A program to run a Schmid-Leiman transformation using SPSS's Matrix language was published in 2005 by Woolf & Preising in Behavior Research Methods volume 37, pages 48 to 58). It is probably not important for you to know what a Schmid-Leiman transformation is, but I'll explain in comments if you feel it is necessary.
In modifying the program for my own data, I'm getting an error I can't figure out:
Error # 12302 in column 12. Text: ,
Syntax error.
Execution of this command stops.
Error in RIGHT HAND SIDE of COMPUTE command.
The MATRIX statement skipped.
Here is the beginning of the code. The error is showing as coming in Line 6:
* Encoding: UTF-8.
* Schmid-Leiman Solution for 2 level higher-order Factor analysis.
Matrix.
* ENTER YOUR SPECIFICATIONS HERE.
* Enter first-order pattern matrix.
Compute F1={.461, .253, -.058, -.069;
.241, .600, .143, .033;
.582, .047, -.077, -.125;
.327, .297, -.120, -.166;
.176, .448, -.240, -.099;
.680, .069, -.036, -.138;
.415, .228, -.091, -.153;
.
.
.
.390, .205, .002, -.098;
.164, .369, -.170, -.047
}.
As shown above, the text generating the error is shown as a comma (,), but the actual text (following the COMPUTE statement) in column 12 is an open bracket ({). So I have no idea what is going on. Can someone help?
For reference, the original code as proposed by Woolf & Preising (2005) is found here;
The Woolf & Preising article is found here
PS: The sample program given in the link above does run on my copy of SPSS. Here's the beginning of that code:
* Schmid-Leiman Solution for 2 level higher-order Factor analysis.
Matrix.
* ENTER YOUR SPECIFICATIONS HERE.
* Enter first-order pattern matrix.
Compute F1={0.099, 0.5647, -0.1521;
0.0124, 0.9419, -0.1535;
-0.1501, 0.6177, 0.4218;
0.7441, -0.0882, 0.1425;
0.6241, 0.2793, -0.1137;
0.8693, -0.0331, 0.0289;
-0.0154, -0.2706, 0.6262;
-0.0914, 0.0995, 0.7216;
0.1502, 0.0835, 0.398}.
Related
a=1
% Construct the trasfer function
num=[a 1 3]
den=[1 2 10]
G=tf(num,den)
% Impulse response
impulse(G)
% Step response
step(G)
When I click on 'run' this error appears "error: Order numerator >= order denominator"
If you follow the error on the terminal, it suggests that line 95 in imp_invar.m of the control package is to blame. (if you don't know where this was installed, you can find out by typing pkg list in your terminal)
If you convert this error to a warning, the code continues. Obviously you do so at your own risk. I would make a backup of the original .m file just in case.
Note that the same code run on matlab does not issue any error or warning (which is odd in itself, given the stark note about invalid impulse invariance in this scenario from octave ... there is a reference quoted inside imp_invar.m if you're interested.)
Screenshot
>>> boxes = tf.random_normal([ 5])
>>> with s.as_default():
... s.run(boxes)
... s.run(keras.backend.argmax(boxes,axis=0))
... s.run(tf.reduce_max(boxes,axis=0))
...
array([ 0.37312034, -0.97431135, 0.44504794, 0.35789603, 1.2461706 ],
dtype=float32)
3
0.856236
.
Why am I getting 0.8564. I expect the value to be 1.2461. since 1.2461 is big.right?
I am getting correct answer if i use tf.constant.
But I am not getting correct answer while using radom_normal
Each time a new boxes is regenerated when you run s.run() with radom_normal. So your three results are different. If you want to get consistent results, you should only run s.run() once.
result = s.run([boxes,keras.backend.argmax(boxes,axis=0),tf.reduce_sum(boxes,axis=0)])
print(result[0])
print(result[1])
print(result[2])
#print
[ 0.69957364 1.3192859 -0.6662426 -0.5895929 0.22300807]
1
0.9860319
In addition, the code should be given in text format rather than picture format.
TensorFlow is different from numpy because TF only uses symbolic operations. That means when you instantiate the random_normal, you don't get numeric values, but a symbolic normal distribution, so each time you evaluate it, you get different numbers.
Each time you operate with this distribution, with any other operation, you are getting different numbers, and that explains the results you see.
I would like to make matrix calculator, but I struggle a little bit, how to make an input of the program. I have commands that user can use in calculator. Some takes 1 argument, 2 arguments or 3 arguments. I was inspired by program on this website http://www.ivank.net/blogspot/matrix_pascal/matrices.pas
But I don't really understand, how the input is made. Program from the website use parse, split procedures, but I don't know, how does it work. Does it exists some website, where it is good explained (Parse in Pascal)? I would like to really understand it.
This is, how it should looks like:
command: sum X Y
command: multiply X
command: transpose X
In the sample which inspired you, all the calculation is realized by the 'procedure parse(command:String);'.
The first step consists to extract the command and all parameters by:
com := Split(command, ' ');
In your case, you will obtain for 'command: sum X Y':
Length(com) = 3
com[0] = 'sum'; com[1] = 'X'; com[2] = 'Y';
But, be carefull, the 'X' and 'Y' parameters shall not have characters between numbers.
DISCLAIMER: This question is only for those who have access to the econometrics toolbox in Matlab.
The Situation: I would like to use Matlab to simulate N observations from an ARIMA(p, d, q) model using the econometrics toolbox. What's the difficulty? I would like the innovations to be simulated with deterministic, time-varying variance.
Question 1) Can I do this using the in-built matlab simulate function without altering it myself? As near as I can tell, this is not possible. From my reading of the docs, the innovations can either be specified to have a constant variance (ie same variance for each innovation), or be specified to be stochastically time-varying (eg a GARCH model), but they cannot be deterministically time-varying, where I, the user, choose their values (except in the trivial constant case).
Question 2) If the answer to question 1 is "No", then does anyone see any reason why I can't edit the simulate function from the econometrics toolbox as follows:
a) Alter the preamble such that the function won't throw an error if the Variance field in the input model is set to a numeric vector instead of a numeric scalar.
b) Alter line 310 of simulate from:
E(:,(maxPQ + 1:end)) = Z * sqrt(variance);
to
E(:,(maxPQ + 1:end)) = (ones(NumPath, 1) * sqrt(variance)) .* Z;
where NumPath is the number of paths to be simulated, and it can be assumed that I've included an error trap to ensure that the (input) deterministic variance path stored in variance is of the right length (ie equal to the number of observations to be simulated per path).
Any help would be most appreciated. Apologies if the question seems basic, I just haven't ever edited one of Mathwork's own functions before and didn't want to do something foolish.
UPDATE (2012-10-18): I'm confident that the code edit I've suggested above is valid, and I'm mostly confident that it won't break anything else. However it turns out that implementing the solution is not trivial due to file permissions. I'm currently talking with Mathworks about the best way to achieve my goal. I'll post the results here once I have them.
It's been a week and a half with no answer, so I think I'm probably okay to post my own answer at this point.
In response to my question 1), no, I have not found anyway to do this with the built-in matlab functions.
In response to my question 2), yes, what I have posted will work. However, it was a little more involved than I imagined due to matlab file permissions. Here is a step-by-step guide:
i) Somewhere in your matlab path, create the directory #arima_Custom.
ii) In the command window, type edit arima. Copy the text of this file into a new m file and save it in the directory #arima_Custom with the filename arima_Custom.m.
iii) Locate the econometrics toolbox on your machine. Once found, look for the directory #arima in the toolbox. This directory will probably be located (on a Linux machine) at something like $MATLAB_ROOT/toolbox/econ/econ/#arima (on my machine, $MATLAB_ROOT is at /usr/local/Matlab/R2012b). Copy the contents of #arima to #arima_Custom, except do NOT copy the file arima.m.
iv) Open arima_Custom for editing, ie edit arima_Custom. In this file change line 1 from:
classdef (Sealed) arima < internal.econ.LagIndexableTimeSeries
to
classdef (Sealed) arima_Custom < internal.econ.LagIndexableTimeSeries
Next, change line 406 from:
function OBJ = arima(varargin)
to
function OBJ = arima_Custom(varargin)
Now, change line 993 from:
if isa(OBJ.Variance, 'double') && (OBJ.Variance <= 0)
to
if isa(OBJ.Variance, 'double') && (sum(OBJ.Variance <= 0) > 0)
v) Open the simulate.m located in #arima_Custom for editing (we copied it there in step iii). It is probably best to open this file by navigating to it manually in the Current Folder window, to ensure the correct simulate.m is opened. In this file, alter line 310 from:
E(:,(maxPQ + 1:end)) = Z * sqrt(variance);
to
%Check that the input variance is of the right length (if it isn't scalar)
if isscalar(variance) == 0
if size(variance, 2) ~= 1
error('Deterministic variance must be a column vector');
end
if size(variance, 1) ~= numObs
error('Deterministic variance vector is incorrect length relative to number of observations');
end
else
variance = variance(ones(numObs, 1));
end
%Scale innovations using deterministic variance
E(:,(maxPQ + 1:end)) = sqrt(ones(numPaths, 1) * variance') .* Z;
And we're done!
You should now be able to simulate with deterministically time-varying variance using the arima_Custom class, for example (for an ARIMA(0,1,0)):
ARIMAModel = arima_Custom('D', 1, 'Variance', ScalarVariance, 'Constant', 0);
ARIMAModel.Variance = TimeVaryingVarianceVector;
[X, e, VarianceVector] = simulate(ARIMAModel, NumObs, 'numPaths', NumPaths);
Further, you should also still be able to use matlab's original arima class, since we didn't alter it.
x(n) is given
need x(-n+3)
so to solve it:
first advance the x(n) signal by 3 units(time)
then fold it, or make a reflection of it
are the above steps correct or is the following correct
first fold the x(n) signal
then advance the signal by 3 units
?
Yes, this is a common source of confusion when learning about signals. Here's what I usually do.
Let y[n] = x[-n+3]. Because of -n, y[n] is obviously a time-reversed version of x[n]. But the question about the shift remains.
Notice that y[3] = x[0]. Therefore, y[n] is achieved by first reflecting x[n] about n=0 and then delaying the reflected signal by 3.
For example, let x[n] be the unit step function u[n]. Draw x[n], then draw y[n].
Actually here is what I do:
Let
x(n) = {1,-1,2,4,-3,0,6,-3,-1,2,7,9,-7,5}
^
Suppose origin or n=0 is 6. Note that the ^ symbol indicates the origin. First, we find the folder sequence of x(-n) from x(n). So first we fold or we can say reverse the form of x(n), we get,
The folder sequence of x(-n) from x(n) is
x(-n) = {5,-7,9,7,2,-1,-3,6,0,-3,4,2,-1,1}
^
then shift the sequence of x(-n) towards right hand side by 3 units, we will get
x(-n+3) = {5,-7,9,7,2-1,-3,6,0,-3,4,2,-1,1}
^
Now, the sample 4 is at the origin.
Above steps are correct.
The following steps can be corrected too if these are followed like:
first fold the x(n) signal
then delay the signal by 3 units this will yield x(-n+3).