I currently need to do something like this attached graph, but my attempts with Surfer (for windows) software only produced something with isolines, and that made it appared rough. I am not very good at writing scripts, and will be happy for suggestions with window based (and free) plotting software. But I have to go for script if I have not better choice. I attach the plot example
Try filled.contour() in R. See example(filled.contour) to see what it can do - and what the code should look like.
Oh yes Richard, the filled.contour provides a very good idea. Thank you on this! But it is suggested that we don't show the contours, and rather made it smooth like an image plot that Xuanlong suggested. I will appreciate if you can advise on this, I am going to play with this though. Xuanlong, Thank you very much. But I am wondering where my data comes in here. I have to put some data. The blue line a threshold set, beyond which is an abnormally high z-value (caused in relation to y and x). Thanks to you, both. I am going to explore the hints further. I hope I will get it rightly done.
filled.contour() will look smooth if you set nlevels to be high.
filled.contour will smooth the data, whereas plotting the raster will show the raw data better. Which is more useful will depend on your problem. You could also use image(), but then it is more difficult to get the scale.
You can import your data with read.table() or read.csv() - probably easier. Save your data as a cvs file in excel. The functions are very sensitive to everything being exactly right.
I use rainbow with the start and end arguments to specify which portion of the rainbow I should use
so instead of rev(rainbow(100))[30:100]
I would use rev(rainbow(70, start=0, end=7/10))
The problem with setting nlevels too high is that the black lines around each bin in the scale plot will overlap.
To avoid this problem, you can easily modify the function filled.contour as follows:
edit(filled.contour) -> my.filled.contour
In the function, change the line #41
rect(0, levels[-length(levels)], 1, levels[-1L], col = col)
to
rect(0, levels[-length(levels)], 1, levels[-1L], col = col, border=NA)
To see the difference, try:
z
It could be easily done in Python (matplotlib), Octave, Scilab or gnuplot.
In Python, something like this will work:
from matplotlib import pyplot as plt
from numpy import linspace, meshgrid, exp
def f(x,y): return (1-x/2+x**5+y**3)*exp(-x**2-y**2)
n = 256
x = linspace(-3,3,n)
y = linspace(-3,3,n)
X,Y = meshgrid(x,y)
plt.contourf(X, Y, f(X,Y), 50)
plt.show()
If you are not comfortable with scripting than you may use R with deducer or commander
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