How to produce an exponentially scaled axis?

Question

Consider the following code:

from numpy import log2
import matplotlib.pyplot as plt

xdata = [log2(x)*(10/log2(10)) for x in range(1,11)]
ydata = range(10)
p         


        

Answer 1

Here is how it is done. A good example to follow. You just subclass the ScaleBase class.

Here's your transform. It's not too complicated when you whittle out all the custom formatters and stuff. Just a little verbose.

from numpy import log2
import matplotlib.pyplot as plt

from matplotlib import scale as mscale
from matplotlib import transforms as mtransforms

class CustomScale(mscale.ScaleBase):
    name = 'custom'

    def __init__(self, axis, **kwargs):
        mscale.ScaleBase.__init__(self)
        self.thresh = None #thresh

    def get_transform(self):
        return self.CustomTransform(self.thresh)

    def set_default_locators_and_formatters(self, axis):
        pass

    class CustomTransform(mtransforms.Transform):
        input_dims = 1
        output_dims = 1
        is_separable = True

        def __init__(self, thresh):
            mtransforms.Transform.__init__(self)
            self.thresh = thresh

        def transform_non_affine(self, a):
            return 10**(a/10)

        def inverted(self):
            return CustomScale.InvertedCustomTransform(self.thresh)

    class InvertedCustomTransform(mtransforms.Transform):
        input_dims = 1
        output_dims = 1
        is_separable = True

        def __init__(self, thresh):
            mtransforms.Transform.__init__(self)
            self.thresh = thresh

        def transform_non_affine(self, a):
            return log2(a)*(10/log2(10))

        def inverted(self):
            return CustomScale.CustomTransform(self.thresh)


mscale.register_scale(CustomScale)

xdata = [log2(x)*(10/log2(10)) for x in range(1,11)]
ydata = range(10)
plt.plot(xdata, ydata)

plt.gca().set_xscale('custom')
plt.show()