[Bug]: secondary_xaxis produces ticks at incorrect locations #29067

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1. jklymak commented on Nov 4, 2024

   

   jklymak

   on Nov 4, 2024

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   Your interpolation won't go lower than zero, but the main axis has negative values so the secondary axes is not one to one. Your forward and reverse should take care of negative values, probably leaving them the same or doing -x**2
2. jklymak commented on Nov 4, 2024

   

   jklymak

   on Nov 4, 2024

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   Or just set_xlim(min=0)...
3. dkweiss31 commented on Nov 4, 2024

   

   dkweiss31

   on Nov 4, 2024

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   Thanks for the quick reply @jklymak ! You're right about the negative values but I was hoping that the top x axis would not go negative and start at 0, same as the bottom x axis. As you suggest the following works code works by specifying the limits (note that if I don't specify the right xlim, the top x axis is still not correct). However it doesn't have the nice buffer of white space on the left and right that we are used to. Is there a way to retain that nice behavior with secondary_xaxis?

   fig, ax = plt.subplots(constrained_layout=True)
   xdata = np.arange(0, 11, 0.4)
   ydata = np.random.randn(len(xdata))
   ax.plot(xdata, ydata, label='Plotted data')
   ax.set_xlim(left=xdata[0], right=xdata[-1])
   ax.axvline(10, color="grey")
   ax.set_xlabel('X [m]')
   ax.legend()
   xnew = xdata**2
   
   def forward(x):
       return np.interp(x, xdata, xnew)
   
   
   def inverse(x):
       return np.interp(x, xnew, xdata)
   
   secax = ax.secondary_xaxis('top', functions=(forward, inverse))
   secax.xaxis.set_minor_locator(AutoMinorLocator())
   secax.set_xlabel('$X_{other}$')
   
   plt.show()

   Producing
   
4. jklymak commented on Nov 4, 2024

   

   jklymak

   on Nov 4, 2024

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   There is an explicit note about this in https://matplotlib.org/stable/gallery/subplots_axes_and_figures/secondary_axis.html#sphx-glr-gallery-subplots-axes-and-figures-secondary-axis-py:

   In order to properly handle the data margins, the mapping functions 
   (forward and inverse in this example) need to be defined beyond the nominal plot limits.
   
   In the specific case of the numpy linear interpolation, 
   [numpy.interp](https://numpy.org/doc/stable/reference/generated/numpy.interp.html#numpy.interp), this 
   condition can be arbitrarily enforced by providing optional keyword arguments left, right such that values 
   outside the data range are mapped well outside the plot limits.
   
5. jklymak

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6. jklymak commented on Nov 4, 2024

   

   jklymak

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   Feel free to carry on the discussion. However, I don't think there is much we can do here - you need to define forward and inverse to handle beyond 0 and 11 if you want margins to work. How exactly you do that is application dependent.
7. jklymak

   closed this as not plannedon Nov 4, 2024
8. dkweiss31 commented on Nov 4, 2024

   

   dkweiss31

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   I'm not sure that's the only issue here: it seems that this functionality doesn't play nice with interp. The following simple example which does away with the negative complications works fine for the first set of functions and doesn't for the second set:

   import matplotlib.pyplot as plt
   import numpy as np
   
   fig, ax = plt.subplots(constrained_layout=True)
   xdata = np.arange(2, 11, 0.4)
   ydata = np.random.randn(len(xdata))
   ax.plot(xdata, ydata, label='Plotted data')
   ax.axvline(10, color="grey")
   ax.set_xlabel('X [m]')
   ax.legend()
   xnew = 2 * xdata
   
   # def forward(x):
   #     return 2 * x
   #
   #
   # def inverse(x):
   #     return x / 2
   
   
   def forward(x):
       return np.interp(x, xdata, xnew)
   
   
   def inverse(x):
       return np.interp(x, xnew, xdata)
   
   secax = ax.secondary_xaxis('top', functions=(forward, inverse))
   secax.set_xlabel('$X_{other}$')
   
   plt.show()

   first set:
   
   second set (interp)
   
9. jklymak commented on Nov 4, 2024

   

   jklymak

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   As above, you need to tell np.interp what to do for x<2 and x>11.

   def forward(x):
       xdatan = np.concatenate(([-50], xdata, [50]))
       xnewn = np.concatenate(([-50*2], xnew, [50*2]))
   
       return np.interp(x, xdatan, xnewn)
   
   
   def inverse(x):
       xdatan = np.concatenate(([-50], xdata, [50]))
       xnewn = np.concatenate(([-50*2], xnew, [50*2]))
       return np.interp(x, xnewn, xdatan)
   
10. dkweiss31 commented on Nov 4, 2024

    

    dkweiss31

    on Nov 4, 2024

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    Ahh I see: that worked, thanks! Unless I'm mistaken about what the tutorial is trying to show, it might be suffering from the same problem I was running into. If there is interest I'd be happy to submit a PR to update the tutorial based on our discussion.
11. jklymak

    reopened this on Nov 4, 2024
12. jklymak commented on Nov 4, 2024

    

    jklymak

    on Nov 4, 2024

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    Ah I see, I think you are correct. Fixing the tutorial would be super helpful!
13. jklymak

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14. timhoffm

    closed this as completedin #29073on Nov 21, 2024
15. QuLogic

    added this to the v3.10.0 milestone on Nov 21, 2024