The variable y may take a value which is in a defined range:
3<y<5
The value of y should be determined introducing a constraint like
|x-y|=min
x is given and should scan a larger range like:
x:=-1000:1:1000
How do I find the exact y-value with a given x?
The results that I consider is like:
x y
-1000 3
. 3
. 3
2.9 3
3 3
3.1 3.1
4 4
5 5
6 5
7 5
. 5
. 5
1000 5
Which means I want to allow a larger "error" but between 3 and 5 it should solve with a very smaller error so that I can resolve this area fine as possible.
What would be the best way to implement something like this in Matlab? Without "IF"-condition and if possible, symbolically. But also numerical alternatives would be interesting.
Based on your comment and example I think you are simply looking for this:
x = -10:0.1:10 %Suppose this is your x
y = max(min(x,5),3) %Force it between 3 and 5 by rounding up or down respectively
Related
My understanding of programming languages is limited, I have started 'coding' with matlab and wanted to transfer a simulation of mine from matlab to julia because there is no licensing fee. What I would like to know is in MATLAB I can auto populate array without ever initializing an array, while I know it is an inefficient way to it, I would like to know if there is similar way to do it on Julia.
Ex in MATLAB
for a in 1:10
x(a)=a;
end
will give me an array x = [ 1 2 3 4 5 6 7 8 9 10], is there a similar way to do that in julia? The way I have been doing it is declaring an empty array using Float64[] and append to it but it doesn't work the same way if the array is multidimensional.
my implementation in Julia for a 1D array
x = Float64[]
for a in 1:10
append!(x,a)
end
Any help regarding this will be greatly appreciated. Thank you!!!
MATLAB explicitly warns against the way you write this code and advises to preallocate for speed.
Julia, OTOH, cares more about performance and prohibits such pattern from the beginning. At the same time, it allows for more convenient and fast alternatives to do the same task.
julia> x = [a for a = 1:10] # OR x = [1:10;]
10-element Vector{Int64}:
1
2
3
4
5
6
7
8
9
10
and for 2D arrays,
julia> x = [i+j-1 for i = 1:5, j = 1:5] # OR x = (1:5) .+ (1:5)' .- 1
5×5 Matrix{Int64}:
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
And you have other convenience functions, like zeros(m,n), fill(val, dims), rand(Type, dims), etc., that you can use to initialize the arrays however you want.
Although #AboAmmar is correct that this generally isn't a good pattern, your code works if you use push! instead of append!. push! adds an element to a vector. append appends 2 vectors.
I am familiar with the command:
(-5:0.1:5)
which creates a vector of values, equally spaced by 0.1, from -5 to 5.
However, is there a way to produce a vector of values, equally spaced from -5 to 5 such that there are, say, 100 values in the vector.
(-5:0.1:5) gives a vector with 101 values, however, is there a way to get a vector of 100 values without manually calculating the step size?
Yes there is. Use function linspace. See documentation here
linspace(1,10,10)
ans =
1 2 3 4 5 6 7 8 9 10
also the question is a duplicate of this question
I am trying to generate all combination of 2 elements in a given range of numbers. I am using 'combnk' function as follows.
combnk(1:4,2)
ans =
3 4
2 4
2 3
1 4
1 3
1 2
combnk(1:6,2)
ans =
1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
4 5
4 6
5 6
The order of combinations returned appears to change. I need to know the order in advance for my program to work properly.
Is there any solution to make sure I get the combinations in a consistent order?
Also, why is MATLAB showing this strange behavior?
The only solution I can think of so far is to first check the 1st entry of the result matrix and flip it up side down using 'flipud' function.
Update: By a little bit of experimenting I noticed the reverse order occurs only when the length of the set of numbers is less than 6. This is why combnk(1:6,2) produce the 'correct' order. Where as combnk(1:5,2) produce the results backwards. This is still big problem.
You could try nchoosek instead of combnk. I don't have the matlab statistics toolbox (only octave), so I don't know if nchoosek has any significant disadvanvatages.
This will solve the ordering issue:
a=combnk(1:4,2);
[~,idx]=sortrows(a);
aNew=a(idx,:);
I don't know why MATLAB is showing this behavior.
I'm using the CLUSTERGRAM object from the Bioinformatics Toolbox (ver 3.7).
MATLAB version R2011a.
I'd like to get permutation vectors for row and columns for clustergram, as I can do with dendrogram function:
x = magic(10);
>> [~,~,permrows] = dendrogram(linkage(x,'average','euc'))
permrows =
9 10 6 7 8 1 2 4 5 3
>> [~,~,permcols] = dendrogram(linkage(x','average','euc'))
permcols =
6 7 8 9 2 1 3 4 5 10
I found that the clustering is not the same from clustergram and dendrogram, most probably due to optimal leaf ordering calculation (I don't want to disable it).
For example, for clustergram from:
clustergram(x)
('average' and 'eucledian' are default methods for clustergram)
the vectors (as on the figure attached) should be:
permrows = [1 2 4 5 3 10 9 6 7 8];
permcols = [1 2 8 9 6 7 10 5 4 3];
So, how to get those vectors programmatically? Anybody well familiar with this object?
Do anyone can suggest a good alternative? I know I can create a similar figure combining imagesc and dendrogram functions, but leaf ordering is much better (optimal) in clustergram, than in dendrogram.
From looking at the documentation, I guess that get(gco,'ColumnLabels') and get(gco,'RowLabels'), where gco is the clustergram object, should give you the reordered labels. Note that the corresponding set-methods take in the labels in original order and internally reorders them.
Consequently, if you have used custom labels (set(gco,'RowLabels',originalLabels))
[~,permrows] = ismember(get(gco,'RowLabels'),originalLabels)
should return the row permutation.
This is a question about Matlab/Octave.
I am seeing some results of the medfilt1(1D Median filter command in matlab) computation by which I am confused.
EDIT:Sorry forgot to mention:I am using Octave for Windows 3.2.4. This is where i see this behavior.
Please see the questions below, and point if I am missing something.
1] I have a 1D data array b=[ 3 5 -8 6 0];
out=medfilt1(b,3);
I expected the output to be [3 3 5 0 0] but it is showing the output as [4 3 5 0 3]
How come? What is wrong here?
FYI-Help says it pads the data at boundaries by 0(zero).
2] How does medfilt2(2D median filter command in matlab) work.
Help says "Each output pixel contains the median value in the m-by-n neighborhood around the corresponding pixel in the input image".
For m=3,n=3, So does it calculate a 3x3 matrix MAT for each of input pixels placed at its center and do median(median(MAT)) to compute its median value in the m-by-n neighbourhood?
Any pointers will help.
thank you. -AD
I was not able to replicate your error with Matlab 7.11.0, but from the information in your question it seems like your version of medfilt1 does not differentiate between an odd or even n.
When finding the median in a vector of even length, one usually take the mean of the two median values,
median([1 3 4 5]) = (3+4)/2 = 3.5
This seems to be what happens in your case. Instead of treating n as odd, and setting the value to be 3, n is treated as even and your first out value is calculated to be
median([0 3 5]) = (3+5)/2 = 4
and so on.. EDIT: This only seems to happen in the endpoints, which suggest that the padding with zeros is not properly working in your Octave code.
For your second question, you are almost right, it calculates a 3x3 matrix in each center, but it does not do median(median(MAT)), but median(MAT(:)). There is a difference!
A = [1 2 3
14 5 33
11 7 13];
median(median(A)) = 11
median(A(:)) = 7