number, the shape of outputs is difference from labels.
i review the tensorflow manual, tf.nn.softmax_cross_entropy_with_logits,
'Logits and labels must have the sameshape [batch_size, num_classes] and the same dtype (either float32 or float64).'
does pytorch support the same function in tensorflow?
many thks
You can convert the targets that you have to a categorical representation.
In the example that you provide, you would have 1 0 0 0.. 0 if the class is 0, 0 1 0 0 ... if the class is 1, 0 0 1 0 0 0... if the class is 2 etc.
One quick way that I can think of is first convert the target Tensor to a numpy array, then convert it from one hot to a categorical array, and convert it back to a pytorch Tensor. Something like this:
targetnp=targets.numpy()
idxs=np.where(targetnp>0)[1]
new_targets=torch.LongTensor(idxs)
loss=criterion(output,new_targets)
CrossEntropyLoss is equivalent to tf.nn.softmax_cross_entropy_with_logits. The input to CrossEntropyLoss is a categorical vector of shape [batch_size]. Use .view() to change the tensor shapes.
labels = labels.view(-1)
output = output.view(labels.size(0), -1)
loss = criterion(output, loss)
calling .view(x, y, -1) causes the tensor to use the remaining datapoints to fill the -1 dimension and will cause an error if there is not enough to make a full dimension
labels.size(0) gives the size of the 0th dimension of the label tensor
Additional
to convert between tensor types you can call the type on the tensor, for example 'labels = labels.long()`
Second Additional
If you unpack the data from a variable like so output.data then you will lose the gradients for that output and be unable to backprop when the time comes

Matlab - read unstructured file

I'm quite new with Matlab and I've been searching, unsucessfully, for the following issue: I have an unstructure txt file, with several rows I don't need, but there are a number of rows inside that file that have an structured format. I've been researching how to "load" the file to edit it, but cannot find anything.
Since i don't know if I was clear, let me show you the content in the file:
8782 PROJCS["UTM-39",GEOGC.......
1 676135.67755473056 2673731.9365976951 -15 0
2 663999.99999999302 2717629.9999999981 -14.00231124135486 3
3 709999.99999999162 2707679.2185399458 -10 2
4 679972.20003752434 2674637.5679516452 0.070000000000000007 1
5 676124.87132483651 2674327.3183533219 -18.94794942571912 0
6 682614.20527054626 2671000.0000000549 -1.6383425512446661 0
...........
8780 682247.4593014461 2676571.1515358146 0.1541080392180566 0
8781 695426.98657108378 2698111.6168302582 -8.5039945992245904 0
8782 674723.80100125563 2675133.5486935056 -19.920312922947179 0
16997 3 21
1 2147 658 590
2 1855 2529 5623
.........
I'd appreciate if someone can just tell me if there is the possibility to open the file to later load only the rows starting with 1 to the one starting with 8782. First row and all the others are not important.
I know than manually copy and paste to a new file would be a solution, but I'd like to know about the possibility to read the file and edit it for other ideas I have.
Thanks!
% Now lines{i} is the string of the i'th line.
lines = strsplit(fileread('filename'), '\n')
% Now elements{i}{j} is the j'th field of the i'th line.
elements = arrayfun(#(x){strsplit(x{1}, ' ')}, lines)
% Remove the first row:
elements(1) = []
% Take the first several rows:
n_rows = 8782
elements = elements(1:n_rows)
Or if the number of rows you need to take is not fixed, you can replace the last two statements above by:
firsts = arrayfun(#(x)str2num(x{1}{1}), elements)
n_rows = find((firsts(2:end) - firsts(1:end-1)) ~= 1, 1, 'first')
elements = elements(1:n_rows)

Insert a row between two known rows in Matlab

I have a set of data shown bellow:
flow Rate (L/min)
Speed(rpm) 1 1.25 1.5 1.75 2 2.25 2.5 2.77 ... 6
Pressure (Pa)
2000 15251.2 15232 15200 15168 15027.2 14912 14752 0 ... 0
2050 16000 15840 15808 15744 15680 15520 15488 15232 ... 0
2100 16384 16256 16217.6 16192 16128 16064 16032 15872 ... 0
2150 17088 17024 16992 16960 16928 16832 16704 16512 ... 0
. . . . . . . . . .
. . . . . . . . . .
. . . . . . . . . .
4250 61120 60800 60768 60736 60672 60736 60608 60416 ... 56960
At a specific speed (from 2000-4250rpm) and flow rate (from 1-6 L/min) as shown there are different pressures.
1) i want to know how can i insert a new row in between two of these speeds for example if i have a speed of 2030rpm i want to be able to find in between which two values the 2030rpm is and insert a row on matlab
demonstration hown below:
2000 15251.2 15232 15200 15168 15027.2 14912 14752 0 ... 0
2030 0 0 0 0 0 0 0 0
2050 16000 15840 15808 15744 15680 15520 15488 15232 ... 0
2) my second problem is how can i interpolate between the two values below (where the zero is and get a value.
15232
0
16000
I really appreciate if any one can answer any of my questions preferably the first one so ic an actually get to the second step lol
Thank you so much
newmat = zeros(size(oldmat,1)+1,size(oldmat,2))
newmat(1:x) = oldmat(1:x)
newmat(x+2:end) = oldmat(x+1:end)
where oldmat, newmat are the old and new versions of your matrix and x+1 the index of the row of 0s inserted into newmat.
Then, supposing that you want linear interpolation, something like:
newmat(x+1,:) = newmat(x,:)+0.6*(newmat(x+2,:)-newmat(x,:))
I expect I've made some small errors, and this is quite specific to your example, if you have trouble fixing and generalising, update your question or comment.
Assuming the data is stored in a matrix called p, for automatically positioning the new row in correct sequence:
Append the new row at the end of p, then:
p = sortrows(p)
Following up on the comments, we have:
newrow = [2130, zeros(1,size(test,2)-1)]
p(size(p,1)+1,:) = newrow
p = sortrows(p)
(if 2130 is the first value of the new row.)
This may help you:
% Matrix dimensions
nCols = 10;
nRows = 8;
% Synthetic data
matrix = [ linspace(2000,4250,nRows)' , 2000*rand(nRows,nCols-1)];
matrix([2,4],2:end) = zeros(2,nCols-1); % where some rows are zeros (2 and 4 on this example)
matrix
matrix =
1.0e+03 *
2.0000 1.7810 1.3674 1.4983 0.7329 1.5439 1.5639 0.2246 0.8653 1.5379
2.3214 0 0 0 0 0 0 0 0 0
2.6429 1.4687 1.4454 1.4801 1.3701 0.7765 0.5881 0.5831 0.2195 0.5459
2.9643 0 0 0 0 0 0 0 0 0
3.2857 0.1458 0.2350 1.4699 1.5787 0.4579 1.0617 1.9288 0.3749 1.3466
3.6071 0.1771 1.2814 1.9412 0.7353 1.2839 0.1830 0.8650 0.5324 0.8591
3.9286 1.5967 0.6576 1.7339 0.4121 0.9690 0.8106 1.3895 1.5957 0.9035
4.2500 1.8860 1.3076 0.1725 0.1733 0.3037 0.2097 1.5162 0.9752 1.2197
If you just want to fill the rows whose elements from the second to the last columns are zeros with the average value of the previous and the next rows.
for i=2:nRows-1
if ( sum(matrix(i,2:end))==0 )
matrix(i,2:end) = mean( matrix([i-1,i+1],2:end) );
end
end
matrix
matrix =
1.0e+03 *
2.0000 1.7810 1.3674 1.4983 0.7329 1.5439 1.5639 0.2246 0.8653 1.5379
2.3214 1.6248 1.4064 1.4892 1.0515 1.1602 1.0760 0.4039 0.5424 1.0419
2.6429 1.4687 1.4454 1.4801 1.3701 0.7765 0.5881 0.5831 0.2195 0.5459
2.9643 0.8072 0.8402 1.4750 1.4744 0.6172 0.8249 1.2560 0.2972 0.9462
3.2857 0.1458 0.2350 1.4699 1.5787 0.4579 1.0617 1.9288 0.3749 1.3466
3.6071 0.1771 1.2814 1.9412 0.7353 1.2839 0.1830 0.8650 0.5324 0.8591
3.9286 1.5967 0.6576 1.7339 0.4121 0.9690 0.8106 1.3895 1.5957 0.9035
4.2500 1.8860 1.3076 0.1725 0.1733 0.3037 0.2097 1.5162 0.9752 1.2197
This code assumes that:
You want to fill rows were only the first column element is non-zero.
You want to replace the zeros with the average between previous and next rows values.
You only interpolate inner rows.
I hope it helps.