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Ilya on 30 Apr 2014
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Edited: Ilya on 2 May 2014
Accepted Answer: W. Owen Brimijoin
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I wonder if it's possible to pad a vector with intermediate points in order to make it more "smooth" in an elegant way (i.e. with no loops)
% Example:
a=[1.1 5.2 9.25 10];
a_desired=[1.1 2.125 3.15 4.175 5.2 5.2 6.55 7.9 9.25 9.25 9.625 10];
a_desired which was generated by:
a_desired=[linspace(a(1), a(2), 5), linspace(a(2), a(3), 4), linspace(a(3), a(4), 3)]
So, what I need is some kind of generalized linspace(), which works not only with begin and end of an interval, but allows the control of the number of points between the "node points" on the interval, where the "node points" are the begin, the end and some points in between.
In the above example, the number of entries in a is known in advance, so it's possible to use linspace() together with concatenation without any loops. But what if a is a generic vector generated by some calculation? Then I end up using loops:
a_desired=[];
for j=1:1:length(a)-1
a_desired=[a_desired, linspace(a(j), a(j+1), NUMBER_OF_PTS)];
end
which affects the performance (even if I preallocate a_desired, it still will be not very nice and not very fast).
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Cedric on 30 Apr 2014
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Edited: Cedric on 30 Apr 2014
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The example is not optimal; you'll get the answer a_desired = min(a):max(a). Let's take another one.. if you have
a = [1, 9, 5, 7] ;
is it
1 2 3 4 5 6 7 8 9 8 7 6 5 6 7
that you want?
Ilya on 30 Apr 2014
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Edited: Ilya on 30 Apr 2014
Sorry, my bad. The question was indeed not very clear, and I've reformulated it. But basically that's it.
Cedric on 30 Apr 2014
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In your first example, the number of points varies with the length of the interval, and not in the example with the FOR loop. How do you define this number of points? It is unclear even in the first example, as the difference between the elements is roughly 4, 4, 1, and you are using numbers of points 5, 4, 3..
Ilya on 30 Apr 2014
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Edited: Ilya on 30 Apr 2014
EMM... The number of pts in the for loop can be set e.g. according to the length of the jth considered interval. But you can also consider that the number of padding points is constant for all intervals. This is not essential (although adjusting the number of padding pts e.g. based on some criterion would be nice). Exactly because of this I'm using arbitrary integers 5, 4, 3.
Cedric on 30 Apr 2014
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Last point, why are values in a duplicated in a_desired? E.g. what don't you have just one 5.2 value but two?
Ilya on 1 May 2014
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Edited: Ilya on 1 May 2014
Sorry once again, this is because of the structure of the for-loop that I'm currently using. If it's possible to implement the same functionality without duplicates (and loops...), then it's even better.
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Accepted Answer
W. Owen Brimijoin on 2 May 2014
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If you find regular array operations more straightforward to follow than cell functions, you could create an array that increases in each column by the amount you need it to, and then unfold it back to a vector like this:
a=[1.1 5.2 9.25 10];
n_pts = 4; %specify the number of intervening points
a_desired = c*msum([a;repmat([diff(a)/n_pts,0],n_pts-1,1)]);
a_desired = a_desired(1:end-n_pts+1);
Something like that?
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Ilya on 2 May 2014
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Edited: Ilya on 2 May 2014
Yep! Actually, I thought in the same direction last time, but you've saved me time that I so desperately need to finish my thesis (and other people will now know the solution too!)
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More Answers (2)
Kelly Kearney on 1 May 2014
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If you have the Mapping Toolbox, interpm will do that:
a2 = interpm(a, ones(size(a)), 1.025)
It's not exactly the same as your linspace example, since you set the minimum interval instead of the number of points.
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Joseph Cheng on 1 May 2014
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you can try to use cellfun
here is a quick thing i created using your example up top. You will have to play around with it a bit to optimize it or see if the for loop is more efficient (memory wise).
a=[1.1 5.2 9.25 10]; %example a value.
%create pairs to be linspaced
a1 = a(1:end-1)';
a2 = a(2:end)';
%make pairs cells
a1 = mat2cell(a1,ones(size(a1)));
a2 = mat2cell(a2,ones(size(a2)));
%use cellfun to do the loops for you
a3=cellfun(@(x,y) linspace(x,y,100)',a1,a2,'UniformOutput',false);
a4 = cell2mat(a3)'
now a quick test shows that the stuff above, for an array that is longer than 500 elements, is faster than using the for loop you have in your question. (for me at least) there will be some variations when you vary the numbers of padding in linspace and number of items in 'a';
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Joseph Cheng on 1 May 2014
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Edited: Joseph Cheng on 1 May 2014
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to get rid of the duplicates
a4 = cell2mat(a3)'
a4(100:100:length(a4)-1)=[]
which should erase the the first duplicate value and the length()-1 should stop it from deleting the last spot.
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