specifying bin increments on a histogram

Hello,

I use the code below to create a plots like this with different data sets. I want all plots to have the same scale, i.e. 0 to 90 and -2 to +2.

Not all data sets have data values at the extreme ends of the scales around 80-90 or 1.8-2 for example.

For example if the data in data set only ranges from 0-75 does the code divide the data up into bin intervals of 75/90 or does it create bins in 1 mph increments so bins 76-90 would have no data?

My question is, how can you specify that you want 90 bins a 1 mph increments and you want 40 bins at 0.1 increments from -2 to +2?

load Data;
% Calculate histogram
nBinsX = 120;
nBinsY = 40;
hist3(Data,[nBinsX,nBinsY]);
% Plot histogram
set(gcf,'renderer','opengl');
s = get(gca,'child');
zData = get(s,'zData');
colormap gray;
set(s,...
  'zData'     ,zData/length(Data),... % normalize frequency
  'FaceColor' ,'interp',...
  'EdgeColor' ,0.3*[1 1 1],...        % make edges visible at all heights
  'CDataMode' ,'auto');               % color tops according to height
% Labels
title('Velocity VS Acceleration Distribution');
xlabel('Velocity (mph)');
ylabel('Acceleration (m/s^2)');
zlabel('Normalized Frequency');

Answer 1

[Walter Roberson]

If you want specific bins, then pass in the list of bin centers if you are using hist(), or pass in the list of bin edges if you are using histc()

[the cyclist]

Note that John is using hist3(), which is just a bit different in syntax from hist() and histc().

Answer 2

[the cyclist]

I suggest you make a careful read of

>> doc hist3

There is a section near the top that describes exactly how to specify either the bin centers or edges that you want.

[John]

Hello,

Thank you for your comments.

I'm going to use this approach: hist3(X,'Edges',edges),

How would you you create a two element cell array?

Would I put the bin edges in a vector like this:

0 1  -2.0-1.9
1 2  -1.9-1.8
2 3  -1.8-1.7
3 4  -1.7-1.6
4 5  -1.6-1.5

Can this be converted to a cell array?

Thanks

[Jan]

edges = {0:4, -2:0.1:-1.6}

[John]

Thanks