A quick question because i fear there may already be an answer (although i cant find it)
i am getting the error: Matrix dimensions must agree.
because i am useing '<'
now with all the other operators there is a way around this either by putting '.' infront or by using a different formula. So what do people do about the less than operator????
i don't see why the greater than or equal to (>=) works but yet less than does not!?
am i being stupid and missed something really obvious??
code snippet
matrix 1 represents an array of 16 numbers
matrix 2 can represents anywhere between 10 and 20 numbers
idx = (matrix2 >= matrix1 * 0.1 & matrix2 < matrix1 * 1.5);
any help guidance or advice on the topic would be much appreciated! thank you!
EDIT
i know the matrices are different sizes but is there a way to use less then with different size arrays? as im not bothered about the size of the array but the numbers within
If you want to compare parts of matrices, like M(1:3,10:12)>A(5:7,1:3), you, probably, have to use the function squeeze():
squeeze(M(1:3,10:12))>squeeze(A(5:7,1:3))
This function remotes singleton dimensions and everything works fine after.
Related
I am working on a code in Least Square Non Negative solution recovery context on Matlab, and I need (with no more details because it's not that important for this question) to know the number of non zero elements in my matrices and arrays.
The function NNZ on matlab does exactly what I want, but it happens that I need more information about what Matlab thinks of a "zero element", it could be 0 itself, or the numerical zero like 1e-16 or less.
Does anybody has this information about the NNZ function, cause I couldn't get the original script
Thanks.
PS : I am not an expert on Matlab, so accept my apologies if it's a really simple task.
I tried "open nnz", on Matlab but I only get a small script of commented code lines...
Since nnz counts everything that isn't an exact zero (i.e. 1e-100 is non-zero), you just have to apply a relational operator to your data first to find how many values exceed some tolerance around zero. For a matrix A:
n = nnz(abs(A) > 1e-16);
Also, this discussion of floating-point comparison might be of interest to you.
You can add in a tolerance by doing something like:
nnz(abs(myarray)>tol);
This will create a binary array that is 1 when abs(myarray)>tol and 0 otherwise and then count the number of non-zero entries.
this is a very basic problem but I didn't find any hints on it. Let's say I have a 2x4 matrix and I want to reduce the dimension of the matrix to only these columns that are in the sum larger than 1:
A=rand(2,4)
ind = sum(A,1).>1
That gives me an indicator of the columns I want to retain. Naively one would assume that I can do that:
A[:,ind]
which doesn't work as ind is a BitArray and only for Bool Arrays this is allowed, i.e., the following works
A[:,[true,true,false,true]]
in return, the following does work:
A[A.>0.5]
But it returns a vector of filtered elements.
What is the logic behind this and how do I solve my problem?
As noted in the comments, this is fixed by using a version of Julia which is >=v0.4.
When I try to do the following command I get an error.
err = sqrt(mean((xi256-xc256).^2))
I am aware that the matrix sizes are different.
whos xi256 xc256` gives:
Name Size Bytes Class Attributes
xc256 27x1 216 double
xi256 513x1 4104 double
I am supposed to negate find the difference of these two matrices. In fact the command given at the top was in the course notes and the course has been running for several years! I have tried googling ways to resolve this error to perform this subtraction regardless but have found no solution. Maybe one of the matrices can be scaled to match the dimensions of the other? However, I have not been able to find any such functions that would let me do this.
I need to find the RMS error in a given set of data. xc256 was calculated through a numerical method, xi256 gives the true value.
Edit: I was able to use another set of results.
First check that xc256 is correctly computed and that you do not have a matrix transposition gone wrong or something like that. Computing the RMS between two vector of different sizes does not make sense, and padding or replicating will get you rid of the error, but is most probably not what you actually want.
There are just two situations that I can think of, I will list them here:
The line is wrong (not likely as it looks pretty normal, but make sure to check the book!)
The input of the line is wrong
Assuming it is in point 2, again there are two possibilities:
xi256 is of incorrect size (likelyhood of this depends on how you got it)
xc256 is of incorrect size
Assuming it is again point 2, there are yet again 2 likely possibilities:
xc256 should be a scalar
xc256 should be a vector with the same size as xi256
From here on there is insufficient information to continue, but check whether you accidentally made it 27 times as long, or 19 times too short somewhere. Just use some breakpoints throughout the code to see how the size develops.
I got the following error while working with a MATLAB program:
Error using - Matrix dimensions must agree
I noticed that the sizes of the matrices I'm trying to subtract from each other were:
firstMatrix --> 425x356
secondMatrix --> 426x356
How can I make them of equal size and go ahead and do my subtraction process?
I tried reshape, but the number of elements here seem to have to be equal.
Thanks.
I think both answers are missing the key point. Blithely subtracting two arrays of different size forgets that those arrays are NOT just numbers. The numbers must mean something. Else, they are just meaningless.
As well, simply deleting a row from the beginning or end may well be wrong, or padding with zeros. Only you know what the numbers mean, and why those arrays are not the same size. So only you can decide what is the proper action.
It might be right to pad, delete, interpolate, do any of these things. Or you might realize there is a bug in your code that created these arrays.
Your matrices have a different number of elements, so there's no point using reshape here (since it maintains the total number of elements). You'll have to discard one of the lines in the larger matrix before doing the subtraction:
For instance, you can discard the last line:
firstMatrix - secondMatrix(1:end - 1, :)
or discard the first line:
firstMatrix - secondMatrix(2:end, :)
Alternatively, you can pad the smaller matrix with default values (e.g NaN or zeroes), as suggested in another answer.
You're missing a row in firstMatrix
So can try:
firstMatrix=[firstMatrix;zeros(1,356)];
This will add a row of zeros at end of firstMatrix making it of 426x356
I have made an array of doubles and when I want to use the find command to search for the indices of specific values in the array, this yields an empty matrix which is not what I want. I assume the problem lies in the precision of the values and/or decimal places that are not shown in the readout of the array.
command:
peaks=find(y1==0.8236)
array readout:
y1 =
Columns 1 through 11
0.2000 0.5280 0.8224 0.4820 0.8239 0.4787 0.8235 0.4796 0.8236 0.4794 0.8236
Columns 12 through 20
0.4794 0.8236 0.4794 0.8236 0.4794 0.8236 0.4794 0.8236 0.4794
output:
peaks =
Empty matrix: 1-by-0
I tried using the command
format short
but I guess this only truncates the displayed values and not the actual values in the array.
How can I used the find command to give an array of indices?
By default, each element of a numerical matrix in Matlab is stored using floating point double precision. As you surmise in the question format short and format long merely alter the displayed format, rather than the actual format of the numbers.
So if y1 is created using something like y1 = rand(100, 1), and you want to find particular elements in y1 using find, you'll need to know the exact value of the element you're looking for to floating point double precision - which depending on your application is probably non-trivial. Certainly, peaks=find(y1==0.8236) will return the empty matrix if y1 only contains values like 0.823622378...
So, how to get around this problem? It depends on your application. One approach is to truncate all the values in y1 to a given precision that you want to work in. Funnily enough, a SO matlab question on exactly this topic attracted two good answers about 12 hours ago, see here for more.
If you do decide to go down this route, I would recommend something like this:
a = 1e-4 %# Define level of precision
y1Round = round((1/a) * y1); %# Round to appropriate precision, and leave y1 in integer form
Index = find(y1Round == SomeValue); %# Perform the find operation
Note that I use the find command with y1Round in integer form. This is because integers are stored exactly when using floating point double, so you won't need to worry about floating point precision.
An alternative approach to this problem would be to use find with some tolerance for error, for example:
Index = find(abs(y1 - SomeValue) < Tolerance);
Which path you choose is up to you. However, before adopting either of these approaches, I would have a good hard look at your application and see if it can be reformulated in some way such that you don't need to search for specific "real" numbers from among a set of "reals". That would be the most ideal outcome.
EDIT: The code advocated in the other two answers to this question is neater than my second approach - so I've altered it accordingly.
Testing for equality with floating-point numbers is almost always a bad idea. What you probably want to do is test to see which numbers are close enough to the target value:
peaks = find( abs( y - .8236 ) < .0001 );
The problem is indeed with the precision. The array that you see displayed is not the actual array, as the actual array has more digits for each of the numbers. Changing the format just changes the way in which the array is displayed, so it doesn't solve the problem.
You have two options, either modify the array or modify what you are looking for. It is probably better to modify what you are looking for, since then you are not changing the actual values.
So instead of looking for equality, you can look for proximity (so the difference between the number you are searching for and the number in the array is at most some small epsilon):
peaks = find( abs(y1-0.8236) < epsilon )
In general, when you are dealing with floats, always try to avoid exact comparisons and use some error thresholds, since the representation of these numbers is limited so they are often stored with small inaccuracies.