Asked by John
on 1 Oct 2012

Hi,

Could somebody advise me on how you would produce a graph like this in matlab

https://dl.dropbox.com/u/54057365/All/effmap.JPG

Thank you

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Answer by per isakson
on 1 Oct 2012

Edited by per isakson
on 31 Oct 2012

Accepted answer

My approach, which requires some work,

- put data in a 2D array, C
- plot with the function, image
- Image type: Indexed (colormap)
- Define a colormap, with a special slot, e.g. 1, for "missing data", white in your case. The length of the colormap = 64 is enough.
- Map C carefully to the colormap, e.g. C(ii,jj)=1 for "missing data"
- User defined tick-labels on the axes
- and more

**--- in response to comments I try to expand a bit ---**

Variables

- cmap is a colormap of size [64x3]. It associates the numbers [1,2,..., 64] to 64 different colors. (64 is an example.)
- D is your 2D data array
- C is an array of the same size as D. The values of C are integers in the interval, [1,64].

You create a suitable colormap, cmap, with the colormapeditor. You add some specific colors to signal special values of D, e.g. "missing data", outliers of various kinds etc.

You define a function, which takes D as input and returns C, e.g.

function C = D2C( D )

C = zeros( size(D) ); % zeros will cause error if not replaced

C( isnan(D) ) = 1; % nans will e displayed with color #1 C( D >= upper ) = 2; % values above upper gets color #2 C( D <= lower ) = 3; % values below lower gets color #3 etc. any command is ok as long as the values of C are integers [1,64]

assert( not( any( C(:)==0 ) ), .... ) end

and to display the graph

colormap( cmap ) image( C )

.

**--- in response to John's comment on 19 Oct 2012 at 16:45 ---**

Here is a start to make a graph that resembles https://dl.dropbox.com/u/54057365/All/eff.JPG

%% Create some data N = 1000; Speed = transpose( logspace( 0, 2, N ) ); Acc = randn( N, 1 ) .* ( exp( - sqrt( 10*Speed/N ) ) ); Efficiency = ( Acc .* Speed );

%% nAccBins = 20; % Set resolution of image nSpeedBins = 30;

Speed_upper = 100; Speed_lower = 0; Acc_upper = 3; Acc_lower = -3;

ix_Speed = ceil( nSpeedBins * ( Speed - Speed_lower ) ... / ( Speed_upper - Speed_lower ) );

ix_Acc = ceil( nAccBins * ( Acc - Acc_lower ) ... / ( Acc_upper - Acc_lower ) );

%% M = accumarray( { ix_Speed, ix_Acc }, Efficiency, [], @mean, nan );

%% imh = imagesc( transpose( M ) ); axh = ancestor ( imh, 'Axes' ); set( axh, 'YDir', 'normal' )

Next step would be to map M to a special colormap with something like

C = D2C( M );

and display with

colormap( cmap ) image( C )

.

**Backlog:**

- axis, ticks and ticklabels
- colorbar

Show 5 older comments

per isakson
on 15 Oct 2012

See above

John
on 19 Oct 2012

Hello Per Isakson,

Thank you for your help with this. Would you mind if I asked you one further question please?

If I was trying to create a graph like this:

https://dl.dropbox.com/u/54057365/All/eff.JPG

So I have 3 sets of data, acceleration, speed and efficiency.

For example Speed =40 and acceleration = 2 has an efficiency of 80%.

Would acceleration and speed be in the 2D array on their own? or would this become a 3D array with acceleration, speed and efficiency?

Thank you

John

per isakson
on 31 Oct 2012

See above

Answer by Image Analyst
on 1 Oct 2012

You would create an image - basically a 2D array. You must have that. If you don't have **that**, then you have nothing at all to display. Then display it with image() or imshow(), and apply a colormap. Not hard at all, assuming you have the data, temperature vs. current, in a 2D array already.

John
on 15 Oct 2012

Hello,

Thanks for your reply. I'm sorry but I don't understand how to get the colours to appear correctly.

This is what I am getting

https://dl.dropbox.com/u/54057365/All/effmat.JPG

but this is what I'm trying to achieve

https://dl.dropbox.com/u/54057365/All/efficiency.JPG

Also how do you switch the axis?

Thank you for your help

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## 1 Comment

## Azzi Abdelmalek (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/49586#comment_102510

plot a graph from what?