Difference between TruncatedSVD and svds - scipy

I see that the documentation for both sklearn.decomposition.TruncatedSVD and scipy.sparse.linalg.svds mention that they both perform SVD for sparse matrices. What is the difference between them?
Thanks.

TruncatedSVD is more feature-rich. It has the scikit-learn API, so you can put it in a sklearn.Pipeline object and call transform on a new matrix instead of having to figure out the matrix multiplications yourself. It offers two algorithms: either a fast randomized SVD solver (the default), or scipy.sparse.svds.
(Full disclosure: I wrote TruncatedSVD.)

There are doing the same
#check how to use TRuncatedSVD
X=[[1,2,3],[1,4,2],[4,1,7],[5,6,8]]
# TRUNCATED SVD
from sklearn.decomposition import TruncatedSVD
svd = TruncatedSVD(n_components=2, n_iter=7, random_state=42)
US=svd.fit_transform(X)
V=svd.components_
S=svd.singular_values_
print('u,s,v', US,S,V)
print('X_restored dot way',np.round(np.dot(US,V),1),'svdinverse way',np.round(svd.inverse_transform(U),1))
# LINALG SVD
U1,S1,V1=np.linalg.svd(X)
print('u1,s1,v1 remark negative mirrored',U1[:,:2]*S1[:2],V1[:2,:])
print('X restored u1,s1,v1, 2 components',np.round( np.dot( U1[:,:2]*S1[:2],V1[:2,:] ),1 ) )
# sparse svd
from scipy.sparse import csc_matrix
from scipy.sparse.linalg import svds, eigs
A = csc_matrix(X, dtype=float)
u2, s2, vt2 = svds(A, k=2)
print('sparse reverses !',u2*s2,vt2)
print('x restored',np.round( np.dot(u2*s2,vt2),1) )
RESULT
u,s,v [[ 3.66997034 -0.34754761]
[ 3.82764223 -2.51681397]
[ 7.61154768 2.83860088]
[11.13470337 -0.96070751]] [14.49264657 3.92883644] [[ 0.44571865 0.46215842 0.76664495]
[ 0.23882889 -0.88677195 0.39572247]]
X_restored dot way
[[1.6 2. 2.7]
[1.1 4. 1.9]
[4.1 1. 7. ]
[4.7 6. 8.2]]
svdinverse way
[[1.6 2. 2.7]
[1.1 4. 1.9]
[4.1 1. 7. ]
[4.7 6. 8.2]]
u1,s1,v1 remark negative mirrored
[[ -3.66997034 0.34754761]
[ -3.82764223 2.51681397]
[ -7.61154768 -2.83860088]
[-11.13470337 0.96070751]] [[-0.44571865 -0.46215842 -0.76664495]
[-0.23882889 0.88677195 -0.39572247]]
X restored u1,s1,v1, 2 components
[[1.6 2. 2.7]
[1.1 4. 1.9]
[4.1 1. 7. ]
[4.7 6. 8.2]]
sparse reverses !
[[-0.34754761 3.66997034]
[-2.51681397 3.82764223]
[ 2.83860088 7.61154768]
[-0.96070751 11.13470337]]
[[ 0.23882889 -0.88677195 0.39572247]
[ 0.44571865 0.46215842 0.76664495]]
x restored
[[1.6 2. 2.7]
[1.1 4. 1.9]
[4.1 1. 7. ]
[4.7 6. 8.2]]
[[-0.25322982 0.0884607 0.88223679]
[-0.26410926 0.64060034 0.16752502]
[-0.52520067 -0.72250421 0.11259767]
[-0.76830021 0.24452723 -0.42534148]]
[14.49264657 3.92883644 0.72625043]
[[-0.44571865 -0.46215842 -0.76664495]
[-0.23882889 0.88677195 -0.39572247]
[-0.86272571 -0.00671608 0.50562757]]
[[1. 2. 3.]
[1. 4. 2.]
[4. 1. 7.]
[5. 6. 8.]]

Related

How to apply best fit distributions in pyspark?

I currently working on a migration from python to pyspark,and I have one step where I find the best fit distribution using a modified function of Fitting empirical distribution to theoretical ones with Scipy (Python)? where I apply best_fit_distribution to each group od Id's, and save the output in a dictionary,there is some way to do that in pyspark? I was doing research about pyspark statistics and I don't find any library that could help me.
For the needs of the development I need to do this part in pyspark, so keep in original python can't be an option.
import scipy.stats as st
import numpy as np
import warnings
def best_fit_distribution(data, bins=200, ax=None)
y, x = np.histogram(data, bins=bins, density=True)
x = (x + np.roll(x, -1))[:-1] / 2.0
# Distributions to check
distribution_list = [st.alpha,st.chi2, st.pearson3] #This is an example
# Best holders
best_distribution = st.norm
best_params = (0.0, 1.0)
best_sse = np.inf
for distribution in distribution_list:
try:
with warnings.catch_warnings():
warnings.filterwarnings("ignore")
params = distribution.fit(data)
arg = params[:-2]
loc = params[-2]
scale = params[-1]
pdf = distribution.pdf(x, loc=loc, scale=scale, *arg)
sse = np.sum(np.power(y - pdf, 2.0))
# if axis pass in add to plot
try:
if ax:
pd.Series(pdf, x).plot(ax=ax)
#end
except Exception:
pass
# identify if this distribution is better
if best_sse > sse > 0:
best_distribution = distribution
best_params = params
best_sse = sse
except Exception:
pass
return (best_distribution.name, best_params)
This is an example and description of my df:
Id
Values
8
59.25
8
25.1
8
39.0333
8
138.3737
8
79.5002
8
52.9
8
0.1674
9
33.8667
9
0.75
9
78.05
9
76.9167
9
14.6667
9
80.3166
9
32.7333
9
0.8333
9
76.95
9
84.4
9
23.1667
9
23.1
9
76.6667
summary
Id
Values
count
34052
1983107
min
8
0.0
max
2558
59646.1712

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

Is there inherent interaction between odes when using matlab's ode45 function?

I am trying to model interaction (diffusion) between population in two blocks using Matlab's ODE45 function.
The population in each block is governed by the same set of ODEs, but has different initial conditions.
However, I find that even when there is no interaction term between the two blocks (such that they should be independent), the difference in initial conditions still seems to influence the populations in them. This means that, when calculated together, the populations in the blocks differ from when calculated separately. Is there an explanation for this?
I tried this with a very simple model for a proof of concept:
Function calls:
[t1,y1] = ode45(#equations, [1:20], [100; 50]); % block 1
[t2,y2] = ode45(#equations, [1:20], [100; 100]); % block 2
[t3,y3] = ode45(#equations2, [1:20], [100;50;100;100]); % (block 1, block 2)
Function modelling population in one block only:
function [dydt] = equations(t,y)
dydt = [(y(1) + 5*y(2)); ... %m
(-y(2) - 2*y(1)); ... %n
];
return
Function modelling both blocks, but without interaction:
function [dydt] = equations2(t,y)
dydt = [(y(1) + 5*y(2)); ... %m, block 1
(-y(2) - 2*y(1)); ... %n, block 1
(y(3) + 5*y(4)); ... %m, block 2
(-y(4) - 2*y(3)); ... %n, block 2
];
return
Result when two blocks calculated separately:
100 50 100 100
-82.5651126482403 -61.2399111160678 -70.8298703784238 -113.079426303262
63.5149326478799 71.2611808361777 40.2624466444427 123.919353592460
-43.1859556122909 -79.8393860152817 -8.90683490348150 -132.277927048621
21.9743397563556 86.8220697463693 -22.6303780909040 137.973908367664
-0.354918519359671 -92.0989416214417 53.7225657700143 -140.919984325956
-21.2629138588043 95.5352361959004 -83.7315196717762 141.098469411677
42.4644938860107 -97.0482785784471 112.050303306456 -138.445708319198
-62.8212058809813 96.6327114704933 -138.138533695681 133.012881410501
81.9247177241794 -94.3111553111835 161.481224015549 -124.963465578347
-99.3812030807478 90.1082810663461 -181.603255611880 114.429891038409
114.845927928453 -84.0902243542573 198.088587603452 -101.593579579227
-128.038489201868 76.4154884940680 -210.635351405432 86.7397437298800
138.675067137440 -67.2257825991193 219.018298709806 -70.1774427449607
-146.535410253369 56.6883955999672 -223.021403427644 52.2241854879404
151.485921591776 -45.0236010893207 222.530966569364 -33.2118012943491
-153.442292768067 32.4726456791801 -217.644899082991 13.5504349990489
152.321501615162 -19.2743265017041 208.457113276080 6.36250787673260
-148.144949801383 5.68381588482233 -195.066043293572 -26.1560882959426
141.024940766196 8.01698112966525 177.792461039602 45.4369626814643
Result when two blocks calculated together:
100 50 100 100
-82.5614380549815 -61.2386250164727 -70.8214933555803 -113.085294273425
63.4808523920134 71.2492107022344 40.2403909177619 123.906052225068
-43.1428254169980 -79.8465560431243 -8.86332933780633 -132.269515222896
21.9512910424390 86.8494877098212 -22.6803808332093 137.993637919124
-0.324547757386938 -92.1128384145929 53.7640196166716 -140.954822929939
-21.3110352832942 95.5328877966081 -83.7748031881502 141.094761269331
42.5230708505071 -97.0466778715188 112.116137495667 -138.418902426909
-62.8798722752104 96.6307163280550 -138.215550196187 132.992890319329
81.9803719854667 -94.2947890971016 161.559769097134 -124.926060504869
-99.4446392159052 90.0812551138450 -181.682036111080 114.372592711550
114.920836766971 -84.0716375263694 198.174987302179 -101.539954624572
-128.100385825296 76.3847285271368 -210.709163142815 86.6824352002572
138.720481894513 -67.1710769635203 219.033021842029 -70.0921219690280
-146.574502140440 56.6199063635580 -222.990477101486 52.1070327582791
151.516503440925 -44.9493339209586 222.520965582938 -33.0950981283072
-153.445319072891 32.3873778023315 -217.627693978053 13.4288556805328
152.307662312383 -19.1817679777499 208.387426508819 6.50027540164903
-148.121448302555 5.59183915228772 -194.975897304652 -26.2998808971022
140.970297553726 8.11485856302741 177.658703561865 45.5804419325655
How much different are the results? Ode45 is a method with step-size control. The computation of the local error that is used to control the step size will certainly be different (not necessarily better or worse) for the joint system. However, the resulting differences in the values should stay inside the relative and absolute tolerances used in the method, which by default are 1e-6. Using smaller tolerances should "decouple" the two systems to an according degree.
PS: Up to changes in the last displayed digit, the correct result should be
1 [ 100. 50. 100. 100. ]
2 [ -82.53524872 -61.2596255 -70.77524805 -113.11125047]
3 [ 63.41855387 71.29313918 40.13392902 123.95857849]
4 [ -43.03253624 -79.8997202 -8.68932914 -132.32487471]
5 [ 21.7852221 86.90710777 -22.92918773 138.04268767]
6 [ -0.101876596 -92.1750490 54.0887768 -140.997575 ]
7 [ -21.58350797 95.59810598 -84.16577854 141.1303955 ]
8 [ 42.83689848 -97.10776622 112.55820169 -138.43848988]
9 [ -63.23290817 96.67381387 -138.69777167 132.97573693]
10 [ 82.36331078 -94.30493447 162.06130481 -124.85147371]
11 [ -99.84521116 90.04854117 -182.18117983 114.2283074 ]
12 [ 115.32870896 -83.9898257 198.6546973 -101.31886073]
13 [-128.50390187 76.2500533 -211.15213965 86.38151638]
14 [ 139.1070883 -66.98413557 219.42337049 -69.71524539]
15 [-146.92604542 56.37752991 -223.30284108 51.65362329]
16 [ 151.80427675 -44.6425276 222.71290379 -32.55815357]
17 [-153.64414443 32.0140048 -217.66536621 12.81103218]
18 [ 152.40882353 -18.74472148 208.26125485 7.1925021 ]
19 [-148.12303898 5.10026242 -194.68879305 -27.05207841]
20 [ 140.87257081 8.64627842 177.21963374 46.37020718]
Both your runs already differ unacceptably in the first line.

Libsvm Classification MATLAB

I used 1~200 data as trainning data, 201~220 as testing data
format likes: 3 class(class 1,class 2, class 3) and 20 features
2 1:100 2:96 3:88 4:94 5:96 6:94 7:72 8:68 9:69 10:70 11:76 12:70 13:73 14:71 15:74 16:76 17:78 18:81 19:76 20:76
2 1:96 2:100 3:88 4:88 5:90 6:98 7:71 8:66 9:63 10:74 11:75 12:66 13:71 14:68 15:74 16:78 17:78 18:85 19:77 20:76
2 1:88 2:88 3:100 4:96 5:91 6:89 7:70 8:70 9:68 10:74 11:76 12:71 13:73 14:74 15:79 16:77 17:73 18:80 19:78 20:78
2 1:94 2:87 3:96 4:100 5:92 6:88 7:76 8:73 9:71 10:70 11:74 12:67 13:71 14:71 15:76 16:77 17:71 18:80 19:73 20:73
2 1:96 2:90 3:91 4:93 5:100 6:92 7:74 8:67 9:67 10:75 11:75 12:67 13:74 14:73 15:77 16:77 17:75 18:82 19:76 20:74
2 1:93 2:98 3:90 4:88 5:92 6:100 7:73 8:66 9:65 10:73 11:78 12:69 13:73 14:72 15:75 16:74 17:75 18:83 19:79 20:77
3 1:73 2:71 3:73 4:76 5:74 6:73 7:100 8:79 9:79 10:71 11:65 12:58 13:67 14:73 15:74 16:72 17:60 18:63 19:64 20:60
3 1:68 2:66 3:70 4:73 5:68 6:67 7:78 8:100 9:85 10:77 11:57 12:57 13:58 14:62 15:68 16:64 17:59 18:57 19:57 20:59
3 1:69 2:64 3:70 4:72 5:69 6:65 7:78 8:85 9:100 10:70 11:56 12:63 13:62 14:61 15:64 16:69 17:56 18:55 19:55 20:51
3 1:71 2:74 3:74 4:70 5:76 6:73 7:71 8:73 9:71 10:100 11:58 12:58 13:59 14:60 15:58 16:65 17:57 18:57 19:63 20:57
1 1:77 2:75 3:76 4:73 5:75 6:79 7:66 8:56 9:56 10:59 11:100 12:77 13:84 14:79 15:82 16:80 17:82 18:82 19:81 20:82
1 1:70 2:66 3:71 4:67 5:67 6:70 7:63 8:57 9:62 10:58 11:77 12:100 13:84 14:75 15:76 16:78 17:73 18:72 19:87 20:80
1 1:73 2:72 3:73 4:71 5:74 6:74 7:68 8:58 9:61 10:59 11:84 12:84 13:100 14:86 15:88 16:91 17:81 18:81 19:84 20:86
1 1:71 2:69 3:75 4:71 5:73 6:73 7:74 8:61 9:61 10:60 11:79 12:75 13:86 14:100 15:90 16:88 17:74 18:79 19:81 20:82
1 1:74 2:74 3:80 4:76 5:78 6:76 7:73 8:66 9:64 10:59 11:81 12:76 13:88 14:90 15:100 16:93 17:74 18:83 19:81 20:85
1 1:76 2:77 3:77 4:76 5:78 6:75 7:73 8:64 9:68 10:65 11:80 12:78 13:91 14:88 15:93 16:100 17:79 18:79 19:82 20:83
1 1:78 2:78 3:73 4:71 5:75 6:75 7:61 8:58 9:57 10:56 11:82 12:73 13:81 14:74 15:74 16:80 17:100 18:85 19:80 20:85
1 1:81 2:85 3:79 4:80 5:82 6:82 7:63 8:56 9:55 10:57 11:82 12:72 13:81 14:79 15:83 16:79 17:85 18:100 19:83 20:79
1 1:76 2:77 3:78 4:75 5:76 6:79 7:65 8:57 9:57 10:63 11:81 12:87 13:84 14:81 15:81 16:82 17:80 18:83 19:100 20:87
1 1:76 2:76 3:78 4:73 5:75 6:78 7:60 8:59 9:51 10:57 11:82 12:80 13:86 14:82 15:85 16:83 17:85 18:79 19:87 20:100
Then, I write code to classify them:
% read the data set
[image_label, image_features] = libsvmread(fullfile('D:\...'));
[N D] = size(image_features);
% Determine the train and test index
trainIndex = zeros(N,1);
trainIndex(1:200) = 1;
testIndex = zeros(N,1);
testIndex(201:N) = 1;
trainData = image_features(trainIndex==1,:);
trainLabel = image_label(trainIndex==1,:);
testData = image_features(testIndex==1,:);
testLabel = image_label(testIndex==1,:);
% Train the SVM
model = svmtrain(trainLabel, trainData, '-c 1 -g 0.05 -b 1');
% Use the SVM model to classify the data
[predict_label, accuracy, prob_values] = svmpredict(testLabel, testData, model, '-b 1');
But the final result for predict_label are all class 1, so the accuracy is 50%, which that it cannot get the correct predict label for class 2 and 3.
Is there something wrong from the format of data, or the code that I implemented?
Please help me, thanks very much.
To elaborate a bit more about the problem, there are at least three problems here:
You just check one values of parameters C (c) and Gamma (g) - behaviour of SVM is heavily dependant on the good choice of these parameters, so it is a common approach to use a grid search using cross validation testing for selecting the best ones.
Data scale also plays an important role here, if some of the dimensions are much bigger then the rest, you will bias the whole classifier, in order to deal with it there are at least two basic approaches: 1. Scale linearly each dimension to some interval (like [0,1] or [-1,1]) or normalize the data by transformation through Sigma^(-1/2) where Sigma is a data covariance matrix
Label imbalance - SVM works best when you have exactly the same amount of points in each class. Once it is not true, you should use the class weighting scheme in order to get valid results.
After fixing these three issues you should get reasonable results.
My guess is that you'd want to tune your parameters.
Make a loop over your -c and -g values (typically logarithimically, eg -c 10^(-3:5) ) and pick the one that is best.
That said, it is advisable to normalize your data, eg. scale it such that all values are between 0 and 1.

How to calculate a Mod b in Casio fx-991ES calculator

Does anyone know how to calculate a Mod b in Casio fx-991ES Calculator. Thanks
This calculator does not have any modulo function. However there is quite simple way how to compute modulo using display mode ab/c (instead of traditional d/c).
How to switch display mode to ab/c:
Go to settings (Shift + Mode).
Press arrow down (to view more settings).
Select ab/c (number 1).
Now do your calculation (in comp mode), like 50 / 3 and you will see 16 2/3, thus, mod is 2. Or try 54 / 7 which is 7 5/7 (mod is 5).
If you don't see any fraction then the mod is 0 like 50 / 5 = 10 (mod is 0).
The remainder fraction is shown in reduced form, so 60 / 8 will result in 7 1/2. Remainder is 1/2 which is 4/8 so mod is 4.
EDIT:
As #lawal correctly pointed out, this method is a little bit tricky for negative numbers because the sign of the result would be negative.
For example -121 / 26 = -4 17/26, thus, mod is -17 which is +9 in mod 26. Alternatively you can add the modulo base to the computation for negative numbers: -121 / 26 + 26 = 21 9/26 (mod is 9).
EDIT2: As #simpatico pointed out, this method will not work for numbers that are out of calculator's precision. If you want to compute say 200^5 mod 391 then some tricks from algebra are needed. For example, using rule
(A * B) mod C = ((A mod C) * B) mod C we can write:
200^5 mod 391 = (200^3 * 200^2) mod 391 = ((200^3 mod 391) * 200^2) mod 391 = 98
As far as I know, that calculator does not offer mod functions.
You can however computer it by hand in a fairly straightforward manner.
Ex.
(1)50 mod 3
(2)50/3 = 16.66666667
(3)16.66666667 - 16 = 0.66666667
(4)0.66666667 * 3 = 2
Therefore 50 mod 3 = 2
Things to Note:
On line 3, we got the "minus 16" by looking at the result from line (2) and ignoring everything after the decimal. The 3 in line (4) is the same 3 from line (1).
Hope that Helped.
Edit
As a result of some trials you may get x.99991 which you will then round up to the number x+1.
You need 10 ÷R 3 = 1
This will display both the reminder and the quoitent
÷R
There is a switch a^b/c
If you want to calculate
491 mod 12
then enter 491 press a^b/c then enter 12. Then you will get 40, 11, 12. Here the middle one will be the answer that is 11.
Similarly if you want to calculate 41 mod 12 then find 41 a^b/c 12. You will get 3, 5, 12 and the answer is 5 (the middle one). The mod is always the middle value.
You can calculate A mod B (for positive numbers) using this:
Pol( -Rec( 1/2πr , 2πr × A/B ) , Y ) ( πr - Y ) B
Then press [CALC], and enter your values for A and B, and any value for Y.
/ indicates using the fraction key, and r means radians ( [SHIFT] [Ans] [2] )
type normal division first and then type shift + S->d
Here's how I usually do it. For example, to calculate 1717 mod 2:
Take 1717 / 2. The answer is 858.5
Now take 858 and multiply it by the mod (2) to get 1716
Finally, subtract the original number (1717) minus the number you got from the previous step (1716) -- 1717-1716=1.
So 1717 mod 2 is 1.
To sum this up all you have to do is multiply the numbers before the decimal point with the mod then subtract it from the original number.
Note: Math error means a mod m = 0
It all falls back to the definition of modulus: It is the remainder, for example, 7 mod 3 = 1.
This because 7 = 3(2) + 1, in which 1 is the remainder.
To do this process on a simple calculator do the following:
Take the dividend (7) and divide by the divisor (3), note the answer and discard all the decimals -> example 7/3 = 2.3333333, only worry about the 2. Now multiply this number by the divisor (3) and subtract the resulting number from the original dividend.
so 2*3 = 6, and 7 - 6 = 1, thus 1 is 7mod3
Calculate x/y (your actual numbers here), and press a b/c key, which is 3rd one below Shift key.
Simply just divide the numbers, it gives yuh the decimal format and even the numerical format. using S<->D
For example: 11/3 gives you 3.666667 and 3 2/3 (Swap using S<->D).
Here the '2' from 2/3 is your mod value.
Similarly 18/6 gives you 14.833333 and 14 5/6 (Swap using S<->D).
Here the '5' from 5/6 is your mod value.