Python-2D直方图以对数比例绘制—错误：无法将float NaN转换为整数

Question

This is my previous plot that i want to convert to 2-d histogram.

mass_bh = (subhalos['SubhaloBHMass'] * 1e10 / 0.704) # in units of M_sol h^-1
vdisp = subhalos['SubhaloVelDisp']

nbins = 200
H, xedges, yedges = np.histogram2d(mass_bh,vdisp,bins=nbins)

fig2 = plt.figure()
plt.pcolormesh(xedges,yedges,Hmasked)
cbar = plt.colorbar()
cbar.ax.set_ylabel('g-r')

plt.ylabel(' $\log(\sigma)\quad$ [km s$^{-1}$] ')
plt.xlabel('$\log(M_{BH})\quad$ [M$_{\odot}$]')
plt.title('$M_{BH}-\sigma$ relation')

This instead, gives me this

My previous plot has both its x and y values converted to logarithmic scaling. But for this histogram conversion, it's not working out so great.

How can I work around this?

Thank you!

Answer 1

@armatita is right about the problem being the data. I think it all comes down to how you do your binning inside histogram2d . See if this example with a random lognormal distribution helps.

import numpy as np
import matplotlib.pyplot as plt

n = 1000

x = np.logspace(2, 10, n)
y = x**1.5
y = y * np.random.lognormal(10, 3, n)

x_bins = np.logspace(np.log10(x.min()), np.log10(x.max()), np.sqrt(n))
y_bins = np.logspace(np.log10(y.min()), np.log10(y.max()), np.sqrt(n))
H, xedges, yedges = np.histogram2d(x, y, bins=[x_bins, y_bins])

fig = plt.figure()
ax1 = fig.add_subplot(211)
ax1.plot(x, y, 'o')
ax1.set_xscale('log')
ax1.set_yscale('log')

ax2 = fig.add_subplot(212)
ax2.pcolormesh(xedges, yedges, H.T)
ax2.set_xscale('log')
ax2.set_yscale('log')

I get the below image, which is what I believe you are looking for. Also note the transpose on H .

Answer 2

Just a suggestion to pick up your curiosity. Although @lanery clearly answers the question, I would like to share a different method of getting a nice 2d histogram in python. Instead of using np.histogram2d, which in general produces quite ugly histograms, I would like to recycle py-sphviewer , a python package for rendering particle simulations using an adaptive smoothing kernel. Consider the following code, which is based on the example of lanery :

import numpy as np import matplotlib.pyplot as plt import sphviewer as sph

def myplot(x, y, extent=None, nb=8, xsize=500, ysize=500):   
    if(extent == None):
        xmin = np.min(x)
        xmax = np.max(x)
        ymin = np.min(y)
        ymax = np.max(y)
    else:
        xmin = extent[0]
        xmax = extent[1]
        ymin = extent[2]
        ymax = extent[3]

    k, = np.where( (x <= xmax) & (x >= xmin) & 
                   (y <= ymax) & (y >= ymin) )

    pos = np.zeros([3, len(k)])
    pos[0,:] = (x[k]-xmin)/(xmax-xmin)
    pos[1,:] = (y[k]-ymin)/(ymax-ymin)
    w = np.ones(len(k))

    P = sph.Particles(pos, w, nb=nb)
    S = sph.Scene(P)
    S.update_camera(r='infinity', x=0.5, y=0.5, z=0, 
                    extent=[-0.5,0.5,-0.5,0.5],
                    xsize=xsize, ysize=ysize)
    R = sph.Render(S)
    R.set_logscale()
    img = R.get_image()

    return img, [xmin,xmax,ymin,ymax]    


n = 1000

x = np.logspace(2, 10, n)
y = x**1.5
y = y * np.random.lognormal(10, 3, n)

H, xedges, yedges = np.histogram2d(x, y, bins=[np.logspace(np.log10(x.min()), np.log10(x.max())),
                                               np.logspace(np.log10(y.min()), np.log10(y.max()))])


img, extent = myplot(np.log10(x), np.log10(y))   #Call the function to make the 2d-histogram

fig = plt.figure()
ax1 = fig.add_subplot(311)
ax1.plot(x, y, 'o')
ax1.set_xscale('log')
ax1.set_yscale('log')

ax2 = fig.add_subplot(312)
ax2.pcolormesh(xedges, yedges, H.T)
ax2.set_xscale('log')
ax2.set_yscale('log')

ax3 = fig.add_subplot(313)
ax3.imshow(img, origin='lower', extent=extent, aspect='auto')

plt.show()

which produces the following output:

The function myplot() is just a very simple function that I've written in order to normalize the data and give it as input of py-sphviewer. The length of the smoothing kernel is typically given by the parameter nb , which specify the number of neighbours over which the smoothing is performed. Although is seems complicated at first sight, the ideas and the implementation are very easy, and the result is by far superior compared to np.histogram2d. But of course, it depends whether you are able to spread out your data or not, and what the meaning and consequence of doing that for your research.