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I have a code that defines a function, then I try to use the variables I defined within that function in another expression. When I do this, I get an error that is:
Undefined function or variable 'phi'.
I'm not sure why phi is undefined, since I have it in the if/else statement.
It might be better to explain with my (shortened) code:
global I11 I22 I33 Mx My Mz w10 w20 w30 eps10 eps20 eps30 eps40...
C110 C120 C130 C210 C220 C230 C310 C320 C330 IC K0
IC = [w10 w20 w30...
eps10 eps20 eps30 eps40...
C110 C120 C130 C210 C220 C230 C310 C320 C330];
opts = odeset('RelTol', 1*10^(-10),'AbsTol', 1*10^(-10));
[t, y] = ode45(#(t,y) DynEqn2(t,y,I11,I22,I33,Mx,My,Mz), [ti tf], IC, opts);
N = sqrt(sum(y(:,4:7).^2,2));
kap = acosd(1-2*y(:,5).^2-2*y(:,7).^2);
phi1 = acosd((2.*(y(:,4).*y(:,5)+y(:,6).*y(:,7)))/sind(kap));
phi2 = asind((2.*(y(:,6).*y(:,4)-y(:,5).*y(:,7)))/sind(kap));
if phi1==phi2
phi = phi1;
elseif phi1==180-phi2
phi = phi1;
elseif -phi1==phi2
phi = -phi1;
elseif -phi1==180-phi2
phi = -phi1;
else
disp('Something is wrong with phi')
end
figure (1)
plot(t,phi)
figure (2)
plot(t,kap)
function soln = DynEqn2(t,y,I11,I22,I33,Mx,My,Mz)
w1 = y(1);
w2 = y(2);
w3 = y(3);
eps1 = y(4);
eps2 = y(5);
eps3 = y(6);
eps4 = y(7);
C11 = y(8);
C12 = y(9);
C13 = y(10);
C21 = y(11);
C22 = y(12);
C23 = y(13);
C31 = y(14);
C32 = y(15);
C33 = y(16);
w1_dot = (Mx - w2*w3*(I33-I22))/I11;
w2_dot = (My - w1*w3*(I11-I33))/I22;
w3_dot = (Mz - w1*w2*(I22-I11))/I33;
eps1_dot = .5*(w1*eps4-w2*eps3+w3*eps2);
eps2_dot = .5*(w1*eps3+w2*eps4-w3*eps1);
eps3_dot = .5*(-w1*eps2+w2*eps1+w3*eps4);
eps4_dot = -.5*(w1*eps1+w2*eps2+w3*eps3);
C11_dot = C12*w3-C13*w2;
C12_dot = C13*w1-C11*w3;
C13_dot = C11*w2-C12*w1;
C21_dot = C22*w3-C23*w2;
C22_dot = C23*w1-C21*w3;
C23_dot = C21*w2-C22*w1;
C31_dot = C32*w3-C33*w2;
C32_dot = C33*w1-C31*w3;
C33_dot = C31*w2-C32*w1;
soln = [w1_dot; w2_dot; w3_dot; ...
eps1_dot; eps2_dot; eps3_dot; eps4_dot; ...
C11_dot; C12_dot; C13_dot; C21_dot; C22_dot; C23_dot; C31_dot; C32_dot; C33_dot];
end
My lines where I calculate phi1, phi2, and then the if/else statement to find phi, are what I am struggling with.
I made sure that the variables defined in the function work, so for example, in the command window I typed in 'y(:,4)' and got the correct output. But whenever I try to use this within the functions i.e. 'phi1', it repeatedly outputs an incorrect value of '90.0000' until I stop it.
Where I define the 'N' variable, it is something similar, yet that one works without errors.
Does anyone have any ideas how to amend this issue?
Any help is appreciated, thanks.
Edit: The complete error message is as follows:
Undefined function or variable 'phi'.
Error in HW6_Q1 (line 85)
plot(t,phi)
I figured out my solution with the help from a colleague not on Stack Overflow.
I forgot the ./, which turned my phi into a matrix, rather than a vector which is what I wanted from it.
I have a function which I have defined as follows, you can see that it clearly requires 7 arguments;
def calc_z(w,S,var,a1,a2,yt1,yt2):
mu = w*S
sigma = mt.sqrt(var)
z = np.random.normal(mu,sigma)
u = [a1,a2,z]
yt = [yt1,yt2,1]
thetaset = np.random.rand(len(u))
m = [i for i in range(len(u))]
max_iter = 30
#Calculate E-step
for i in range(max_iter):
print 'Iteration:', i
print 'z:', z
print 'thetaset', thetaset
devLz = eq6(var,w,S,z,yt,u,thetaset,m)
dev2Lz2 = eq9(var,thetaset,u)
#Calculate M-Step
z = z - (devLz / dev2Lz2)
w = lambdaw * z
for i in range(len(thetaset)):
devLTheta = eq7(yt,u,thetaset,lambdatheta)
dev2LTheta2 = eq10(thetaset,u,lambdatheta)
thetaset = thetaset - (devLTheta / dev2LTheta2)
return z
I am using pyspark so I convert this to a udf
calc_z_udf = udf(calc_z,FloatType())
and then run it as follows (where I am clearly passing in 7 arguments - Or am I going mad!?);
data = data.withColumn('z', calc_z_udf(data['w'],data['Org_Depth_Diff_S'],data['var'],data['proximity_rank_a1'],data['cotravel_count_a2'],data['cotravel_yt1'],data['proximity_yt2']))
When I run this however I am getting an error which states:
TypeError: calc_z() takes exactly 7 arguments (6 given)
Could anyone help me with why this might be as it is clear that when I am running the function I am infact passing in 7 arguments and not 6 as the error states?
I am not sure its is the reason no need to send column objects you can just pass strings:
data = data.withColumn('z', calc_z_udf('w', 'Org_Depth_Diff_S','var', 'proximity_rank_a1', 'cotravel_count_a2', 'cotravel_yt1', 'proximity_yt2'))
I want to remove the directory part from a cell array of strings with filenames. Of course one way would be to loop over the cell arrray and use fileparts but I have over 1e5 files and speed really matters.
My current approach is:
fns = {"/usr/local/foo.lib", "~/baz.m", "home/rms/eula.txt", "bar.m"}
filenames = cellfun (#(fn, s) fn(s+1:end), fns,
num2cell (rindex (fns, filesep())),
"UniformOutput", false)
which gives the desired output:
fns =
{
[1,1] = /usr/local/foo.lib
[1,2] = ~/baz.m
[1,3] = home/rms/eula.txt
[1,4] = bar.m
}
filenames =
{
[1,1] = foo.lib
[1,2] = baz.m
[1,3] = eula.txt
[1,4] = bar.m
}
and takes approx 2e-5s per file. Is there a better (faster, more readable) way to do this?
EDIT I've added Sardars solution and my previous attempt with regex and some benchmark results:
fns = {"/usr/local/foo.lib", "~/baz.m", "home/rms/eula.txt", "bar.m"};
fns = repmat (fns, 1, 1e4);
tic
f1 = cellfun (#(fn, s) fn(s+1:end), fns,
num2cell (rindex (fns, "/")),
"UniformOutput", false);
toc
tic
[~, ~, ~, M] = regexp (fns, "[^\/]+$", "lineanchors");
f2 = cell2mat (M);
toc
tic
## Asnwer from Sardar Usama
f3 = regexprep(fns, '.*/', '');
toc
assert (f1, f2)
assert (f1, f3)
which gives
Elapsed time is 0.729995 seconds. (Original code with cellfun)
Elapsed time is 0.67545 seconds. (using regexp)
Elapsed time is 0.230487 seconds. (using regexprep)
Use regexprep to search the strings till the last / and replace the occurrences with an empty string.
filenames = regexprep(fns, '.*/', '');
I have a struct in Octave that contains some big arrays.
I'd like to know the names of the fields in this struct without having to look at all these big arrays.
For instance, if I have:
x.a=1;
x.b=rand(3);
x.c=1;
The obvious way to take a gander at the structure is as follows:
octave:12> x
x =
scalar structure containing the fields:
a = 1
b =
0.7195967 0.9026158 0.8946427
0.4647287 0.9561791 0.5932929
0.3013618 0.2243270 0.5308220
c = 1
In Matlab, this would appear as the more succinct:
>> x
x =
a: 1
b: [3x3 double]
c: 1
How can I see the fields/field names without seeing all these big arrays?
Is there a way to display a succinct overview (like Matlab's) inside Octave?
Thanks!
You might want to take a look at Basic Usage & Examples. There's several functions mentioned that sound like they'll control the displaying in the terminal.
struct_levels_to_print
print_struct_array_contents
These two functions sound like they're do what you want. I tried both and couldn't get the 2nd one to work. The 1st function changed the terminal output like so:
octave:1> x.a=1;
octave:2> x.b=rand(3);
octave:3> x.c=1;
octave:4> struct_levels_to_print
ans = 2
octave:5> x
x =
{
a = 1
b =
0.153420 0.587895 0.290646
0.050167 0.381663 0.330054
0.026161 0.036876 0.818034
c = 1
}
octave:6> struct_levels_to_print(0)
octave:7> x
x =
{
1x1 struct array containing the fields:
a: 1x1 scalar
b: 3x3 matrix
c: 1x1 scalar
}
I'm running a older version of Octave.
octave:8> version
ans = 3.2.4
If I get a chance I'll check that other function, print_struct_array_contents, to see if it does what you want. Octave 3.6.2 looks to be the latest version as of 11/2012.
Use fieldnames ()
octave:33> x.a = 1;
octave:34> x.b = rand(3);
octave:35> x.c = 1;
octave:36> fieldnames (x)
ans =
{
[1,1] = a
[2,1] = b
[3,1] = c
}
Or you you want it to be recursive, add the following to your .octaverc file (you may want to adjust it to your preferences)
function displayfields (x, indent = "")
if (isempty (indent))
printf ("%s: ", inputname (1))
endif
if (isstruct (x))
printf ("structure containing the fields:\n");
indent = [indent " "];
nn = fieldnames (x);
for ii = 1:numel(nn)
if (isstruct (x.(nn{ii})))
printf ("%s %s: ", indent, nn{ii});
displayfields (x.(nn{ii}), indent)
else
printf ("%s %s\n", indent, nn{ii})
endif
endfor
else
display ("not a structure");
endif
endfunction
You can then use it in the following way:
octave> x.a=1;
octave> x.b=rand(3);
octave> x.c.stuff = {2, 3, 45};
octave> x.c.stuff2 = {"some", "other"};
octave> x.d=1;
octave> displayfields (x)
x: structure containing the fields:
a
b
c: structure containing the fields:
stuff
stuff2
d
in Octave, version 4.0.0 configured for "x86_64-pc-linux-gnu".(Ubuntu 16.04)
I did this on the command line:
print_struct_array_contents(true)
sampleFactorList % example, my nested structure array
Output: (shortened):
sampleFactorList =
scalar structure containing the fields:
sampleFactorList =
1x6 struct array containing the fields:
var =
{
[1,1] = 1
[1,2] =
2 1 3
}
card =
{
[1,1] = 3
[1,2] =
3 3 3
}
To disable/get back to the old behaviour
print_struct_array_contents(false)
sampleFactorList
sampleFactorList =
scalar structure containing the fields:
sampleFactorList =
1x6 struct array containing the fields:
var
card
val
I've put this print_struct_array_contents(true) also into the .octaverc file.
I'm trying to plot a function that contains a definite integral. My code uses all anonymous functions. When I run the file, it gives me an error. My code is below:
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
Q = #(y) integral(#(x)Q_integral(x,y),-3000,0);
P1 = #(y) -1./(1i.*(gamma_sp + cap_gamma)).*(1./(y + 2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1./y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))));
P2 = #(y) conj(P1(y));
P = #(y) P1(y) - P2(y);
sig = #(y) abs(P(y)).^2;
rng = -2000:0.05:1000;
plot(rng,sig(rng))
It seems to me that when the program runs the plot command, it should put each value of rng into sig(y), and that value will be used as the y value in A_integral and Q_integral. However, matlab throws an error when I try to run the program.
Error using -
Matrix dimensions must agree.
Error in #(x,y)G(x)./(y-x+1i*gamma_t)
Error in #(x)A_integral(x,y)
Error in integralCalc/iterateScalarValued (line 314)
fx = FUN(t);
Error in integralCalc/vadapt (line 133)
[q,errbnd] = iterateScalarValued(u,tinterval,pathlen);
Error in integralCalc (line 76)
[q,errbnd] = vadapt(#AtoBInvTransform,interval);
Error in integral (line 89)
Q = integralCalc(fun,a,b,opstruct);
Error in #(y)integral(#(x)A_integral(x,y),-1000,1000)
Error in
#(y)-1./(1i.*(gamma_sp+cap_gamma)).*(1./(y+2.*1i.*gamma_t)*(A(y)-conj(A(0)))-1. /y.*(A(y)-A(0))+cap_gamma./gamma_sp.*Q(y).*(A(0)-conj(A(0))))
Error in #(y)P1(y)-P2(y)
Error in #(y)abs(P(y)).^2
Error in fwm_spec_diff_paper_eqn (line 26)
plot(rng,sig(rng))
Any ideas about what I'm doing wrong?
You have
>> rng = -2000:0.05:1000;
>> numel(rng)
ans =
60001
all 60001 elements get passed down to
A = #(y) integral(#(x)A_integral(x,y),-1000,1000);
which calls
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
(similar for Q). The thing is, integral is an adaptive quadrature method, meaning (roughly) that the amount of x's it will insert into A_integral varies with how A_integral behaves at certain x.
Therefore, the amount of elements in y will generally be different from the elements in x in the call to A_integral. This is why y-x +1i*gamma_t fails.
Considering the complexity of what you're trying to do, I think it is best to re-define all anonymous functions as proper functions, and integrate a few of them into single functions. Look into the documentation of bsxfun to see if that can help (e.g., bsxfun(#minus, y.', x) instead of y-x could perhaps fix a few of these issues), otherwise, vectorize only in x and loop over y.
Thanks Rody, that made sense to me. I keep trying to use matlab like mathematica and I forget how matlab does things. I modified the code a bit, and it produces the right result. The integrals are evaluated very roughly, but it should be easy to fix that. I've posted my modified code below.
%%% List of Parameters %%%
gamma_sp = 1;
cap_gamma = 15;
gamma_ph = 0;
omega_0 = -750;
d_omega_0 = 400;
omega_inh = 100;
d_omega_inh = 1000;
%%% Formulae %%%
gamma_t = gamma_sp/2 + cap_gamma/2 + gamma_ph;
G = #(x) exp(-(x-omega_inh).^2./(2*d_omega_inh.^2))./(sqrt(2*pi)*d_omega_inh);
F = #(x) exp(-(x-omega_0).^2./(2*d_omega_0.^2))./(sqrt(2*pi)*d_omega_0);
A_integral = #(x,y) G(x)./(y - x + 1i*gamma_t);
Q_integral = #(x,y) F(x)./(y - x + 1i*gamma_t);
w = -2000:0.05:1000;
sigplot = zeros(size(w));
P1plot = zeros(size(w));
P2plot = zeros(size(w));
Pplot = zeros(size(w));
aInt_range = -1000:0.1:1200;
qInt_range = -2000:0.1:100;
A_0 = sum(A_integral(aInt_range,0).*0.1);
for k=1:size(w,2)
P1plot(k) = -1./(1i*(gamma_sp + cap_gamma)).*(1./(w(k)+2.*1i.*gamma_t).*(sum(A_integral(aInt_range,w(k)).*0.1)-conj(A_0))-1./w(k).*(sum(A_integral(aInt_range,w(k)).*0.1)-A_0)+cap_gamma./gamma_sp.*sum(Q_integral(qInt_range,w(k)).*0.1).*(A_0-conj(A_0)));
P2plot(k) = conj(P1plot(k));
Pplot(k) = P1plot(k) - P2plot(k);
sigplot(k) = abs(Pplot(k)).^2;
end
plot(w,sigplot)