Hi, everybody:

I don't have experience with images or contours and need some help plotting

a 'z' quantity for given x,y coordinates.

What are the choices?

Here is a small sample of the data:

The first row has the i-th x-coordinate at which the field starts to have

the value [i,j].

The first column has the j-th y-coordinate at which the field starts to

have the value [i,j].

The coordinate steps are not constant, nor the same for both dimensions;

they can be anything because they come from some odd finite difference

program.

My first shot at this is getting combersome, I am hoping for a better way.

So far, because the second decimal place in the x,y coordinate is alwasy

zero, I simply turned those coordinates into integers by multiplying by ten

and truncating; then by subtracting the first value from the rest, they look

very much like matrix indeces (except for the missing ones):

Then, because I don't know any better, I broadcast the values onto another

matrix, to fill in the in-between values:

Now, I have a matrix where every i,j has its own z-value and I am supposed

to be able to plot it with ax.contourf(mymatrix)...which I can, up until

about mymatrix[:,:1900] or so, afte that, I get the following error:

Traceback (most recent call last):

File "C:\findiff\t1.py", line 73, in <module>

ax.contourf(full[:,:2000])

File "C:\Python26\lib\site-packages\matplotlib\axes.py", line 7322, in

contourf

return mcontour.QuadContourSet(self, *args, **kwargs)

File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1106, in

__init__

ContourSet.__init__(self, ax, *args, **kwargs)

File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 700, in

__init__

self._process_args(*args, **kwargs)

File "C:\Python26\lib\site-packages\matplotlib\contour.py", line 1130, in

_process_args

C = _cntr.Cntr(x, y, z.filled(), _mask)

ValueError: Arguments x, y, z, mask (if present) must be 2D arrays.

x, y, z must be castable to double.

My current matrix is about 12000x5000.

Any asistance would be greatly appreciated.

Thanks,

Germán

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