How To Produce An Exponentially Scaled Axis?
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) plt.plot(xdata, ydata) p
Solution 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
classCustomScale(mscale.ScaleBase):
name = 'custom'def__init__(self, axis, **kwargs):
mscale.ScaleBase.__init__(self)
self.thresh = None#threshdefget_transform(self):
return self.CustomTransform(self.thresh)
defset_default_locators_and_formatters(self, axis):
passclassCustomTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = Truedef__init__(self, thresh):
mtransforms.Transform.__init__(self)
self.thresh = thresh
deftransform_non_affine(self, a):
return10**(a/10)
definverted(self):
return CustomScale.InvertedCustomTransform(self.thresh)
classInvertedCustomTransform(mtransforms.Transform):
input_dims = 1
output_dims = 1
is_separable = Truedef__init__(self, thresh):
mtransforms.Transform.__init__(self)
self.thresh = thresh
deftransform_non_affine(self, a):
return log2(a)*(10/log2(10))
definverted(self):
return CustomScale.CustomTransform(self.thresh)
mscale.register_scale(CustomScale)
xdata = [log2(x)*(10/log2(10)) for x inrange(1,11)]
ydata = range(10)
plt.plot(xdata, ydata)
plt.gca().set_xscale('custom')
plt.show()
Solution 2:
The easiest way is to use semilogy
from numpy import log2
import matplotlib.pyplot as plt
xdata = log2(range(1,11)) * (10/log2(10))
ydata = range(10)
plt.semilogy(xdata, ydata)
plt.show()
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