# convert a range vector of bin centers to bin edges. Bin Centers and edges are non-uniform

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Is there a trivial way of converting a vector that contain bin centers to bin edges. The caveat here is that they are unequal. Does matlab provide a trivial way to do it?

Thanks

##### 2 Comments

Peter Kövesdi
on 9 Apr 2024

Moved: Voss
on 9 Apr 2024

I have exactly the same question: I have bin centers and want to have edges, that way, that the centers stay centered between the resulting edges.

So, is there any solution?

### Answers (5)

Image Analyst
on 22 Oct 2013

Where do you want the edges? Exactly half-way between the centers? If so you simply use conv():

% Make up some non-uniformly spaced bin centers.

binCenters = unique(sort(randi(99, 1,10)))

% Get edges half way in between.

binEdges = conv(binCenters, [0.5, 0.5], 'valid')

% Define end points outside of bin centers.

binEdges = [0 binEdges inf]

##### 0 Comments

W. Owen Brimijoin
on 22 Oct 2013

If we assume the simplest case, that the edges should be precisely between the centres (i.e., not geometric, or logarithmic, etc), and that the first and last edge should be -inf and +inf, respectively, then this is a simple solution:

%example of random bin centres:

centers = cumsum(randi(10,20,1))

%create equidistant edges:

edges = [-inf;centers(:)+[diff(centers(:))/2;inf]];

##### 0 Comments

tsan toso
on 23 Oct 2013

##### 3 Comments

Image Analyst
on 23 Oct 2013

Peter Kövesdi
on 9 Apr 2024

Ok, I come up with this solution:

binedges = zeros(size(binmids));

binedges(1)=2*binmids(1);

for i=2:length(binmids)

binedges(i) = 2*binmids(i)-binedges(i-1);

end

But maybe, there's a more Matlab way of doing it?

##### 0 Comments

Steven Lord
on 9 Apr 2024

If you know one of the edges of the region you're trying to bin, either the left-most or right-most edge, you can do this. Let's say you had bin centers at 1, 3.5, 7, and 11.5.

c = [1; 3.5; 7; 11.5];

Since we know these are the bin centers, the distances from each center to both edges of the bin are the same. So bin 1 is the interval [1-d1, 1+d1] for some distance d1. Let's ignore which edge is contained in each bin. Similarly bin 2 is [3.5-d2, 3.5+d2], etc. Because bin 1 ends at the start of bin 2, we have 1+d1 = 3.5-d2 or d1+d2 = 3.5-1. We can do the same for bins 2 and 3 and bins 3 and 4. This gives us a coefficient matrix and a right-hand side.

M = [1 1 0 0; 0 1 1 0; 0 0 1 1]

R = diff(c)

With one more piece of information (either the left edge of the interval being binned or the right edge) we can fix either d1 or d4. Let's say we know d1 is 1 (the leftmost bin starts at 0.)

d1 = 1;

Md1 = [1 0 0 0; M];

Rd1 = [d1; R];

d = Md1\Rd1

This puts the bin edges at 0 (c(1)-d(1)), 2 (c(1)+d(1) or c(2)-d(2)), 5 (c(2)+d(2) or c(3)-d(3)), 9 (c(3)+d(3) or c(4)-d(4)), and 14 (c(4)+d(4)).

edges = [c-d; c(end)+d(end)]

edges = [c(1)-d(1); c+d]

Are these bins centered at the values in c?

centers = [c, edges(1:end-1)+diff(edges)/2]

xline(edges)

hold on

plot(c, 1, 'x')

xlim([-1 15])

xticks(edges)

Sure looks like it.

The same holds if you know d4. [I'm not going to plot this; note that the plot above doesn't depend on d1, just the result d, and the backslash call below gives the same d vector as the backslash call above.]

d4 = 2.5;

Md4 = [M; 0 0 0 1];

Rd4 = [R; d4];

d = Md4\Rd4

Generalizing the creation of M to any number of bins is easy with two calls to diag.

What if we'd chosen a different d1 or d4 value?

d1 = 0.5;

Rd1 = [d1; R];

d = Md1\Rd1;

edges = [c-d; c(end)+d(end)]

figure

xline(edges)

hold on

plot(c, 1, 'x')

xlim([-1 15])

xticks(edges)

Those x markers do appear to be centered between the edges. But again, you do need one of the bin widths (easiest to write is either the first or last) which is equivalent to knowing one of the edges of one of the bins. Otherwise how would I distinguish the d1 = 1 case from the d1 = 0.5 case? The two plots are different.

##### 1 Comment

Peter Kövesdi
on 10 Apr 2024

Thank you very much for this more Matlab-like solution!

> (the leftmost bin starts at 0.)

Sorry, in my solution I assumed that silently in Line 2.

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