Plot a function with matplotlib?

S

Steven D'Aprano

I have matplotlib and iPython, and want to plot a function over an
equally-spaced range of points.

That is to say, I want to say something like this:

plot(func, start, end)

rather than generating the X and Y values by hand, and plotting a scatter
graph. All the examples I've seen look something like this:

from pylab import *
import numpy as np
t = arange(0.0, 2.0+0.01, 0.01) # generate x-values
s = sin(t*pi) # and y-values
plot(t, s)
show()


which is fine for what it is, but I'm looking for an interface closer to
what my HP graphing calculator would use, i.e. something like this:


plot(lambda x: sin(x*pi), # function or expression to plot,
start=0.0,
end=2.0,
)

and have step size taken either from some default, or better still,
automatically calculated so one point is calculated per pixel.

Is there a way to do this in iPython or matplotlib?
 
A

Alex van der Spek

I have matplotlib and iPython, and want to plot a function over an
equally-spaced range of points.

That is to say, I want to say something like this:

plot(func, start, end)

rather than generating the X and Y values by hand, and plotting a
scatter graph. All the examples I've seen look something like this:

from pylab import *
import numpy as np
t = arange(0.0, 2.0+0.01, 0.01) # generate x-values s = sin(t*pi) #
and y-values
plot(t, s)
show()


which is fine for what it is, but I'm looking for an interface closer to
what my HP graphing calculator would use, i.e. something like this:


plot(lambda x: sin(x*pi), # function or expression to plot,
start=0.0,
end=2.0,
)

and have step size taken either from some default, or better still,
automatically calculated so one point is calculated per pixel.

Is there a way to do this in iPython or matplotlib?

Not to my knowledge unless you code it yourself.

However in gnuplot (www.gnuplot.info)

gnuplot>>> set xrange[start:end]
gnuplot>>> foo(x)=mycomplicatedfunction(x)
gnuplot>>> plot foo(x)

or shorter still

gnuplot>>> plot [start:end] foo(x)

without the need to set the xrange in advance.
 
M

Miki Tebeka

I'm looking for an interface closer to
what my HP graphing calculator would use, i.e. something like this:


plot(lambda x: sin(x*pi), # function or expression to plot,
start=0.0,
end=2.0,
)

and have step size taken either from some default, or better still,
automatically calculated so one point is calculated per pixel.

Is there a way to do this in iPython or matplotlib?
I don't think there is, but using range and list comprehension you can write a little utility function that does that:

HTH
--
Miki Tebeka <[email protected]>
http://pythonwise.blogspot.com

def simplot(fn, start, end):
xs = range(start, end+1)
plot(xs, [fn(x) for x in xs)])
 

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