我建议生成一个伪色条如下(见解释说明)：

import matplotlib.pyplot as plt

import numpy as np

from matplotlib.colors import LogNorm

import matplotlib.gridspec as gridspec

delta = 0.025

x = y = np.arange(0, 3.01, delta)

X, Y = np.meshgrid(x, y)

Z1 = plt.mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)

Z2 = plt.mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)

Z = 1e6 * (Z1 * Z2)

fig=plt.figure()

#

# define 2 subplots, using gridspec to control the

# width ratios:

#

# note: you have to import matplotlib.gridspec for this

#

gs = gridspec.GridSpec(1, 2,width_ratios=[15,1])

# the 1st subplot

ax1 = plt.subplot(gs[0])

lvls = np.logspace(0,4,20)

CF = ax1.contourf(X,Y,Z,

norm = LogNorm(),

levels = lvls

)

CS = ax1.contour(X,Y,Z,

norm = LogNorm(),

colors = 'k',

levels = lvls

)

#

# the pseudo-colorbar

#

# the 2nd subplot

ax2 = plt.subplot(gs[1])

#

# new levels!

#

# np.logspace gives you logarithmically spaced levels -

# this, however, is not what you want in your colorbar

#

# you want equally spaced labels for each exponential group:

#

levls = np.linspace(1,10,10)

levls = np.concatenate((levls[:-1],np.linspace(10,100,10)))

levls = np.concatenate((levls[:-1],np.linspace(100,1000,10)))

levls = np.concatenate((levls[:-1],np.linspace(1000,10000,10)))

#

# simple x,y setup for a contourf plot to serve as colorbar

#

XC = [np.zeros(len(levls)), np.ones(len(levls))]

YC = [levls, levls]

CM = ax2.contourf(XC,YC,YC, levels=levls, norm = LogNorm())

# log y-scale

ax2.set_yscale('log')

# y-labels on the right

ax2.yaxis.tick_right()

# no x-ticks

ax2.set_xticks([])

plt.show()

这将给你一个这样的情节：

编辑

或者,在调用颜色栏时,使用新的级别和间距=“比例”选项：

>替换这行：

lvls = np.logspace(0,4,20)

用这些：

lvls = np.linspace(1,10,5)

lvls = np.concatenate((lvls[:-1],np.linspace(10,100,5)))

lvls = np.concatenate((lvls[:-1],np.linspace(100,1000,5)))

lvls = np.concatenate((lvls[:-1],np.linspace(1000,10000,5)))

>替换这行：

cbar = plt.colorbar(CF, ticks=lvls, format='%.4f')

有了这个：

cbar = plt.colorbar(CF, ticks=lvls, format='%.2f', spacing='proportional')

你会结束这个情节：

(格式仅更改,因为新的刻度不需要4位小数)

编辑2

如果你想自动生成像我使用的级别,你可以考虑这段代码：

levels = []

LAST_EXP = 4

N_LEVELS = 5

for E in range(0,LAST_EXP):

levels = np.concatenate((levels[:-1],np.linspace(10**E,10**(E+1),N_LEVELS)))